## AP SSC 10th Class Physics Solutions Chapter 7 Human Eye and Colourful World

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 7 Human Eye and Colourful World Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Physics Solutions 7th Lesson Human Eye and Colourful World

### 10th Class Physics 7th Lesson Human Eye and Colourful World Textbook Questions and Answers

Question 1.
What is the function of lens in human eye?
The eye lens forms a real image on retina then we can see the objects.

Question 2.
How does it help to see objects at long distances and short distances?
Eye lens can adjust itself in shape, so that it helps to see the objects at long and short distances.

Question 3.
How is it possible to get the image at the same distance on the retina?
When the eye is focussed at distant object, the ciliary muscles are relaxed. So the focal length of eye lens adjusted itself which is equal to the distance of object from the retina. Then we can see the object clearly.
When the eye is focussed on a closer object the ciliary muscles adjust the focal length in such a way that the image is formed on retina and we see the object clearly.

Question 4.
Are we able to see all objects in front of our eye clearly?
Yes. By maintaining the 25 cm distance of the object from our eye, we can see the objects in front of our eye.

Question 5.
How do the lenses used in spectacles correct defects of vision?
To form image on retina.
To correct
myopia- Concave lens
hypermetropia i- Convex lens

Question 1.
How do you correct the eye defect Myopia? (AS1)

• Some people cannot see objects at long distances but can see nearby objects clearly. This type of defect in vision is called ‘Myopia’ or ‘Near sightedness’.
• Myopia is corrected by using a con-cave lens of focal length equal to the distance of the far point F from the eye.
• This lens diverges the parallel rays from distant object as if they are coming from the far point.
• Finally the eye lens forms a clear im-age at the retina.

Question 2.
Explain the correction of the eye defect Hypermetropia. (AS1)

• A person with hypermetropia can see distant objects clearly but cannot see objects at near distances. This is also known as ‘far sightedness’.
• Eye lens can form a clear image on the retina when any object is placed beyond near point.
• To correct the defect of hypermetropia, we need to use a lens which forms an image of an object beyond near point at H, when the object is between H and L. This is possible only when a double con¬vex lens is used.
• The image acts like an object for the eye lens. Hence final image due to eye is formed at retina.

Question 3.
How do you find experimentally the refractive index of material of a prism? (AS1)
(OR)
Write the experimental procedure in finding the refractive index of a prism.
(OR)
Which quantity will decide whether a given medium is denser or rarer? How do you find that quantity of prism experimentally?
Refractive index decides whether a medium is denser or rarer.
Aim :
To find the refractive index of a prism.

Materials required :
Glass prism, white chart of size 20 x 20 cm, pencil, pins, scale and protractor.

Procedure :

1. Keep a prism on white chart.
2. Draw the triangular base of the prism with pencil.
3. Remove the prism.
4. The shape of the outline drawn prism is triangle and name its vertices as P, Q and R.
5. PQ and PR be the refracting surfaces.
6. Find the angle between PQ and PR. This is the angle of the prism (A) (or) Refracting angle.
7. Mark M on PQ and draw a perpendicular line to PQ at M.
8. Place the centre of the protractor at M, along the normal and mark an angle of 30° and then draw a line up to M. This line denotes incident ray. This angle is called angle of incidence.
9. Place the prism in its position (triangle) again.
10. Now fix two pins vertically on the line at points A and B .
11. See the images of pins through the 2nd refracting side (PR).
12. Fix another two pins at points C and D such that all the four pins appear to lie along the same line.
13. Remove the pins and prism, join the pin-holes. Draw the incident and emergent rays.
14. The angle between the normal and the emergent ray at N is the angle of emergence.
15. The line passing through the points A, B, M, N, C and D represents the path of light when it suffers refraction through prism.

The angle of deviation :

• Extend incident and emergent rays are intercept at a point ‘O’.
• The angle between these two rays is the angle of deviation (d).
• Note the emergent deviation angles for different values of i, in the given table.

• Draw the graph between angle of incidence on X – axis and the angle of deviation on Y – axis.
• We notice that the angle of deviation decreases first and then increases with increase of the angle of incidence.
• Mark points on a graph paper and join the points to obtain a graph (smooth curve).
• Draw a tangent line to the curve, which parallel to X-axis, at the lowest point of the graph.

• The point where this line cuts the Y- axis gives the angle of minimum deviation. It is denoted by D.
• From the graph, we notice that, at angle of minimum deviation, the angle of incidence is equal to the angle of emergence.
• By finding A and D we can find refractive index of prism by using formula .

Question 4.
Explain the formation of rainbow. (AS1)
(OR)
What is the natural spectrum occuring in sky? Explain the formation of that spectrum.

• A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on rain drops.
• It is combined result of reflection, refraction and dispersion of sunlight from water droplets, in atmosphere.
• It always forms in the direction opposite to the sun.
• To see a rainbow, the sun be must behind us and the water droplets falls in front of us.
• When a sunlight enters a spherical raindrop, it is refracted and dispersed. The different colours of light are bent in different angles.
• When different colours of light falls on the back inner surface of drop, it (Water drop) reflects (different colours of light) interwnally (total internal reflection).
• The water drops again refract the different colours, when it comes out from the raindrop.
• After leaving this different colours from the raindrop as rainbow, reach our eye. Thus, we see a rainbow.

Question 5.
Explain briefly the reason for the blue of the sky. (AS1)

• The reason for blue sky is due to the molecules N2 and O2.
• The size of these molecules are comparable to the wavelength of blue light.
• These molecules act as scattering centres for scattering of blue light.
• So scattering of blue light by molecules of N2 and O2 is responsible for blue of the sky.

Question 6.
Explain two activities for the formation of artificial rainbow. (AS1)
(OR)
Give two activities for the formation of the artificial rainbow? And explain it.
(OR)
Suggest an experiment to create a rainbow in your classroom and explain the procedure.
Activity -1 :

• Select a white coated wall on which the sun rays fdll.
• Stand in front of a wall in such a way that the sun rays fall on your back.
• Hold a tube through which water is flowing.
• Place your finger in the tube to obstruct the flow of water.
• Water comes out from small gaps between the tube and finger like a fountain.
• Observe the changes on wall while showering the water.

Activity – II : (Activity – 4)

• Take a metal tray and fill with water.
• Place a mirror in water such that it makes an angle to the water surface.
• Now focus white light on the mirror through the water.
• Keep a white card board sheet above the water surface.
• We may observe the colours VIBGYOR on the board.
• The splitting of white light into different colours (VIBGYOR) is called dispersion.
• So consider a white light is a collection of waves with different wavelengths.
• Violet has shortest wavelength, and red has longest wavelength.

Question 7.
Derive an expression for the refractive index of the material of a prism. (AS1)
Derivation of formula for refractive index of a prism :
PQ and PR are refracting surfaces of prism, i, is an angle of incidence, i2 is an angle of emergence, is an angle of refraction of first surface and r2 is an angle of incidence on second surface.

1. From triangle OMN, we get
d = i1 – r1 + i2 – r2
∴ d = (i1 + i2)-(r1+ r2) ……… (1)
2. From triangle PMN, we have
∠A + (90° – r1) + (90° – r2) = 180°
⇒ r1 + r2 = A ……… (2)
3. From (1) and (2),
⇒ d = (i1 + i2) – A
⇒ A + d = i1 + i2
4. This is the relation between angle of incidence, angle of emergence, angle of deviation and angle of prism.

5. From Snell’s law, we know that
n1 sin i = n2 sin r

6. Let ‘n’ be the refractive index of the prism.
7. Using Snell’s law at M, refractive index of air n1 = 1; i = i1; n2 = n; r = r1.
⇒ sin i1 = n sin r1 …………. (4)
8. Similarly, at N, n1 = n; i = r2; n2 = 1; i1 = i2
⇒ n sin r2 = n sin i2 ………… (5)
9. We know that at the angle of minimum deviation position (D), i.e. i1 = i2
10. We will notice that MN is parallel to the base of prism QR.

14. This is the formula for the refractive index of the prism.

Question 8.
Light of wavelength λ1 enters a medium with refractive index n2 from a medium with refractive index n1 What is the wavelength of light in second medium? (AS1)
Wavelength of first medium =λ1
Refractive index of first medium = n1
Let the wavelength of second medium = λ2,
Refractive index of second medium = n2
Light enters from first medium to second medium = $$\Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}$$ (∵ υ = const)
$$\Rightarrow \lambda_{2}=\frac{\lambda_{1} \mathrm{n}_{1}}{\mathrm{n}_{2}}$$

Note:
For the below questions the following options are given. Choose the correct option by making hypothesis based on given assertion and reason. Give an explanation.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true and R is not the correct explanation of A.
c) A is true but R is false.
d) Both A and R are false.
e) A is false but R is true.

Question 9.
Assertion (A) : The refractive index of a prism depends only on the kind of glass of which it is made of and the colour of light. (AS2)
Reason (R) : The refractive index of a prism depends on the refracting angle of the prism and the angle of minimum deviation.
(b) Both A and R are true and R is not the correct explanation of A.

Question 10.
Assertion (A) : Blue colour of sky appears due to scattering of light.
Reason (R) : Blue colour has shortest wavelength among all colours of white light. (AS2)
(C) A is true but R is false.

Question 11.
Suggest an experiment to produce a rainbow in your classroom and explain the procedure. (AS3)
Activity to produce a rainbow in classroom :

• Take a prism and place it on the table near a vertical white wall.
• Take a thin wooden plank, make a small hole in it and fix it vertically on the table.
• Place the prism between the wooden plank and wall.
• Place a white light source behind the hole of the wooden plank. Switch on the light.
• Adjust the height of the prism such that the light falls on one of the lateral surfaces.
• Observe the changes in the emerged ray of the prism.
• Adjust the prism by slightly rotating it till you get an image on the wall.
• We observe a band of different colours on the wall.
• These colours are nearly equal to the colours of the rainbow, i.e., VIBGYOR.

Question 12.
Prisms are used in binoculars. Collect information why prisms are used in binoculars. (AS4)

• Binoculars consists of a pair of identical or mirror symmetrical telescope mounted side by side and aligned to point accurately in the same direction, allowing the viewer to use both eyes, when viewing distant objects.
• The size of binoculars is reduced by using prisms.
• We get good image with more brightness.
• Objective size and optical quality should be increased by using prisms in binoculars.

Question 13.
Incident ray on one of the face (AB) of prism and emergent ray from the face AC are given in figure. Complete the ray diagram. (AS5)

Question 14.
How do you appreciate the role of molecules in the atmosphere for the blue colour of the sky? (AS6)
(OR)
How do you appreciate the nature of molecules, responsible for the blue of the sky?

• The sky appear blue due to atmospheric refraction and scattering of light through molecules.
• Molecules are scattering centres.
• The reason to blue sky is due to the molecules N2 and 02.
• The sizes of these molecules are comparable to the wavelength of blue light.
• In the absence of molecules there will be no scattering of sunlight and the sky will appear dark.
• We should appreciate the molecules which are scattering centres.

Question 15.
Eye is the only organ to visualise the colourful world around us. This is possible due to accommodation of eye lens. Prepare a six line stanza expressing your wonderful feelings. (AS6)
“Many people simply see
The world in black and white
But through my eyes the world’s alive
So colourful and bright
Don’t close your mind to the sights
That light up the night and day
There’s so much to see here on this earth
And not a rupee do you have to pay”
The most obvious things aren’t the ones
There is beauty in the unknown
with willing eyes and an open mind
The true wonders you will be shown

Question 16.
How do you appreciate the working of ciliary muscles in the eye? (AS6)
(OR)
Which muscles are helpful in changing the focal length of eye lens? How do you appreciate those muscles?

• Ciliary muscles are helpful in changing focal length of eye lens.
• The ciliary muscles which attached with eye lens help to change the focal length of eye lens.
• When the eye is focussed on a distant object, these are relaxed. So the focal length of eye lens increases to its maximum value.
• The parallel rays coming into the eye are focussed on the retina then we can see the object clearly.
• When the eye is focussed on a nearer object the muscles are strained so the focal length of eye-lens decreases. The ciliary muscles adjust the focal length and the image is formed on retina then we can see the object.
• Their process of adjusting focal length of eve lens is called accomodation.
• So ciliary muscles must be appreciated for its accomodation of eye lens.

Question 17.
Why does the sky sometimes appear white? (AS7)
(OR)
The sky sometimes appears white. What is the reason behind it?

• Our atmosphere contains atoms and molecules of different sizes.
• According to their sizes, they are able to scatter different wavelengths of light.
• For example, the size of the water molecule is greater than the size of the N2 or O2.
• It acts as a scattering centre for light other frequencies which are lower than the frequency of blue light.
• On a hot day due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere. These water molecules scatter the colours of other frequencies (other than blue).
• All such colours of other frequencies reach our eye and white colour is appeared to us.

Question 18.
Glass is known to be transparent material. But ground glass is opaque and white in colour. Why? (AS7)

• Ground glass is glass whose surface has been ground that produces a flat but rough finish.
• Ground glass has the effect of rendering the glass translucent by scattering of light during transmission thus blurring visibility while still transmitting light.
• To get more permanent frost, the glass may be ground by rubbing with some gritty substance.

Question 19.
If a white sheet of paper is stained with oil, the paper turns transparent. Why? (AS7)
(OR)
What is the reason behind the paper turns transparent when it is immersed in oil?

• The paper fibres have higher index of refraction probably much greater than 1.5.
• The oil or fat also has a high index of refraction so that it nearly matches the index of refraction of the paper fibres and it reduces the scattering significantly.
• The fat adhering to the cellulose fibers lowers the index of refraction of the cellulose and also fills in air voids, so that visible light passes through the bag with significantly less scattering.
• The oil connects the fibres in the paper with a liquid which can transmit by refraction (rather than scatter) light that falls upon it. As a result, the paper stained with oil is turned transparent.

Question 20.
A light ray falls on one of the faces of a prism at an angle 40° so that it suffers angle of minimum deviation of 30°. Find the angle of prism and angle of refraction at the given surface. (AS7)
Given that, incident ray on one of the prisms (i1) = 40°
Angle of minimum deviation (D) = 30°
Angle of prism (A) = ?
A+D = 2i1
⇒ A = 2i1 – D
⇒ A = 2(40°) -30°
= 80° – 30°
= 50°
∴ Angle of prims (A) = 50°
Angle of refraction (r1) = ?

Question 21.
The focal length of a lens suggested to a person with Hypermetropia is 100 cm. Find the distance of near point and power of the lens. (AS7)
i) The distance of near point :
If the distance of near point is ‘d’ and focal length is ‘f’ then the relation between

Question 22.
A person is viewing an extended object. If a converging lens is placed in front of his eye, will he feel that the size of object has increased? Why? (AS7)

• A simple magnifier allows us to put the object closer to the eye than we could normally focus and forms an enlarged virtual images.
• The principle behind this is angular magnification.
• The magnification is Ma = $$\frac{25}{f}$$ for the close focus point, but since that causes eye strain, it is usually desirable to put the images at infinity giving Ma = $$\frac{25}{f}$$
• So he fees the size of object is increased. The reason is mentioned.
• Angular magnification: ‘The ratio of the angle substended at the eye by the image formed by an optical instrument to the angle substended at the eye by the object being viewed.”

Fill In The Blanks

1. The value of least distance of distinct vision is about ……………………… .
2. The distance between the eye lens and retina is about ……………………… .
3. The maximum focal length of the eye lens is about ……………………… .
4. The eye lens can change its focal length due to working of ……………………… muscles.
5. The power of lens is ID then focal length is ……………………… .
6. Myopia can be corrected by using ……………………… lens.
7. Hypermetropia can be corrected by using ……………………… lens.
8. In minimum deviation position of prism, the angle of incidence is equal to angle of ……………………… .
9. The splitting of white light into different colours (VIBGYOR) is called ……………………… .
10. During refraction of light, the character of light which does not change is ……………………… .

1. 25 cm
2. 2.5 cm
3. 2.5 cm
4. ciliary
5. 100 cm
6. bi-concave
7. bi-convex
8. emergence
9. dispersion of light
10. frequency

Multiple Choice Questions

1. The size of an object as perceived by an eye depends primarily on
A) actual size of the object
B) distance of the object from the eye
C) aperture of the pupil
D) size if the image formed on the retina
B) distance of the object from the eye

2. When objects at different distances are seen by the eye which of the following remains constant?
A) focal length of eye-lens
B) object distance from eye-lens
C) the radii of curvature of eye-lens
D) image distance from eye-lens
D) image distance from eye-lens

3. During refraction, …………….. will not change.
A) wavelength
B) frequency
C) speed of light
D) all the above
B) frequency

4. A ray of light falls on one of the lateral surfaces of an equilateral glass prism placed on the horizontal surface of a table as shown in figure. For minimum deviation of ray, which of the following is true?

A) PQ is horizontal
B) ‘QR’ is horizontal
C) ‘RS’ is horizontal
D) Either ‘PQ’ or ‘RS’ is horizontal
B) ‘QR’ is horizontal

5. Far point of a person is 5 m. In order that he has normal vision what ki nd of spectacles should he use?
A) concave lens with focal length 5 m
B) concave lens with focal length 10 m
C) convex lens with focal length 5 m
D) convex lens with focal length 2.5 mm
A) concave lens with focal length 5 m

6. The process of re-emission of absorbed light in all directions with different intensi¬ties by the atom or molecule is called ……………
A) scattering of light
B) dispersion of light
C) reflection of light
D) refraction of light
A) scattering of light

### 10th Class Physics 7th Lesson Human Eye and Colourful World InText Questions and Answers

10th Class Physics Textbook Page No. 104

Question 1.
Can you imagine the shape of rainbow when observed during travel in an airplane?

The shape of rainbow, when observed during travel in an airplane is circular as shown in the following figure.

10th Class Physics Textbook Page No. 88

Question 2.
Why do the values of least distance of distinct vision and angle of vision change with person and age?

• The ciliary muscle helps the eye lens to change its focal length by changing radii of curvature of eye lens.
• When the eye is focussed on a distant object, the ciliary muscles are relaxed so that the focal length of eye lens has its maximum value which is equal to its distance from the retina.
• The working of ciliary muscle in eye changes from person to person.
• So, the values of least distance of distinct vision and angle of vision changes with person and age.

10th Class Physics Textbook Page No. 89

Question 3.
How can we get same image distance for various positions of objects?
For different positions of object, the image distance remains constant only when focal length of lens changes.

Question 4.
Can you answer above question using concepts of refraction through lenses?
The focal length of a lens depends on the material by which it has been made and radii of curvatures of lens. We need to change focal length of eye lens to get same image distance for various positions of object.

10th Class Physics Textbook Page No. 90

Question 5.
How does eye lens change its focal length?
(OR)
What is the role of ciliary muscles in the eye ? Write the answer in one or two sentences only.
The ciliary muscles to which eye lens is attached help the eye lens to change its focal length by changing the radii of curvature of eye lens.

Question 6.
How does this change (eye lens changes its focal lengths) take place in eye ball?
When the eye is focussed on a distant object the ciliary muscles are relaxed. So the focal length of eye lens has its maximum value which is equal to its distance from the retina. The parallel rays coming into the eye are then focussed on to the retina and we see the object clearly.

Question 7.
Does eye lens form a real image or virtual image?
The eye lens forms a real and inverted image of an object on the retina.

Question 8.
How does the image formed on retina help us to perceive the object without change in its shape, size and colour?

• The eye lens forms a real and inverted image of an object on the retina.
• This retina is a delicate membrane, which contains large number of receptors (125 million) called ‘rods’ and ‘cones’.
• They receive the light signals and identify the colour, and the intensity of light.
• These signals are transmitted to the brain through optic-nerve fibres.
• The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 9.
Is there any limit to change of focal length of eye lens?
Yes. When the object is at infinity, the parallel rays from the object falling on the eye lens are refracted. They form a point-sized image on retina. In this situation, eye lens has a maximum focal length.

Question 10.
What are the maximum and minimum focal lengths of the eye lens?
Maximum focal length is 2.5 cm and minimum focal length is 2.27 cm.

Question 11.
How can we find the maximum and minimum focal lengths of the eye lens?
(OR)
Calculate the maximum and minimum focal lengths of the eye lens.
a) When the object is at infinity,
u = – ∞ ; v = 2.5 cm (image distance which is equal to distance between eye lens and retina)

b) Object at a distance of 25 cm from eye. In this case, eye has minimum focal length.
Here u = -25 cm ; v 2.5 cm

10th Class Physics Textbook Page No. 91

Question 12.
What happens if eye lens is not able to adjust its focal length?
In this case the person cannot see the object clearly and comfortably.

Question 13.
What happens if the focal length of eye lens is beyond the range of 2.5 cm to 2.27 cm?
The vision (image) becomes blurred due to defects of eye lens.

10th Class Physics Textbook Page No. 92

Question 14.
What can we do to correct myopia?
To correct myopia, we use concave lens.

10th Class Physics Textbook Page No. 93

Question 15.
How can you decide the focal length of the lens to be used to correct myopia?
Assume that the object distance (u) is infinity and image distance (v) is equal to distance of far point.
u = – ∞ ; v = distance of far point = – D
Let ‘f be the focal length of bi-concave lens.

Here ‘f is negative showing that it is a concave lens.

Question 16.
What happens when the eye has a minimum focal length greater than 2.27 cm?
In this case, the rays coming from the nearby object after refraction at eye lens, forms image beyond the retina.

10th Class Physics Textbook Page No. 94

Question 17.
How can you decide the focal length of convex lens to be used?
Here object distance (u) = -25 cm
Image distance (v) = – d (distance of near point)
Let ‘f be the focal length of bi-convex lens.

If d > 25 cm, then ‘f’ becomes positive then we need to use bi-convex lens to correct hypermetropia.

10th Class Physics Textbook Page No. 95

Question 18.
Have you ever observed details in the prescription?
A prescription that contains some information regarding type of lens to be used to correct vision.

Question 19.
What does it (“my sight is increased or decreased”) mean?
Usually doctors, after testing the defects of vision prescribe corrective lenses indicating
their power which determines the type of lens to be used and its focal length.

Question 20.
What do you mean by power of lens ? flftorH)
The reciprocal of focal length is called power of lens.

Question 21.
Doctor advised to use 2D lens. What is its focal length?
Given power of lens P = 2D

∴ Focal length of lens (f) = 50 cm.

10th Class Physics Textbook Page No. 96

Question 22.
How could the white light of the sun give us various colours of the rainbow?
Due to reflection, refraction and dispersion of sunlight.

Question 23.
What happens to a light ray when it passes through a transparent medium bounded by plane surfaces which are inclined to each other?
When light incident on one of the plane surfaces and emerges from the other.

Question 24.
What is a prism?
A prism is a transparent medium separated from the surrounding medium by consisting two refracting plane surfaces which are inclined.

10th Class Physics Textbook Page No. 97

Question 25.
What is the shape of the outline drawn for a prism?
The outline drawn for a prism is in a triangle shape.

10th Class Physics Textbook Page No. 98

Question 26.
How do you find the angle of deviation?
The angle between the extended incident and emergent rays is called angle of deviation.
(OR)
Extend both incident and emergent rays till they meet at a point ‘O’. Measure the angle between these two rays. This is the angle of deviation.

Question 27.
What do you notice from the angle of deviation?
The angle of deviation decreases first and then increases with increase in the angle of incidence.

Question 28.
Can you draw a graph between angle of incidence and angle of deviation?

Yes, we can draw the graph between angle of incidence and angle of deviation.

Question 29.
From the graph, can you find the minimum of the angle of deviation?
Yes, we can. Draw a tangent line to the curve, parallel to X – axis, at the lowest point of the graph. The point where this line cuts Y – axis gives the angle of minimum deviation.

Question 30.
Is there any relation between the angle of incidence (i) angle of emergence (r) and piangle of deviation (d)?
(i1 + i2)= A + D
i + r = A + D

10th Class Physics Textbook Page No. 101

Question 31.
In activity-3,we noticed that light has chosen different paths. Does this mean that the refractive index of the prism varies from colour to colour?
Yes, refractive index of the prism varies from colour to colour.

Question 32.
Is the speed of light of each colour different?
In vacuum – No, the speed of each colour is constant.
In medium – Speed is different for different colours.

Question 33.
Can you guess now, why light splits into different colours when it passes through a prism?
Due to dispersion of light and different wavelength of colours in medium.

10th Class Physics Textbook Page No. 102

Question 34.
Does it (light passing through a prism) split into more colours? Why?
We know the frequency of light is the properly of the source and it is equal to number of waves leaving the source per second. This cannot be changed by any medium. Hence frequency doesn’t change due to refraction. The coloured light passing through any transparent medium retains its colour.

Question 35.
Can you give an example in nature, where you observe colours as seen in activity 3?
Yes, in rainbow. It is a good example of dispersion of light.

Question 36.
When do you see a rainbow in the sky?
Dut to the refraction, reflection and dispersion of sunlight, when the sunlight passes through the rain drops, we can see the rainbow in the sky.

Question 37.
Can we create a rainbow artificially?
Yes, we can create a rainbow artificially.

10th Class Physics Textbook Page No. 104

Question 38.
Why does the light dispersed by the raindrops appear as a bow?

• A rainbow is not the flat two dimensional arc as it appears to us.
• The rainbow we see is actually a three dimensional cone with the tip of our eye.
• All the drops that disperse the light towards us lie in the shape of the cone – a cone of different layers.
• The drops that disperse red colour to our eye are on the outermost layer of the cone, similarly, the drops that disperse orange colour to our eye are on the layer of the cone beneath the red colour cone.
• In this way the cone responsible for yellow lies beneath orange and so on it till the violet colour cone becomes the innermost cone.

Question 39.
Why is the sky blue?
A clear cloudless day – time sky is blue because molecules in the air scatter blue light from the sun more than thev scatter red liuht.

Question 40.
What is scattering?
A. Atoms or molecules which are exposed to absorb light energy and emit some part of the light energy in different directions is called scattering of light.

10th Class Physics Textbook Page No. 106

Question 41.
Why is that the sky appears white sometimes when you view it in certain direction on hot days?

• On a hot day, due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere.
• These water molecules scatter the colours of other frequencies (other than blue).
• All such colours of other frequencies reach our eye and the sky appears white.

Question 42.
Do you know the reasons for appearance the red colour of sun during sunrise and at sunset?
(OR)
Sun appears red in colour during sunrise and sunset. Give reason.
(OR)
Why does sky appear red at Sunshine and Sunset?

• The light rays from the sun travel more distance in atmosphere to reach our eye in morning and evening times.
• During sunrise and sunset except red light all colours scatter more and vanish before they reach us.
• Since scattering of red light is very less, it reaches us.
• As a result sun appears red in colour during sunrise and sunset.

Question 43.
Can you guess the reason why sun does not appear red during noon hours?
During noon hours, the distance to be travelled by the sun rays in atmosphere is less than when compared to morning and evening hours. Therefore all colours reach our eye without scattering. Hence the sunlight appears white in noon hours.

10th Class Physics 7th Lesson Human Eye and Colourful World Activities

Activity – 1

Question 1.
How do you find least distance of distinct vision?
(OR)
What is the least distance a person can see an object comfortably and distinctly known as ? Write an activity to find that (least, distance of distinct vision) distance.

• The least distance a person can see an object comfortably and distinctly is known as least distance of distinct vision.
• Hold the textbook at certain distance with your hands.
• Try to read the contents on the page.
• Gradually move the book towards eye, till it reaches very close to your eyes.
• You may see that printed letters on the page appear blurred or you feel strain to read.
• Now move the book backwards to a position where you can see clear printed letters without strain.
• Note down its value.
• Repeat this activity with other friends and note down the distances for distinct vision in each case.
• Find the average of all these distances of clear vision.
• You will notice that to see an object comfortably and distinctly, you must hold it at a distance about 25 cm from your eyes.
• This 25 cm distance is called least distance of distinct vision.
• This value varies from person to person and with age.

Activity – 2

Question 2.
How can you find angle of vision?
(OR)
What is maximun angle a person is able to see whole object? Write an activity to find that angle.

• The maximum angle, at which we are able to see the whole object is called angle of vision.
• Arrange a retort stand.
• Collect a few wooden sticks (or) PVC pipes that are used for electric wiring.
• Prepare sticks or pipes of 20 cm, 30 cm, 35 cm, 40 cm, 50 cm from them.
• Keep the retort stand on a table and stand near the table such that vour head is beside the vertical stand.
• Adjust the clamp on horizontal rod and fix at a distance of 25 cm from the eyes.
• Ask one of your friends to fix a wooden stick of 30 cm height to the clamp in vertical position.
• Now keeping your vision parallel to horizontal rod of the stand, try to see the top and bottom of wooden stick kept in vertical position.
• If you are not able to see both ends of stick at this distance (25 cm), adjust the vertical stick on the horizontal rod till you are able to see both ends of the stick at nearest possible distance from your eye.
• Fix the clamp to the vertical stick at this position.
• Without changing the position of the clamp on horizontal rod, replace this stick of 30 cm length.
• Try to see the top and bottom of the stick simultaneously without any change in the position of eye.
• Try the same activity with various lengths of the sticks.
• You can see the whole object AB which is at a distance of 25 cm (least distance of clear vision) because the rays coming from the ends A and B of the object will enter the eye.
• Similarly we can also see complete object CD with eye as explained above.
• Let us assume that AB is moving closer to eye to a position A B .
• You notice that you will be able to see only the part (EF) of the object A B because the rays coming from E and F enter your eye.

• The rays coming from A and B cannot enter your eye.
• The rays coming from the extreme ends of an object forms an angle at the eye.
• If this angle is below 60°, we can see whole object.
• If this angle is above 60°, then we see only the part of the object.
• This maximum angle, at which we are able to see the whole object is called angle of vision.
• The angle of vision for a healthy human being is about 60°.
• It varies from person to person and with age.

Activity – 3

Question 3.
Describe an activity for dispersion of light.
(OR)
What is the name given to process when white light passes through a prism it splits into different colours ? Explain the process with an activity.

• The splitting of white light into different colours is called despersion of light.
• Do this experiment in the dark room.
• Take a prism and place it on the table near a vertical white wall.
• Take a thin wooden plank.
• Make a small hole in it and fix it vertically on the table.
• Place the prism between the wooden plank and wall.
• Place a white light source behind the hole of wooden plank.
• Switch on the light.
• The rays coming out of the hole of plank become a narrow beam of light.
• Adjust the height of the prism such that the light falls on one of the lateral surfaces.
• Observe the changes in emerged rays of the prism.
• Adjust the prism by slightly rotating it till you get an image on the wall.
• You will observe that white light is splitting into certain different colours.

Activity – 6

Question 4.
Describe an experiment for scattering of light.
(OR)
What is the principle involved in blue of the sky ? Explain the principle with an experiment?

• Take the solution of sodium-thio-Sulphate (hypo) with sulphuric acid in a glass beaker.
• Place the beaker in which reaction is taking place in an open place where abundant sunlight is available.
• Watch the formation of grains of sulphur and observe changes in beaker.
• You will notice that sulphur precipitates as the reaction is in progress.
• At the beginning, the grains of sulphur are smaller in size as the reaction progresses, their size increases due to precipitation.
• Sulphur grains appear blue in colour at beginning and slowly their colour becomes white as their size increases.
• The reason for this is scattering of light.
• At the beginning, the size of grains is small and almost comparable to the wavelength of blue light.
• Hence they appear blue in the beginning.
• As the size of grains increases, their size becomes comparable to the wavelengths of other colours.
• As a result of this, they act as scattering centres for other colours.
• The combination of all these colours appears as white.

## AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Physics Solutions 6th Lesson Refraction of Light at Curved Surfaces

### 10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Textbook Questions and Answers

Question 1.
Have you ever touched a magnifying glass with your hand?
Yes.

Question 2.
Have you touched the glass in the spectacles used for reading with your hand?
A. Yes.

Question 3.
Is it a plane or curved surface?
Curved surface.

Question 4.
Is it thicker in the middle or at the edge?
Answeer:
Magnifying glass and some spectacle are thicker in middle whereas some spectacles are thicker at edge.

Question 1.
A man wants to get a picture of a zebra. He photographed a white donkey after fitting a glass, with black stripes, on to the lens of his camera. What photo will he get? Explain. (AS1)
(OR)
A person wants to get a picture of zebra and he photographed a white donkey fitting a glass with black stripes. Does he get photo of zebra? Explain.
The person was unable to gel the picture <>l zebra because only two rays are enough to form complete image after convergence. So he will get the image of white donkey but the intensity may be less.
(OR)
He will get a picture of while donkey because e\ery part of lens forms an image so if you cover lens with stripes still it forms a complete image. However, the intensity of the image will be reduced.

Question 2.
Two converging lenses are to be placed in the path of parallel rays so that the rays remain parallel after passing through both lenses. How should the lenses be arranged? Explain with a neat ray diagram. (AS1)

• Two lenses are placed in the path of parallel rays as shown in figure.
• The first lens is placed in the direction of parallel lines, which converges at focus.
• The second lens is arranged so that it is the focus of 2nd then emerging rays will be parallel.

Question 3.
The focal length of a converging lens is 20 cm. An object is 60 cm from the lens. Where will the image be formed and what kind of image is it? (AS1)
f = 20 cm (by sign conversion f = + 20 cm)
u = 60 cm (by sign conversion u = – 60 cm)

Image will be formed at 30cm in between F1, and 2F1. Image is real, inverted and diminished.

Question 4.
A double convex lens has two surfaces of equal radii ‘R’ and refractive index n = 1.5. Find the focal length ‘f’. (AS1)
(OR)
What is the focal length ‘f, when its double convex lens has two surfaces of equal radii ‘R’ and refractive index n = 1.5?
R1 = R2 = R (suppose)
Focal length (f) = ?; Refractive index (n) = 1.5

∴ Focal length of lens = Radius of curvature of surface.

Question 5.
Write the lens maker’s formula and explain the terms in it. (AS1)
(OR)
Ravi wants to make a lens. Which formula he has to follow ? Write the formula and explain the terms in it.
(OR)
Write lens formula.
Lens maker’s formula:
$$\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$$
n = Refractive index of the medium
R1 = Radius of curvature of 1 st surface
R2 = Radius of curvature of 2nd surface
f = Focal length

Question 6.
How do you verify experimentally that the focal length of a convex lens is increased when it is kept in water? (AS1)
(OR)
Write an activity to show that the focal length of a lens depends on its surrounding medium.
Aim :
To prove focal length of convex lens is increased when it is kept in water.

Apparatus :
Convex lens, water, cylindrical vessel, circular lens holder, stone.

Procedure :

1. Take a cylindrical vessel like glass tumbler.
2. Its height must be greater than the focal length of lens, (the around four times focal length of lens).
3. Keep a black stone inside the vessel at its bottom.
4. Pour the water into the vessel such that the height of the water level from the top of the stone is greater than the focal length of lens.
5. Now dip the lens horizontally using a circular lens holder.
6. Set the distance between stone and lens that is equal to or less than focal length of lens.
7. Now see the stone through the lens.
8. We can see the image of the stone.
9. If we dip the lens to a certain height which is greater than the focal length of lens in air, still we can see the image.
10. This shows that the focal length of convex lens has increased in water.
11. Thus we conclude that the focal length of lens depends upon the surrounding medium.

Note : For convenience, use 5 or 10 cm focal length convex lens.

Question 7.
How do you find the focal length of a lens experimentally? (AS1)

• Take the lens (Ex : Convex), which focused towards the distant object.
• A white coated screen (Ex : White paper) is placed on the other side of the lens.
• Adjust the screen till you get a clear image of the object.
• At this position measure the distance between the lens and screen which is equal to the focal length of the lens.

Question 8.
Harsha tells Siddhu that the double convex lens always behaves like a convergent lens. But Siddhu knows that Harsha’s assertion is wrong and corrected Harsha by asking some questions. What are the questions asked by Siddhu? (As2)
The questions asked by Siddhu :

1. Is the object placed beyond 2f point?
2. Is the object located at 2f point?
3. Is the object located in between the 2f and the focal point?
4. Is the object located at the focal point?
5. Is the object located in front of the focal point?
6. Is the lens kept in a medium with refractive index less than lens or more than lens?

Question 9.
Assertion (A): A person standing on the land appears taller than his actual height to a fish inside a pond. (AS2)
Reason (R) : Light bends away from the normal as it enters air from water.
Which of the following is correct? Explain.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true and R is not the correct explanation of A.
c) A is true but R is false.
d) Both A and R are false.
e) A is false but R is true.
Explanation :
Because the light travelling from water to air it bends away from the normal so the fish observes the apparent image of the person, appears taller than his original.

Question 10.
A convex lens is made up of three different materials as shown in the figure Q-10. How many of images does it form? (AS2)

• A lens made of three different materials of refractive indices say n1, n2 and n3.
• These three materials will have three different refractive indices. Thus for a given object it forms three images.

Question 11.
Can a virtual image be photographed by a camera? (AS2)
Yes, we can.
Ex : – A plane mirror forms a virtual image, we can able to take photograph of that image in plane mirror.

Question 12.
You have a lens. Suggest an experiment to find out the focal length of the lens. (AS3)
(OR)
Through an experiment, find out the focal length of the lens.
Aim :
To find focal length of given lens.

Apparatus :
Object (candle), convex lens, v – stand, screen.

Procedure :

• Take a v-stand and place it on a long table at the middle.
Place a convex lens on the v-stand. Imagine the principal axis of the lens.
• Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
• Adjust a screen (a sheet of white paper placed perpendicular to the axis) which is on other side of the lens until you get an image on it.
• Measure the distance of the image from the v-stand of lens (image distance V) and also measure the distance between the candle and stand of lens (object distance ‘u’). Record the values in the table.

• Now place the candle at a distance of 60 cm from the lens, try to get an image of the candle flame on the other side on a screen. Adjust the screen till you get a clear image.
• Measure the image distance V and object distance ‘u’ and record the values in table.
• Repeat the experiment lor various object distances like 50 cm, 40 cm, 30 cm, etc. Measure the image distances in all cases and note them in table.
• Using the formula $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$, find f in all the cases. We will observe the value ‘f is equal in all cases. This value off is the focal length of the given lens.

Question 13.
Let us assume a system that consists of two lenses with focal length f1, and f2 respectively. How do you find the focal length of the system experimentally, when
i) two lenses are touching each other
ii) they are separated by a distance ‘d’ with common principal axis? (AS3)
Experimental Proof:
i) Two lenses are touching each other :
Aim :
To find focal length of combination of two convex lenses, touching each other. Material required : Convex lenses – 2 (with known focal lengths say f, and f2); V-stands – 2, candle, screen scale.

Procedure:

• Place two V-stands with two convex lenses as they touch each other on a table.
• Place a candle (object) far away from the lenses.
• Adjust a screen, which is placed other side of the lenses until we get a clear image on it.
• At that position, measure the image distance (v) and object distance (u).
• Do this experiment for several object distances and record in the given table.

ii) They are separated by a distance of ‘d’ :
Procedure :

• Now place v-stands along with lenses with distance’d’.
• Do the same procedure again.
• Record the observations in the given table.
• Find the average of the ‘f’comb.

Question 14.
Collect the information about the lenses available in an optical shop. Find out how the focal length of a lens may be determined by the given power’ of the lens. (AS4)
I had collected the information regarding different lenses available at optical shops.
The relationship between power and focal length is power (D) = $$\frac{1}{f}$$. f is in meters.

 Power of lens in diopters Type of lens Focal length 0.25 Convex 400 cm 0.5 Convex 200 cm 1 Convex 100 cm -2 Concave 50 cm – 1 Concave – 100 cm -0.5 Concave – 200 cm -0.25 Concave – 400 cm

Question 15.
Collect the information about lenses used by Galileo in his telescope. (AS4)
(OR)
What lenses are used by Galileo in his telescope?

A Galilean telescope is defined as having one convex lens and one concave lens. The concave lens serves as the ocular lens or the eye piece, while the convex lens serves as the objective. The lens are situated on either side of a tube such that the focal point of the ocular lens is the same as the focal point for the objective lens.

Question 16.
Use the data obtained by activity – 2 in table-1 of this lesson and draw the graphs of u vs v and $$\frac{1}{u}$$ vs $$\frac{1}{v}$$ (AS5)
(OR)
By obtaining data from activity – 2 in table – 1 of this lesson, draw the graphs of u vs v and $$\frac{1}{u}$$ vs $$\frac{1}{v}$$
Graph of u – v using data obtained by activity – 2. Take lens with focal length 30 cm.

 Object distance (u) Image distance (v) Focal length (f) 60 cm 60 cm 30 cm 50 cm 75 cm 30 cm 40 cm 120 cm 30 cm

The graph looks like this

The shape of the graph is rectangular hyperbola.

Graph of $$\frac{1}{u}$$ – $$\frac{1}{v}$$

For these values the graph is straight line which touches the axis as shown in figure.

Question 17.
Figure shows ray AB that has passed through a divergent lens. Construct the path of the ray up to the lens if the F position of its foci is known. (AS5)

The path of the ray up to the lens if the position of foci is known for ray AB is diverging lens or concave lens path.

Question 18.
Figure shows a point light source and its image produced by a lens with an optical axis N1, N2. Find the position of the lens and its foci using a ray diagram. (AS5)

1. The object is in between focus and optic centre.
2. The image is virtual, erect and magnified. Nv
3. l is the lens, ‘O’ is the object and T is the image.

Question 19.
Find the focus by drawing a ray diagram using the position of source S and the image S’ given in the figure. (AS5)

1.  Image is real.
2. l’ is lens, ‘O’ is object and T is image.
3.  Lens is convex.

(Or)

1. Image is real.
2. l’ is lens, ‘O’ is object and ‘I’ is image.
3. Lens is convex.

Question 20.
A parallel beam of rays is incident on a convergent lens with a focal length of 40 cm. Where should a divergent lens with a focal length of 15 cm be placed for the beam of rays to remain parallel after passing through the two lenses? Draw a ray diagram. (AS5)

1. A parallel beam of rays when incident on a convergent lens, after refraction they meet at the focus of the lens.

2. A beam of rays which is incident on a divergent lens, after refraction, pass parallel to the principal axis. If we extend these incident rays, they seems to meet at focus of the lens.

3. Hence the divergent lens should be kept at 25 cm distance from convergent lens (40 – 15 = 25 cm) as shown in the figure.

PF = 40 cm (Focal length of convergent lens)
P’F = 15 cm (Focal length of divergent lens)
PP’ = 40 – 15 = 25 cm (Position of divergent lens)

Question 21.
Draw ray diagrams for the following positions and explain the nature and position of image.
i) Object is placed at 2F2
ii) Object is placed between F2 and optic centre P. (AS5)
i) Object is placed at 2F2:

Nature : Real, inverted and diminished.
Position : Image is formed on the principal axis between the points F1, and 2F1.

ii) Object is placed between F2 and optic centre P :

Nature : Virtual, erect and magnified.
Position : Same side of the lens where object is placed.

Question 22.
How do you appreciate the coincidence of the experimental facts with the results obtained by a ray diagram in terms of behaviour of images formed by lenses? (AS6)

• Ray diagrams are very useful in optics.
• By the ray diagrams, we can easily find the values of image distance, object distance, focal length, radius of curvature, magnification, etc.
• These results are exactly equal to the result gotten by an experiment.
• For example : In the experiment, with a convex lens, we get clear image of an object, on a screen by adjusting the screen.

Then, we measure the image distane (v) practically. This takes more time and requires equipped lab also.

But, by simply draw a ray diagram on a paper, we can get exact image distance (v) very easily, without lab.

• So, ray diagrams are very useful in the construction of microscopes, telescopes, etc.
• Hence, one can trust and depend on the result of ray diagrams instead of several lab experiments.
• So, I appreciate the ray diagrams.

Question 23.
Find the refractive index of the glass which is a symmetrical convergent lens if its focal length is equal to the radius of curvature of its surface. (AS7)
Given that lens is convergent symmetrical

Question 24.
Find the radii of curvature of a convexo – concave convergent lens made of glass with refractive index n = 1.5 having focal length of 24 cm. One of the radii of curvature is double the other. (AS7)

Question 25.
The distance between two point sources of light is 24 cm. Where should a convergent lens with a focal length of f = 9 cm be placed between them to obtain the images of both sources at the same point? (AS7)
For Source S1 :

∴ The convex lens may be placed between the two sources, such that a distance of 18 cm from one source, and 6 cm from other source.

Question 26.
Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height? Why? (AS7)
(OR)
If your friend is standing near an edge of the swimming pool and you are in the water, do you find he is taller or shorter than his usual height?

1. My friend appears to be taller because the light is travelling from rarer to denser.
2. The rays bend in such away that they seems to be coming from long distance.
3. So it is actually apparent image of my friend which appears to be taller due to refraction.

Fill in the Blanks

1. The rays from the distant object, falling on the convex lens pass through ……………….. .
2. The ray passing through the ……………….. of the lens is not deviated.
3. Lens formula is given by ……………….. .
4. The focal length of the plano-convex lens is 2R where R is the radius of curvature of the surface. Then the refractive index of the material of the lens is ……………….. .
5. The lens which can form real and virtual images is ……………….. .

1. Tocus
2. optical centre
3. $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$
4. 1.5
5. convex lens

Multiple Choice Questions

1. Which one of the following materials cannot be used to make a lens?
A) water
B) glass
C) plastic
D) clay
D) clay

2. Which of the following is true?
A) The distance of virtual image is always greater than the object distance for convex lens.
B) The distance of virtual image is not greater than the object distance for convex lens.
C) Convex lens always forms a real image.
D) Convex lens always forms a virtual image.
B) The distance of virtual image is not greater than the object distance for convex lens.

3. Focal length of the plano-convex lens is when its radius of curvature of the surface is R and n is the refractive index of the lens.

4. The value of the focal length of the lens is equal to the value of the image distance when the rays are
A) passing through the optic centre
B) parallel to the principal axis
C) passing through the focus
D) in all the cases
D) in all the cases

5. Which of the following is the lens maker’s formula?

C

### 10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Additional Questions and Answers

Question 1.
Derive a relation between refractive indices of two media (n1, n2), object distance (u), image distance (v) and radius of curvature (R) for a curved surface.
(OR)
Derive $$\frac{\mathbf{n}_{2}}{\mathbf{v}}-\frac{\mathbf{n}_{1}}{\mathbf{u}}=\frac{\mathbf{n}_{2}-\mathbf{n}_{1}}{\mathbf{R}}$$
(OR)
Derive curved surface formula.

• Consider a curved surface separating two media of refractive indices n1, and n2.
• A point object is placed on the principal axis at point ‘O’.
• The ray which travels along the principal axis passes through the pole undeviated.
• The second ray, which forms an angle with a princi¬pal axis, meets the interface at A. The angle of incidence is Q1. The ray bends and passes through the second medium along the line AI. The angle of refraction is Q2.
• The two refracted rays meet at I and the image is formed there.
• 6) Let the angle made by the second refracted ray with principal axis be γ and the angle between the normal and principal axis be β.
• From figure,

PO = u (object distance), PI = v (Image distance),
PC = R (radius of curvature) and n1, n2 are refractive indices of the media.
From ∆ACO, θ1 = α + β
∆ACI, β = θ2 + γ
⇒ θ2 = β – γ
According to Snell’s law, n1sin θ1 = n2 sin θ2.
∴ n1 sin (α + β) = n2 sin (β – γ) …………….. (1)
As per paraxial approximation,
sin (α + β) = α + β and sin (β – γ) = β – γ.
∴ (1) ⇒ n1(α + β) = n2 (β – γ)
⇒ n1 α + n1β = n2 β – n2 γ — (2)
Since all angles are small, we can write

∴ This is the required relation for curved surfaces.

Question 2.
Derive expression for lens maker’s formula.
(OR)
Prove $$\frac{1}{\mathbf{f}}=(\mathbf{n}-\mathbf{1})\left(\frac{1}{\mathbf{R}_{1}}-\frac{1}{\mathbf{R}_{2}}\right)$$.

Procedure :

• Imagine a point object ‘O’ placed on the principal axis of the thin lens
• Let this lens be placed in a medium of refractive index na and let refractive index of lens be nb.
• Consider a ray, from ‘O’ which is incident on the convex surface of the lens with radius of curvature R1 at A.
• The incident ray refracts at A.
• It forms image at Q, if there were no concave surface.
• From figure Object distance PO = – u;

Image distance PQ = v = x
Radius of curvature R = R1
n1 = na and n2 = nb.

• But the ray that has refracted at A suffers another refraction at B on the concave surface with radius of curvature (R2).
• At B the ray is refracted and reaches I.
• The image Q of the object due to the convex surface. So I is the image of Q for concave surface.
• Object distance u = PQ = + x
Image distance PI = v
Radius of curvature R = – R2
• The refraction of the concave surface of lens is medium -1 and surrounding is medium – 2.
∴ n1 = nb and n2 = na

Question 3.
Derive the lens formula.
1. Consider an object 00′ placed on the principal axis in front of a convex lens as shown in the figure. Let II’ be the real image formed by the lens, i.e. the other side of it.

2. From the figure : PO, PI, PFt are the object distance, image distance and focal length respectively.

### 10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 64

Question 1.
What happens to a ray that is incident on a curved interface separating the two media? Are the laws of refraction still valid?
It undergoes deviation from its path. Yes, the laws of reflection are still valid.

Question 2.
How do rays betid when they are incident on a curved surface?
A ray will bend towards the normal when it travels from rarer to denser medium and bends away from the normal when it travels from denser to a rarer medium.

10th Class Physics Textbook Page No. 65

Question 3.
What happens to ray that travels along the principal axis? Similarly, a ray that travels through the centre of curvature?

According to Snell’s law the ray which travels along the normal drawn to the surface does not deviate from its path. Hence both rays in the given condition travel along normal, so they do not deviate.

Question 4.
What difference do you notice in the refracted rays in 4 (a) and 4 (b)? What could be the reason for that difference?

• In figure 4 (a) ray travelling parallel to the principal axis strikes a convex surface and passes from a rarer medium to a denser medium.
• In figure 4 (b) a ray travelling parallel to the principal axis strikes a convex surface passes from a denser medium to a rarer medium.
• Figure 4 (a) : The refracted ray moves towards the normal.
• Figure 4 (b) : The refracted ray moves away from the normal.
Reason : The main reason is that light passes through different media.

10th Class Physics Textbook Page No. 66

Question 5.
What difference do you notice in refracted rays in 4 (c) and 4 (d)? What could be the reasons for that difference?
(OR)
Draw the ray diagrams when the incident ray passes through the curved surfaces.
a) Rarer medium to denser medium.
b) Denser medium to rarer medium.

• In figure 4 (c) a ray travelling parallel to the principal axis strikes a concave surface and passes from a denser medium to a rarer medium.
• In figure 4 (d) a ray travelling parallel to the principal axis strikes a concave surface and passes from a rarer medium to a denser medium.

Reasons :

• Figure 4 (c) :The refracted ray reaches a particular point on the principal axis.
• Figure 4 (d) : The refracted ray moves away from the principal axis.
• The main reason is that light passes through different media.

Question 6.
You might have observed that a lemon in the water of a glass tumbler appears bigger than its actual size, when viewed from the sides of tumbler.
1) How can you explain this (appeared) change in size of lemon?
It can be explained by using refraction. When light travels from one medium to another medium it undergoes refraction.

2) Is the lemon that appears bigger in size an image of lemon or is it the real lemon?
That is image of lemon.

3) Can you draw a ray diagram to explain this phenomenon?

10th Class Physics Textbook Page No. 70

Question 7.
What happens to the light ray when a transparent material with two curved surfaces is placed in its path?
The light ray undergoes refraction.

Question 8.
Yes, we have heard about lenses. A transparent material bounded by two spherical v surfaces is called lens.

Question 9.
How does a light ray behave when it is passed through a lens?
A light ray will deviate from its path in some cases and does not deviate in some other cases.

10th Class Physics Textbook Page No. 72

Question 10.
How does the lens form an image?
Lens forms an image through converging light rays or diverging light rays.

Question 11.
If we allow a light ray to pass through the focus, which path does it take?
The ray passing through the focus takes a parallel path to principal axis after refraction.

10th Class Physics Textbook Page No. 73

Question 12.
What happens when parallel rays of light fall on a lens making some angle with the principal axis?
The rays converge at a point (or) appear to diverge from a point lying on the focal plane.

Question 13.
What do you mean by an object at infinity? What type of rays fall on the lens?
The distance between the lens and the object is very much greater than when compared to object size is known as object at infinity. Parallel rays fall on the lens.
The object at infinity means distant object. The rays falling on the lens from an object at infinity are parallel to principal axis.

10th Class Physics Textbook Page No. 77

Question 14.
Could you get an image on the screen for every object distance with a convex lens?
No, when the object is placed between pole and focus we will get virtual, erect and enlarged image on the other side of the- object.

Question 15.
Why don’t you get an image for certain object distances?
Because at those distances the light rays diverge each other.

Question 16.
Can you find the minimum limiting object distance for obtaining a real image? What do you call this minimum limiting object distance?
Yes, this minimum limiting object distance is called focal length.

Question 17.
When you do not get an image on the screen, try to see the image with your eye directly from the place of the screen. Could you see the image? What type of image do you see?
Yes, we can see the image. This is a virtual image which we cannot capture on screen.

Question 18.
Can you find the image distance of a virtual image? How could you do it?
We can find the image distance of virtual image by using lens formula $$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$$ (if we know the focal length of lens and object distance.)

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Activities

Activity – 1

Question 1.
Write an activity to observe the light refraction at curved surface.
Procedure and observation :

• Draw an arrow of length 4 cm usfng a black sketch pen on a thick sheet of paper.
• Take an empty cylindrical-shaped transparent vessel.
• Keep it on the table.
• Ask your friend to bring the sheet of paper on which arrow was drawn behind the vessel while you look at it from the other side.
• We will see a diminished image of the arrow.
• Look at the arrow from the same position as before.
• We can observe an inverted image.

Explanation :

• In the first case, when the vessel is empty, light from the arrow refracts at the curved interface, moves through the glass, enters in to air then it again undergoes refraction on the opposite curved surface of vessel and comes out into the air.
• In this way light travels through two media, comes out of the vessel and forms a diminished image.
• In the second case, light enters the curved surface, moves through water, comes out of the glass and forms an inverted image.

Lab Activity

Question 2.
Write an activity to know the characteristics of image due to convex lens at various distances.
Aim:
Determination of focal length of bi-convex lens using UV method.

Material Required :
V Stand, convex lens, light source, screen, meter scale. Take a V-stand and place it on a long (nearly 2m) table at the middle. Place a convex lens on the v-stand. Imagine the principal axis of the lens. Light a candle and ask your friend to take the candle far away from the lens along the principal axis. Adjust a screen (a sheet of white paper placed perpendicular to the axis) which is on other side of the lens until you get an image on it.

Procedure :

1. Take a V-stand and place a convex lens on this stand.
2. Imagine the principal axis of the lens.
3. Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
4. We use a screen because it forms a real image generally which will form on a screen. Real images cannot be seen with an eye.
5. Adjust the screen, on other side of lens until clear image forms on it.
6. Measure the distance of the image from the stand and also measure the distance between the candle and stand of lens.
7. Now place the candle at a distance of 60 cm from the lens such as the flame of the candle lies on the principal axis of the lens.
8. Try to get an image of candle flame on the other side on a screen.
9. Adjust the screen till you get a clear image.
10. Measure the distance of image (v) from lens and record the value of’u’ and V in the table.
11. Repeat this for various distances of images; in all cases note them in the table.

Observation :

Conclusion : From this we conclude that a convex lens forms both real and virtual images when object is placed at various positions.

## AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Physics Solutions 5th Lesson Refraction of Light at Plane Surfaces

### 10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Textbook Questions and Answers

Question 1.
Why is it difficult to shoot a fish swimming in water? (AS1)
(OR)
If the fish is swimming in water, why it is difficult to shoot?
(OR)
A shooter finds it difficult to shoot a fish swimming in water. Why?
Due to refraction of light, it is difficult to shoot a fish swimming in water.

Reason :
The light rays coming from the fish towards shooter, bend at water-air interface. So, shooter sees only image of the fish, but not actual fish.

Question 2.
The speed of light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. (AS1)
Speed of light in diamond = 1,24,000 km/s
Speed of light in vacuum = 3,00,000 km/s

Question 3.
Refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass? (AS1)
Refractive index of glass relative to water = $$\frac{n_{g}}{n_{w}}=\frac{9}{8}$$
∴ Refractive index of water relative to glass = $$\frac{\mathrm{n}_{\mathrm{w}}}{\mathrm{n}_{\mathrm{g}}}=\frac{8}{9} \cdot\left[\because \mathrm{n}_{12}=\frac{1}{\mathrm{n}_{21}}\right]$$

Question 4.
The absolute refractive index of water is 4/3. What is the critical angle ? (AS1)
Absolute refractive index of water = 4/3

Critical angle of water = C = 48°5′ = 48.5°.

Question 5.
Determine the refractive index of benzene if the critical angle is 42°. (AS1)
Critical angle of benzene = 42°.

Question 6.
Explain the formation of mirage. (AS1)
(OR)
How is the mirage formed? Explain.
(OR)
A person walking on a road observed some water being present on the road but when he went there actually he did not find water. What is that actually formed called? Explain that process.
(OR)
Sometimes during the hot summer at noon time on tar roads, it appears that there is water on the road, but there would really be no water. What do you call this phenomenon? Explain why it happens.
(OR)
Why do you see a mirage the road on a hot summer day?

• During hot summer day, air just above the road surface is very hot and the air at higher altitudes is cool.
• We know that refractive index of air increases with density.
• So, the cooler air at the top has greater refractive index than hotter air just above the road.

• Light travels faster through the thinner hot air than the denser cool air above it.
• On hot days, the temperature decreases with height.
• Thus the refractive index of air increases with height.
• When the light from a tall object such as tree or from the sky passes, through a medium just above the road whose refractive index decreases towards ground, suffers refraction and takes a curved path because of total internal reflection.
• This refracted light reaches the observer in a direction shown as in second figure.
• This appears to the observer that the ray is reflected from ground.
• Hence we will see water on road, which is the virtual image of sky and an inverted image of tree on the road.
• Such virtual images of distant high objects cause the optical illusion called ‘mirage’.

Question 7.
How do you verify experimentally that $$\frac{\sin i}{\sin r}$$ is a constant? (AS1)
(OR)
Explain the experiment that shows the relation between angle of incidence and angle of refraction through figure.
(OR)
Write an experiment to obtain the relation between angle of incidence and angle of refraction.
Aim:
Identifying relation between angle of incidence and angle of refraction.

Materials required :
A plank, white chart, protractor, scale, small black painted plank, a semi-circular glass disc of thickness nearly 2 cm pencil and laser light.

Procedure :

1. Take a wooden plank which is covered with white chart.
2. Draw two perpendicular lines, passing through the middle of the paper as shown in the figure (a).
3. Let the intersecting point be O.
4. Mark one line as NN which is normal to the another line marked as MM.
5. Here MM represents the line drawn along the interface of two media and NN represents the normal drawn to this line at ‘O’.
6. Take a protractor and place it along NN in such a way that its centre coincides with ‘O’ as shown in figure (b).
7. Then mark the angles from 0° to 90° on both sides of the line NN.
8. Repeat the same on the other side of the line NN.
9. The angles should be represented on circular line.
10. Now place semi circular glass disc so that its diameter coincides with the interface line (MM) and its centre coincides with the point O.
11. Take the laser light and send it along NN in such a way that the laser propagates from air to glass through the interface at point O and observe the way of laser light coming from other side of disc.
12. There is no deviation.
13. Send Laser light along a line which makes 15° (angle of incidence) with NN and see that it must pass through point O.
14. Measure its corresponding angle of refraction.
15. Repeat the experiment with angle of incidences of 20°, 30°, 40°, 50° and 60° and note the corresponding angles of refraction.

Observation :

• Find sin i, sin r for every i and r note down the values in table.

• Evaluate $$\frac{\sin i}{\sin r}$$ for every incident angle i.
• We get $$\frac{\sin i}{\sin r}$$ as constant.
• That is the relationship between angle of incidence and angle of refraction.
• The ratio of sin i and sin r is called refractive index.

Question 8.
Explain the phenomenon of total internal reflection with one or two activities. (AS1)
Procedure :

1. Place the semi-circular glass disc in such a way that its diameter coincides with interlace line MM and its center coincides with point O’.
2. Now send light from the curved side of the semicircular glass disc.
3. The light travels from denser medium to rarer medium.
4. Start with angle of incidence (i), equals to 0° and observe for refracted on other side of the disc.
5. It does not deviate into its path when entering rarer medium.
6. Send laser light along angles of incidence 5°, 10°, 15°, etc. and measure the angle of refraction.
7. And tabulate the results in the given table.

Observation :

• Make a table shown below and note the values ‘i’ and ‘r’.

• At particular angle of incidence, the refracted ray does not come out but grazes the interface separating the air and glass. This angle is called critical angle.
• When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e. light never enters rarer medium. This phenomenon is called total internal reflection.

Question 9.
How do you verify experimentally that the angle of refraction is more than angle of incidence when light rays travel from denser to rarer medium? (AS1)
(OR)
When the light rays travel from denser to rarer medium, how can you explain, the angle of refraction is more than angle of incidence experimentally?
Procedure :

• Take a metal disc. Use a protractor and mark angles along its edge as shown in the figure.
• Arrange two straws at the centre of the disc, in such a way that they can be rotated freely about the centre of the disc.
• Adjust one of the straws to make an angle 10°.
• Immerse half of the disc vertically into the water, filled in a transparent vessel. While dipping, verify that the straw at 10° must be inside the water.
• From the top of the vessel, try to view the straw which is inside the water as shown in the figure.
• Then adjust the other straw which is outside of the water until both straws look like they are in a single straight line.
• Then take the disc out of the water and observe the two straws on it. You will find that they are not in a single straight line.
• Measure the angle between the normal and second straw. Note the values in the following table.

• Do the same for various angles and find the corresponding angles of refraction and note them in the table.

Observation :
We will find the angle of refraction is more than angle of incidence.
i. e., r > i.

Conclusion :
When light travels from denser (water) to rarer (air) it bends away from the normal.

Question 10.
Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water. How does it appear and why? (Make hypothesis and do the above experiment) (AS2)

• The black metallic .ball appears to be raised up in the water because the path of the ray changes its direction at the interface, separating the two media, i.e., water and air.
• This path is chosen by light ray so as to minimize time of travel between ball and eye.
• This can be possible only when the speed of light changes at interface of two media.
• In another way the speed of light is different in different media.

Hypothesis :
Speed of light changes when it travels from one medium to another medium.

Question 11.
Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.
1) What changes do you notice?
2) What could be the reasons for these changes? (AS2)

1. We cannot see the glass rod in glycerine but we can see the rod in water.
2. We can also observe an apparent image of glass rod in water.
3. Reasons:
i) Glycerine has essentially same refractive index as glass.
ii) So, any light passing through these is bent equally.
iii) Since both are transparent, it is not possible for our eye to distinguish the boundary by a change in the angle of reflection, and the glass seems to vanish.
iv) But, the refractive index of glass and water are different.
v) So the glass rod is visible to our eye in water. .

Question 12.
Do Activity-7 again. How can you find critical angle of water? Explain your steps briefly. (AS3)

Procedure:

1. Take a cylindrical transparent vessel.
2. Place a coin at the bottom of the vessel.
3. Now pour water until you get the image of the coin on the water surface.
4. This is due to total internal reflection.

Critical angle of water :

1. Refractive index of water = 1.33
2. The sine of critical angle of water = $$\frac{1}{\text { Refractive index }}$$
3. Sin C = $$\frac{1}{\text { 1.33 }}$$ ⇒ sin C = 0.7518.
∴ C = 8.7°
4. ∴ The critical angle of water = 48.7°.

Question 13.
Collect the values of refractive index of the following media. (AS4)

 Water, coconut oil, flint glass, crown glass, diamond, benzene and hydrogen gas.

 Medium Refractive Index 1. Water 1.33 2. Coconut oil 1.445 3. Flint glass 1.65 4. Crown glass 1.52 5. Diamond 2.42 6. Benzene 1.50 7. Hydrogen gas 1.000132

Question 14.
Collect information on working of optical fibres. Prepare a report about various uses of optical fibres in our daily life. (AS4)
(OR)
What do you know about the working of optical fibres and make a report of various uses of optical fibres in our daily life?
(OR)
How are the optical fibres working? What are the various uses of optical fibres in our daily life?
1) An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer (10-6 m).
2) A bunch of such thin fibres form a light pipe.

Working :
1. Optical fibre having three parts-namely core (n = 1, 7), clading (n = 1, 6) and shielding.
2. The ray of light AB gets refracted at point ‘B’ into core and incident at ‘C’ with angle of incidence i (i > c).

3. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
4. The light is thus transmitted along the fibre.
5. The optical fibre is also based on ‘Fermat’s principle.

Uses :

1. Optical fibres are used in ‘endoscopy’ to see the internal organs like throat, stomach, intestines, etc.
2. Optical fibres are used in transmitting communication signals through light pipes.
3. Optical fibres are used in international telephone cables laid under the sea, in large computer networks, etc.
4. In optical fibre about 2000 telephone signals appropriately mixed with light waves may be simultaneously transmitted through a typical optical fibre.

Question 15.
Take a thin thermocol sheet. Cut it in circular discs of different radii like 2 cm, 3 cm, 4 cm, 4.5 cm, 5 cm etc. and mark centers with sketch pen. Now take needles of length nearly 6 cm. Pin a needle to each disc at its centre vertically. Take water in a large opaque tray and place the disc with 2 cm radius in such a way that the needle is inside the water as shown in figure. Now try to view the free end (head) of the needle from surface of the water.

1) Are you able to see the head of the needle?
Now do the same with other discs of different radii. Try to see the head of the needle, each time.
Note : The position of your eye and the position of the disc on water surface should not be changed while repeating the Activity with’other discs.
2) At what maximum radius of disc, were you not able to see the free end of the needle ?
3) Why were you not able to view the head of the nail for certain radii of the discs ?
4) Does this Activity help you to find the critical angle of the medium (water) ?
5) Draw a diagram to show the passage of light ray from the head of the nail in different situations. (AS4)

1. Yes, we can see head of the needle.

2. Height of the pin = 6 cm
Radius of the disc = x cm
Critical angle of water = C = 48.7°
Tan C = 48.7°
$$\frac{x}{6}$$ = 1.138 ⇒ x = 6.828 cm
So, at radius of 6.8 cm we cannot see the free end of the needle.

3. Because the light rays coming from object undergoing total internal reflection by touching the surface of disc.

4. Yes, we can find critical angle.
Refractive index of air (n2) = 1.003 ; Refractive index of water (n1) = 1.33
Sin C = $$\frac{n_{2}}{n_{1}}=\frac{1.003}{1.33}$$ = 0.7541 ⇒ C = 48.7°

5.

Question 16.
Explain the refraction of light through the glass slab with a neat ray diagram. (AS5)
(OR)
Draw a glass slab diagram and explain the refraction of light through glass slab.
(OR)
Write the procedure of a lab Activity to understand lateral shift of light rays through a glass slab.
(OR)
How can you find lateral shift using glass slab?
Aim :
A) Determination of position and nature of image formed by a glass slab.
B) Understanding lateral and vertical shift.
C) Determination of refractive index of given glass slab.

Materials required :
Plank, chart paper, clamps, scale, pencil, thin glass slab and pins.

Procedure :

1. Place a piece of chart on a plank. Clamp it. Place a glass slab in the middle of the paper.
2. Draw border line along the edges of the slab by using a pencil. Remove it. You will get a figure of a rectangle.
3. Name the vertices of the rectangle as A, B, C and D.
4. Draw a perpendicular at a point on the longer sides (AB) of the rectangle.
5. Now draw a line, from the point of intersection where side AB of rectangle and perpendicular meet, in such a way that it makes 30° angle with the normal.
6. This line represents the incident ray falling on the slab and the angle it makes with normal represents angle of incidence.
7. Now place the slab on the paper in such a way that it fits in the rectangle drawn. Fix two identical pins on the line making 30° angle with normal, such that they stand vertically with same height.
8. By looking at the two pins from the other side of the slab, fix two pins in such a way that all pins appear to be along a straight line.
9. Remove the slab and take out pins. Draw a straight line by joining the dots formed by the pins up to the edge CD of the rectangle.
10. This line represents emergent ray of the light.
11. Draw a perpendicular to the line CD where our last line drawn meets the line CD.
12. Measure the angle between emergent ray and normal.
13. This is called angle of emergence.
14. The angle of incidence and angle of emergence are equal.
15. Incident emergent rays are parallel.
16. The distance between the parallel rays is called shift.

Question 17.
Place an object on the table. Look at the object through the transparent glass slab. You will observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray in this situation. (AS5)

Question 18.
What is the reason behind the shining of diamond and how do you appreciate it? (AS6)
(OR)
For which reason is the diamond shining and how is it appreciable?

• The critical angle of diamonds is very low, i.e., 24.4°.
• So if a light ray enters diamond, it undergoes total internal reflection.
• It makes the diamond shine brilliant.
• So total internal reflection is main cause of brilliance of diamonds.
• Majority of people are attracted towards diamonds due to this property.
• So we have to thoroughly appreciate total internal reflection for brilliance of diamonds.

Question 19.
How do you appreciate the role of Fermat’s principle in drawing ray diagrams? (AS6)
(OR)
How do you admire the role of Fermat’s principle in drawing ray diagrams? Fermat’s principle: The light ray always travels in a path which needs shortest possible time to cover distance between two points.
This principle has lot of importance on optics. This is used in

1. Laws of reflection (i.e., angle of incidence = angle of reflection)
2. Laws of refraction (Snell’s law)
3. To derive refractive index of a medium.
4. To derive refractive index of glass slab.
So, I appreciate the Fermat’s principle.

Question 20.
A light ray is incident on air-liquid interface at 45° and is refracted at 30°. What is the refractive index of the liquid? For what angle of incidence will the angle between reflected ray and refracted ray be 90°? (AS7)
i) Given that angle of incidence (i) = 4S°
angle of refraction (r) = 30°

ii) Given that angle between reflected and refracted ray is 90°.
We know angle of incidence = angle of reflection
∴ Angle of refraction (r) = 90 – angle of incidence
= 90 – i

Critical angle = 54.7°. This angle is also known as polarising angle.

Question 21.
Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface from a certain viewing position. (AS7)
When a test tube is immersed at a certain angle in a tumbler of water appears to have
a mirror surface from a certain viewing positions due to total internal reflection.

Explanation :

• The critical angle for glass is 42°.
• The glass and air in test tube works as denser and rarer mediums.
• The rays of light while travelling through water strike glass – air interface of test tube at an angle of more than 42° (i > c) they get totally internal reflected as shown figure.
• When these reflected rays reach the eye, they appear to come from the surface of test tube itself.
• Now the test tube appears like silvary.

Question 23.
In what cases does a light ray not deviate at the interface of two media? (AS7)

1. When a light ray incident is perpendicular to the interface of surface, it does not undergo deviation.
2. When a light ray incident is more than critical angle, it does not undergo deviation (refraction) but it undergoes reflection to come back into the original medium.

Question 25.
When we sit at camp fire, objects beyond the fire seen swaying. Give the reason for it. (AS7)
(OR)
What are the reasons for the objects beyond the fire seen swaying, when we sit at camp fire?

• The temperature of the surrounding air changes due to convection of heat by the camp fire.
• This leads to chang in density and refractive index of air, continuously.
• The continuous change in refractive index of air changes the refracted path of the light ray.
• This is the cause for swaying of an object.

Question 26.
Why do stars appear twinkling? (AS7)
(OR)
What is the reason for the appearance of stars like twinkling?

• The twinkling of a star is due to atmospheric refraction of star light.
• The atmosphere consists of a number of layers of varying densities.
• When light rays coming from a star pass through this layers and undergo refraction for several times.
• Thats why stars appear twinkling.

Question 27.
Why does a diamond shine more than a glass piece cut to the same shape? (AS7)
(OR)
What is the reason for shining of diamond brightly as compared to glass piece cut?

• The critical angle of a diamond is very low (i.e., 24.4°).
• So if a light ray enters a diamond it definitely undergoes total internal reflection.
• Whereas it is not possible with glass piece cut to the same shape.
• So diamond shines more than a glass piece.

Fill In The Blanks

1. At critical angle of incidence, the angle of refraction is ……………… .
2. n1 sin i = n2 sin r, is called ……………… .
3. Speed of light in vacuum is ……………… .
4. Total internal reflection takes place when a light ray propagates from …………. to …………… medium.
5. The refractive index of a transparent material is 3/2. The speed of the light in that medium is …………… .
6. Mirage is an example of ……………… .

1. 90°
2. Snell’s law
3. 3 × 108 m/s
4. denser, rarer
5. 2 × 108 m/s
6. optical illusion / total internal reflection

Multiple Choice Questions

1. Which of the following is Snell’s law?

B)

2. The refractive index of glass with respect to air is 2. Then the critical angle of glass air interface is ………………….
A) 0°
B) 45°
C) 30°
D) 60°
C) 30°

3. Total internal reflection takes place when the light ray travels from …………….. .
A) rarer to denser medium
B) rarer to rarer medium
C) denser to rarer medium
D) denser to denser medium
C) denser to rarer medium

4. The angle of deviation produced by the glass slab is …………… .
A) 0°
B) 20°
C) 90°
D) depends on the angle formed by the light ray and normal to the slab
D) depends on the angle formed by the light ray and normal to the slab

### 10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Additional Questions and Answers

Question 1.
Derive Snell’s law.
(OR)
Prove n1 sin i = n2 sin r.
(OR)
Derive the Snell’s formula from Fermat’s principle.
(OR)
Derive the formula in realtion with and of incidence and angle of refraction.
Let X be the path and A be the point above X and B be the point below X.
Now, we have to find the way from A to B.

1. Let us try to calculate how long it would take to go from A to B by the two paths through point D and another through point C.
2. If we draw a perpendicular DE, between two paths at D, we see that the path on line is shortened by the amount EC.
3. On the other hand, in the water, by drawing corresponding perpendicular CF we find that we have to go to the extra distance DF in water. These times must be equal since we assumed there was no change in time between two paths.
4. Let the time taken by the man to travel from E to C and D to F be ∆t and v1 and v2 be the speeds of the running and swimming. From figure we get,
EC = v1 ∆t and DF = v2 ∆t
⇒ $$\frac{\mathrm{EC}}{\mathrm{DF}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}$$ ………….. (1)
5. Let i and r be the angles measured between the path ACB and normal NN, perpendicular to shore line X.

### 10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 56

Question 1.
Why should you see a mirage as a flowing water?

• A mirage is a naturally occuring optical phenomenon, in which light rays are bent to produce a displaced image of distant objects or the sky.
• As light passes from colder air (higher place) to warmer air (lower place), the light ray bends away from the direction of the temperature gradient.
• Once the rays reach the viewer’s eye, the visual cortex interprets it as if it traces back along a perfectly straight “line of sight”. However this line is at a tangent to the path the ray takes at the point it reaches the eye.
• The result is that an “inferior image ” of the sky above appears on the ground.
• The viewer may incorrectly interprets this sight as water that is reflecting the sky, which is to the brain, a more reasonable and common occurrence.

Question 2.
Can you take a photo of a mirage?

• Yes, I can take a photo of a mirage.
• Our eye can catches the total internal reflected rays.
• So, camera lens also catches the same.

10th Class Physics Textbook Page No. 46

Question 3.
What difference do you notice in fig 2(a) and Fig 2(b) with the respect to refracted rays?
(OR)
Draw the ray diagram of refraction in between denser and rarer medium.

In figure 2(a) the light ray bends towards normal whereas in 2(b) the light ray bends away from the normal.

Question 4.
Is there any relation between behaviour of refracted rays and speed of the light?
Yes. The speed of light changes when it travels from one medium to another medium. So the light may bend towards normal or away from normal.

10th Class Physics Textbook Page No. 47

Question 5.
Why db different material media possess different values of refractive indices?
Refractive index depends on nature of material. So different media have different values of refractive indices.

10th Class Physics Textbook Page No. 48

Question 6.
On what factors does the refractive index of a medium depend?
Refractive index depends on (1) Nature of material and (2) Wavelength of light used.

10th Class Physics Textbook Page No. 49

Question 7.
Can we derive the relation between the angle of incidence and the angles of refraction theoretically?
Yes, we can derive the relation between angle of incidence and angles of refraction theoretically. We can get nt sin i = n2 sin r.

10th Class Physics Textbook Page No. 53

Question 8.
Is there any chance that angle of refraction is equal to 90° ? When does this happen?
Yes, when angle of incidence is equal to critical angle then angle of refraction is equal to 90°.

10th Class Physics Textbook Page No. 54

Question 9.
What happens to light when the angle of incidence is greater than critical angle?
When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e., light never enters rarer medium. This phenomenon is called total internal reflection.

10th Class Physics Textbook Page No. 57

Question 10.
How does light behave when a glass slab is introduced in its path?
The light ray undergoes refraction two times.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Activities

Activity – 1

Question 1.
Procedure :
Take some water in a glass tumbler. Keep a pencil in it. See the pencil from one side of glass and also from the top of the glass.
Observation:

1. How does it look?
From the side it appears to be bent. From the top it appears as it is raising.

2. Do you find any difference between two views?
Yes, the position of pencil is different.

Activity – 2

Question 2.
Procedure :

1. Go to a long wall (of length of 30 feet) facing the Sun. Go to the one end of a wall and ask someone to bring a bright metal object near the other end of the wall.
2. When the object is few inches from the wall, it will distort and we will see a reflected image on the wall as though the wall were a mirror.

Observation:

Why is there an image of the object on the wall?
The image is due to refraction of light.

Activity-3 Refraction

Question 3.
Procedure: –

1. Take a shallow vessel with opaque walls such as a mug, a tin or a pan.
2. Place a coin at the bottom of the vessel.
3. Move away from the vessel until we cannot see the coin (fig. 2). Ask someone to fill the vessel with water. When the vessel is filled with water the coin comes back into view (fig. 3).

1. Why are you able to see the coin when the vessel is filled with water?
The ray of light originated from the coin does not reach your eye when the vessel is empty. Hence you are not able to see the coin. But the coin becomes visible after the vessel is filled with water.

2. How is it possible? Do you think that the ray reaches your eye when the vessel is filled with water?
Yes, it reaches the second instance.

3. What happens to the light ray at interface between water and air?
It bends towards the normal.

4. What could be the reason for this bending of the light ray in the second instance?
It is based on Fermat’s principle, which states that the light ray always travels in a path which needs shortest possible time to cover the distance between the two points.

Activity – 4

Question 4.
Prove that when light ray travels from denser to rarer medium it bends away from the normal.

Procedure :

1. Take a metal disc. Use protractor and mark angles along its edge as shown in the figure.
2. Arrange two straws from the centre of the disk.
3. Adjust one of the straws to the angle 10°.
4. Immerse half of the disc vertically into the water, filled in a transparent vessel.
5. Inside the water the angle of straw should be at 10°.
6. From the top of the vessel try to view the straw which is inside the water.
7. Then adjust the other straw which is outside the water until both straws are in a single straight line.
8. Then take the disc out of the water and observe the two straws on it.
9. We will find that they are not in a single straight line.
10. It could be seen from the side view while half of the disc is inside the water.
11. Measure the angle between the normal and second straw. Draw table for various angles and corresponding angles of refraction.

Observation :
We observe that ‘r’ is greater than ‘i’ in all cases and when light travels from
denser to rarer medium it bends away from the normal.

Activity – 6

Question 5.
Why can we not see a coin placed in water from the side of glass?
Procedure :

1. Take a transparent glass tumbler and coin.
2. Place a coin on a table and place glass on the coin.
3. Observe the coin from the side of the glass. We can see the coin.
4. Now fill the glass with water and observe the coin from the side of the glass tumber.
5. Now we cannot see the coin because the coin rises up due to refraction.

Activity – 7

Question 7.
Why can we see the coin in water from top? What is the phenomenon behind that?

Procedure :

1. Take a cylindrical transparent vessel. Place a coin at the bottom of the vessel.
2. Now pour water until we will get the image of the coin on the water surface.
3. This is due to total internal reflection.

Activity – 8

Question 8.
Write an Activity to find refractive index of glass slab by calculating vertical shift.
(OR)
Explain the experiment with glass slab in determination of refraction through vertical shift.

Procedure :

1. Take a glass slab and measure the thickness of the slab.
2. Take a white chart and fix it on the table.
3. Place the slab in the middle of the chart.
4. Draw line around it.
5. Remove the slab from its place.
6. The lines form a rectangle. Name the vertices of it as A. B, C and D. ‘
7. Draw a perpendicular to the longer line AB of the rectangle at any point on it.
8. Place slab again in the rectangle ABCD.
9. Take a needle. Place at a point P in such a way that its length is parallel to the AB on the perpendicular line at a distance of 20 cm from the slab.
10. Now take another needle and by seeing at the first needle from the other side of the slab, try to keep the needle so that it forms a straight line with the first needle. 1 1)
11. Remove the slab and observe the positions of the needles.
12. They are not in same line.
13. Draw a perpendicular line from the second needle to the line on which the first needle is placed.
14. Take the intersection point as Q.
15. The distance between P and Q is vertical shift.
16. We will get the same vertical shift placing needle at different distances.

## AP SSC 10th Class Physics Solutions Chapter 1 Heat

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 1 Heat Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Physics Solutions 1st Lesson Heat

### 10th Class Physics 1st Lesson Heat Textbook Questions and Answers

Question 1.
What would be the final temperature of a mixture of 50 g of water at 20° C temperature and 50 g of water at 40° C temperature? (AS1)
In CGS system :
Mass m1 = 50 g
Higher temperature = T1 = 40° C
Mass m2 = 50 g
Lower temperature = T2 = 20° C

Question 2.
Explain, why dogs pant during hot summer days using the concept of evaporation. (AS1)
(OR)
How do dogs cool their body? Explain by using the process of evaporation.

• Dogs pant during hot summer days and get their body cooled. This cooling effect is due to evaporation.
• Evaporation is a surface phenomenon. Temperature of a system falls during evaporation.
• During summer the temperature in the human body increases.
• The temperature of the skin becomes higher and the water in the sweat glands starts evaporating. Since evaporation is a cooling process human body becomes cool.
• Dogs don’t have sweat glands. Their body is covered with hair. They have sweat glands only in their feet.
• So by panting the water on the tongue undergoes evaporation resulting in the cooling of the dog’s body.

Question 3.
Why do we get dew on the surface of a cold soft drink bottle kept in open air? (AS1)
(OR)
Raju observed small droplets of water outside a cold soft drink bottle kept in open air. What is the reason for the formation of droplets?

• When cold soft drink bottle is kept in open air, the temperature of surrounding air is higher than the temperature of cold drink bottle.
• Air contains molecules in the form of vapour.
• During the motion of water molecules in air strike the surface of cold drink bottle.
• Then the molecules of air lose their kinetic energy which leads to lower the temperature and they convert into droplets.
• So dew is formed on the surface of cold soft drink bottle.

Question 4.
Write the differences between evaporation and boiling. (AS1)
(OR)
Sita observed decrease in quantity of spirit kept in a vessel placed in open air. Whereas Ramu observed formation of bubbles on a water surface when it is heated. What are those two processes? Distinguish between those two processes.

 Evaporation Boiling 1. The process of escaping of molecules from the surface of a liquid at any temperature is called evaporation. 1. The process in which the liquid phase changes to gaseous phase at constant temperature. This temperature is called boiling point of liquid. 2. Evaporation takes place at any temperature. 2. Boiling takes place at a definite temperature. 3. The temperature of liquid gets down. 3. The temperature of liquids increases up to a constant temperature. 4. The kinetic energy does not change. 4. The kinetic energy of the molecules increases with the increase of temperature. 5. The evaporation depends on surface area, wind, speed, humidity. 5. The boiling depends on atmospheric pressure. 6. It is surface phenomenon. 6. It is bulk phenomenon. 7. Eg : 1) Wet clothes dries. 2) Sea water evaporates to form clouds. 7. Eg : 1) Water boils at 100° C.

Question 5.
Does the surrounding air become warm or cool when vapour phase of H2O condenses? Explain. (AS1)

• Gases have more higher energy than liquids and solids.
• When vapour condenses, it changes from gas to liquid.
• Therefore there is a drop in energy.
• This energy has to go (somewhere) to the surroundings.
• So surrounding air becomes warm when vapour phase of H20 condenses.

Question 6.
a) How much energy is transferred when 1 gm of boiling water at 100°C condenses to water at 100°C?
CGS system :
Mass of water = m = 1 gm
Latent heat of vapourisation = 540 cal/gm.
The amount of heat energy released when 1 gm of boiling water at 100°C condenses to water at 100°C
Q = mLvapour = 1 × 540 = 540 cal.

(OR)

In SI system :
Mass of water = m = 1 gm = 1 × 10-3 kg
Latent heat of vapourisation = 540 cal/gm.
The amount of heat energy released when 1 gm of boiling water at 100°C condenses to water at 100°C
Q = mLvapour – 1 × 540 = 540 cal.
In SI, Q = 540 × 4.18 = 2257 J.

b) How much energy is transferred when 1 gm of boiling water at 100° C cools to water 0° C?
CGS system :
Latent heat of vapourisation = 540 cal/gm
The amount of heat energy released when 1 gm of boiling water at 100°C condenses to water at 100 °C. m = 1 gm.
Q1 = mLvapour = 1 × 540 = 540 cal.
The specific heat of water = 1 cal/gm-°C
Difference in temperature = 100-0 = 100°C.
The heat released to cool water to 0°C is
Q2 = mS∆T = 1 × 1 × 100 = 100 cal.
∴ Total energy released = 540 + 100 = 640 cal.

c) How much energy is released or absorbed when 1 gm of water at 0° C freezes to ice at 0° C?
In CGS system :
Mass of water = m = 1 gm
Latent heat of fusion of ice (L) = 80 cal/gm
The energy transferred or released when 1 gm of water at 0° C freezes to ice at 0° C.
Q = mLfreeze = 1 × 80 = 80 Cal.

(OR)

In SI system :
Mass of water = m = 1 gm = $$\frac{1}{1000} \mathrm{~kg}$$
Latent heat of fusion = L = 3.36 × 105 J/kg.
Amount of heat released or transferred when lgm of water at 0°C freezes to ice at 0°C.
Q = mLfusion = $$\frac{1}{1000} \mathrm{~kg}$$ × 3.36 × 105 = 3.36 × 102 = 336J.

(OR)

In CGS system :
Mass of water = m = 1 gm
Latent heat of fusion of ice (L) = 80 cal/gm
The energy transferred or released when 1 gm of water at 0° C freezes to ice at 0°C.
Q = mLfreeze = 1 × 80 – 80 cal.
(Or)
In SI system : In SI, Q = 80 × 4.2 [1 cal = 4.2 J]
Q = 1 x 10-3 × 3.36 × 105 = 3 36 J.

d) How much energy is released or absorbed when 1 gm of steam at 100°C turns to ice at 0°C?
In CGS system :
Mass of water = m = 1 gm
Latent heat of vapourisation = Lvapour = 540 cal/gm
Latent heat of fusion of ice = Lfusion = 80 cal
Specific heat of water = S = 1 cal/gm-0°C
Difference in temperature
∆T = 100 – 0 = 100°C.
The energy transferred when 1 gram of steam at 100°C turns to ice at 0°C
Q = mLvapour + mS∆T + mLfusion
= 1 × 540 + 1 × 1 × 100 + 1 × 80 = 540 + 100 + 80 = 720 cal.

(OR)

Mass of water = m = 1 gm = $$\frac{1}{1000} \mathrm{~kg}$$
Latent heat of vapourisation = Lvapour = 2.25 × 106 J/kg
Latent heat of fusion = Lfusion = 3.36 × 105 J/kg
Difference in temperature
∆T = 373 – 273 – 100 K.
Specific heat of water = 4180 J/kg-K
The energy transferred when 1 gram of steam at 100° C turns to ice at 0°C =

(OR)

In CGS system :
Conversion : Steam at 100°C → Water at 100°C → Water at 0°C → Ice at 0°C.
Mass of water = m = 1 gm
Latent heat of vapourisation = Lvapour = 540 cal/gm
Latent heat of fusion of ice = Lfusion = 80 cal
Specific heat of water = S = 1 cal/gm-0°C
Difference in temperature
∆T= 100-0 = 100°C.
The energy transferred when 1 gram of steam at 100°C turns to ice at 0°C
Q = mLvapour + mS∆T + mLfusion
= 1 × 540 + 1 × 1 × 100 + 1 × 80
= 540 + 100 + 80 = 720 cal. = 720 × 4.18 = 3009.6 J.

Question 7.
Explain the procedure of finding specific heat of solid experimentally. (AS1)
(OR)
Determination of specific heat of solid experimentally.
(OR)
Ravi wanted to prepare solid with high specific heat to use on cooking utensil. What fool does he need to find the specific heat of aluminium and copper? How should he conduct the experiment?
Aim : To find the specific heat of given solid.
Apparatus : Calorimeter, thermometer, stirrer, water, steam heater, wooden box and lead shots.

Procedure:

• Measure the mass of the calorimeter with stirrer = m1 gm
• Fill water one third volume of calorimeter and measure the mass = m2 gm.
• At this time initial temperature = T1.
• Mass of the water = m2 – m1 gm.
• Take a few lead shots and place them in steam heater and heat up to 100° C. Let this temperature be T2.
• Transfer the lead shots into calorimeter and measure the final (or) resultant temperature T3.
• Mass of calorimeter with contents = m3 gm and mass of lead shots = m3 – m2 gm.
• If the specific heats of the calorimeter, lead shots and water are Sc, Sl and Sw respectively, by using method of mixtures we have
Heat lost by the solid = Heat gained by the calorimeter + water
• Knowing the specific heats of calorimeter and water we can calculate specific heat of solid (lead shots).

Question 8.
Convert 20° C into Kelvin scale. (or) Change 20°C into absolute scale. (AS1)
T = t°C + 273 = 20 + 273 = 293
⇒ T = 293 K.

Question 9.
Your friend is asked to differentiate between evaporation and boiling. What questions could you ask to make him to know the differences between evaporation and boiling? (AS2)
(OR)
Veena found that the water kept in a pot is cool and Siva observed when water is heated the temperature remains constant for some time until water turns into vapour. What are the processes involved in these two aspects? Ask some questions to understand these aspects.
The questions asked by me are :

• How do wet clothes get dried without heating?
• Are boiling and evaporation one and same or different?
• Is there any difference in kinetic energy if it boils?
• Is the temperature the main cause for boiling and evaporation?
• What are the factors which influence evaporation?
• Is boiling temperature for water always 100° C?

Question 10.
What happens to the water when wet clothes dry? (AS3)

1. When wet clothes dry, the water present in the clothes is evaporated.
2. So that the process of evaporation causes the wet clothes dry.

Question 11.
Equal amounts of water are kept in a cap and in a dish. Which will evaporate faster? Why? (AS3)
(OR)
Srinu kept. equal amounts of water in a cap and in a dish in open air. What is his observation? Explain the experiment.
Aim : To show the evaporation of equal amounts of water in cap and dish.
Apparatus : Cap, dish, water.

Procedure :

• Take equal amounts of water in cap and dish. Keep them in open air for two hours. Now weigh the water in the cap and the dish.
• We can observe that the weight of water in dish is less than that of water in cap.
• This shows that the water in dish has more evaporation than the water in cap.
• It is due to more surface area of dish.
• As the surface area increases rate of evaporation also increases.

Question 12.
Suggest an experiment to prove that rate of evaporation of a liquid depends on its surface area and vapour already present in surrounding air. (AS3)
Aim: The rate of evaporation of liquid depends on its surface area and vapour already present in surrounding air.
Apparatus : Two dishes of different surface areas and water.

Procedure :

• Take two dishes of different surface area.
• Pour equal amounts of water in the both dishes.
• Keep aside for two to three hours.
• Observe them after some time.
Dish with more surface area has less quantity of water than the dish having less surface area. ,
• This shows evaporation increases with increasing of surface area.
• Take two dishes of equal surface area containing water.
• This experiment should be conducted on more humid day and less humid day.
• We will find that evaporation is less on more humid day due to more vapour in the air.
• So evaporation decreases with vapour in the air.

Question 13.
Place a Pyrex funnel with its mouth-down in a sauce pan full of water, in such a way that the stem tube of the funnel is above the water or pointing upward into air. Rest the edge of the bottom portion of the funnel on a nail or on a coin so that water can get under it. Place the pan on a stove and heat it till it begins to boil. Where do the bubbles form first? Why? Can you explain how a natural geyser works using this experience? (AS4)

• When Pyrex funnel with its mouth down in a sauce pan then the bubbles formed by the heat energy come from the top of the funnel.
• That is from stem tube.
• This is because of pressure inside mouth of funnel increases rapidly due to increasing of heat energy.
• Pressure inside the funnel rs more than outside the funnel and very high at stem.
• Hence, bubbles come from stem of the funnel and escapes through stem tube with force, like a geyser.

Working of natural geyser by using this experience :

• Geysers are the fountains of hot water coming under the layers of the earth.
• It is a hole with narrow and deep from the bottom of the earth layers.
• It contains water.
• Water heats up due to high temperatures of the inner layers of the earth.
• As by the pressure of water at top layers of the hole, temperature rises, water boils.
• This hot water comes with narrow vent with high pressure, like Lava from the Volcano.

Question 14.
Collect the information about working of natural geyser and prepare a report. (AS4)
Natural Geysers :

• Geysers are the fountains of hot water coming under the layers of the earth.
• It is a hole with narrow and deep from the bottom of the earth layers.
• It contains water.
• Water heats up due to high temperatures of the inner layers of the earth.
• As by the pressure of water at top layers of the hole, temperature rises, water boils.
• This hot water comes with narrow vent with high pressure, like Lava from the Volcano.
• This looks like a water fountain at the surface of the earth.

Question 15.
Assume that heat is being supplied continuously to 2 kg of ice at – 5°C. You know that ice melts at 0°C and boils at 100°C. Continue the heating till it starts boiling. Note the temperature for every minute. Draw a graph between temperature and time using the values you get. What do you understand from the graph ? Write the conclusions. (AS5)
Graph between time and temperature from ice melting at 5° C to boils at 100° C.

Understanding from the graph :

• $$\overline{\mathrm{AB}}$$ = Ice warms up from – 5°C to 0°C
• $$\overline{\mathrm{BC}}$$ = Ice melts at 0°C for a certain time period. So $$\overline{\mathrm{BC}}$$ indicates no rising in temperature.
• $$\overline{\mathrm{CD}}$$ = Water warms up from 0°C to 100° C, $$\overline{\mathrm{CD}}$$ indicates rising in temperature.
• $$\overline{\mathrm{DE}}$$ = Water boils at 100° C for a certain time period. So $$\overline{\mathrm{DE}}$$ indicates no rising in temperature.

Conclusion :

• The temperature remains same at 0° C until all the ice converted into water. So, 0° C is the melting point of water.
• The temperature remains constant at 100° C until all the water converted into water vapour. So, 100° C is the boiling point of the water.

Question 16.
How do you appreciate the role of the higher specific heat of water in stabilising atmospheric temperature during winter and summer seasons? (AS6)

• Due to higher specific heat of water oceans absorb the solar energy for maintaining a relatively constant temperature.
• Oceans absorb large amounts of heat at the equator.
• The oceans moderate the surrounding temperature near the equator.
• Ocean water transports the heat away from the equator to areas closer to the north and south pole.
• This transported heat helps moderate the climate in parts of the Earth that are far from the equator.
• So higher specific heat of water is stabilising atmospheric temperature.
• So we have to extremely appreciate the role of higher specific heat of water to stabilise the atmospheric temperature.

Question 17.
Suppose that 1 / of water is heated for a certain time to rise and its temperature by 2°C. If 2 l of water is heated for the same time, by how much will its temperature. (AS7)
Mass of 1 litre of water (m1) = 1 kg ; ∆T1 = 2°C
Mass of 2 litres of water (m2) = 2 kg ; ∆ T2 = ?
Time duration is same. So same heat is absorbed by water in both the cases
⇒ Q1 = Q2
m1S(∆T1) = m2S (∆T2)

So the rise in temperature for 2 kg of water = 1°C.

Question 18.
What role does specific heat play in keeping a watermelon cool for a long time after removing it from a fridge on a hot day? (AS7)

• Generally, watermelon contains large percentage of water.
• Water has high specific heat value than other substances.
• High specific heat substances oppose the increase of temperature. Hence they continuous of the coolingness.
• So watermelon retains coolness after removing from fridge on a hot day due to the high specific cheat of water.

Question 19.
If you are chilly outside the shower stall, why do you feel warm after the bath if you stay in bathroom? (AS7)

• In the bathroom, the number of vapour molecules per unit volume is greater than the number of vapour molecules per unit volume outside the room.
• When we try to dry ourselves with a towel, the vapour molecules surrounding you condense on your skin.
• Condensation is a warming process.
• Because of the condensation, you feel warm outside the shower stall when it is chilly.

Question 20.
Three objects A at 30°C, B at 303K and C at 420 K are in thermal contact. Then answer the follwing questions.
(i) Which are in “Thermal equibrium” among A, B and C?
(ii) From which object to another heat transferred? (2 Marks)
i) 303K – 273K + 30K = 0°C + 30°C = 30°C.
∴ A and B objects are in ‘Thermal equibrium”.
ii) From object ‘C’ to objects ‘A’ and ‘B’ heat transferred.

Fill in the Blanks

1. The SI unit of specific heat is …………………. .
2. …………………. flows from a body at higher temperature to a body at lower temperature.
3. …………………. is a cooling process.
4. An object A at 10° C and another object B at 10 K are kept in contact, then heat will flow from …………………. to …………………. .
5. The latent heat of fusion of ice is …………………. .
6. Temperature of a body is directly proportional to …………………. .
7. According to the principle of method of mixtures, the net heat lost by the hot bodies is equal to …………………. by the cold bodies.
8. The sultryness in summer days is due to
9. …………………. is used as a coolant.
10. Ice floats on water because …………………. .

1. J/kg – K
2. Heat
3. Evaporation
4. A, B
5. 80 cal/gm
6. Average kinetic energy of the molecules of the body.
7. net heat gained
8. high humidity
9. Water
10. the density of ice is less than that of water

Multiple Choice Questions

1. Which of the following is a warming process?
A) evaporation
B) condensation
C) boiling
D) all the above
B) condensation

2. Melting is a process in which solid phase changes to ………………. .
A) liquid phase
B) liquid phase at constant temperature
C) gaseous phase
D) any phase
B) liquid phase at constant temperature

3. Three bodies A, B and C are in thermal equilibrium. The temperature of B is 45° C. Then the temperature of C is ……………… .
A) 45° C
B) 50° C
C) 40° C
D) any temperature
A) 45° C

4. The temperature of a steel rod is 330 K. Its temperature in ° C is ……………… .
A) 55° C
B) 57° C
C) 59° C
D) 53° C
B) 57° C

5. Specific heat S =

6. Boiling point of water at normal atmospheric pressure is ……………… .
A) 0° C
B) 100° C
C) 110° C
D) -5° C
B) 100° C

7. When ice melts, its temperature ……………… .
A) remains constant
B) increases
C) decreases
D) cannot say
A) remains constant

### 10th Class Physics 1st Lesson Heat InText Questions and Answers

10th Class Physics Textbook Page No. 1

Question 1.
Take a piece of wood and a piece of metal and keep them in a fridge or ice box. After 15 minutes, take them out and ask your friend to touch them. Which is colder? Why?
1) The metal piece is colder than the wooden piece.
2) Because more heat energy flows out of our body so metal piece gives coldness to our body, than wooden piece.

Question 2.
What could be the reason for difference in coldness of metal and wood?

• Due to more heat energy loss by our body when touches the metal piece compared to the wooden piece.
• In other way we say degree of coldness of the metal piece is greater than that of. the wooden piece.

Question 3.
Does it have any relation to the transfer of heat energy from our body to the object?

• Yes, the principle of calorimetry, means heat loss by hot body is equal to heat gained by cold body.
• This means that when heat energy flows out of our body we feel the coldness and when heat energy enters our body we feel hotness.

10th Class Physics Textbook Page No. 2

Question 4.
Why does transfer of heat energy take place between objects?

• Due to the temperature difference between the two bodies which are in thermal contact.
• Now heat energy transfers from hot body to cold body until they attain same temperature.

Question 5.
Does transfer of heat take place in all situations?
No, when the bodies are in thermal equilibrium there is no transfer of heat energy.

Question 6.
What are the conditions for transfer of heat energy?

• Two bodies should have difference in temperature.
• They (two bodies) are in thermal contact with each other.
• When the bodies have equal temperature there is no transfer of heat energy.

Question 7.
What is temperature?
(OR)
Define temperature.
Temperature : The measure of hotness or coldness of a body is called temperature.

Question 8.
How can you differentiate temperature from heat?

• Heat is a thermal energy that flows from hot body to cold body. Temperature is measure of the hotness or coldness of a body.
• Temperature decides direction of heat (energy) flow, whereas heat is energy itself that flows.

10th Class Physics Textbook Page No. 2 & 3

Question 9.
Place a laboratory thermometer in a glass tumbler containing hot water. Observe the change in mercury level. Wffet change did you notice in mercury level? Did mercury level increase or decrease?
The mercury level rises up that means temperature of the mercury level increases.

10th Class Physics Textbook Page No. 3

Question 10.
Place a laboratory thermometer in a glass tumbler containing cold water. Observe the change in mercury level. Did mercury level decrease or increase?
The mercury level falls down that shows temperature of the mercury level decreases.

Question 11.
If two different systems A and B in thermal contact, are in thermal equilibrium individually with another system C (thermal contact with A and B), will the systems A and B be in thermal equilibrium with each other?
Yes, A and B will be in thermal equilibrium with each other that means A and B will have equal temperatures.

Question 12.
How would you convert degree Celsius to Kelvin?
Temperature in Kelvin = 273 + Temperature in degree Celsius. [K = t°C + 273]

10th Class Physics Textbook Page No. 4

Question 13.
Take two bowls one with hot water and second with cold water. Gently sprinkle food colour on the surface of the water in both bowls. How do food grains move? Why do they move randomly?
We will notice that the grains of food colour move randomly (jiggle). This happens because of the molecules of water on both bowls are in random motion.

Question 14.
Why do the grains in hot water move more rapidly than the grains in cold water?

• Temperature kinetic energy. So molecules in hot water have more KE than molecules in cold water.
• As water molecules in hot water move rapidly, grains in hot water move more rapidly than the grains in cold water.

10th Class Physics Textbook Page No. 4 & 5

Question 15.
a) Take a cylindrical jar and pour hot water and then coconut oil in the vessel (do not mix them). Keep thermometers in hot water and coconut oil as shown in figure. The reading of thermometer in hot water decreases, at the same time reading of the thermometer, kept in oil increases. Why does this happen?

• Heat transfers from hot water to oil.
• So, water loses heat and shows downfall in temperature.
• Oil takes the heat and shows increasing in the temperature.

b) Can you say that water loses energy’?

• Yes. Due to the temperature difference between the water and oil, water loses energy and oil gains energy.
• Thus some heat energy flows from water to oil.
• This means, the kinetic energy of the molecules of water decreases while the kinetic energy of molecules of oil increases.

c) Can you differentiate between heat and temperature based on the heat transmit activity?
Heat is the energy that flows from a hotter body to a colder body. Temperature denotes which body is hotter and which is colder. So, temperature determines direction of heat (energy) flow, whereas heat is the energy that flows.

10th Class Physics Textbook Page No. 5 & 6

Question 16.
Place two test tubes containing 50 gm of water, 50 gm of oil in boiling water for same time.
a) In which material does the temperature rise quickly? Are the amounts of heat given to the water and oil same? How can you assume this?

• Rise in temperature of oil is faster than the water.
• Yes, same amount of heat energy given to both the oil and water through boiling water.

b) Why does this happen in specific heat?
This happens because rise in temperature depends on the nature of substance.

10th Class Physics Textbook Page No. 7

Question 17.
How much heat energy is required to rise the temperature of unit mass of substance (material) by 1°C?
Energy equal to its specific heat.

Question 18.
Why is the specific heat different for different substances?
(OR)
Explain why specific heat values are different for different materials.

• We know that the temperature of the body is directly proportional to the average kinetic energy of particle of the body.
• The molecules of the system have different forms of energies such as linear, rotational kinetic energy, vibrational energy and potential energy.
• When we supply heat energy, it will be shared in different forms and increase the energy in the system.
• This sharing will vary from material to material.
• If the maximum share of heat energy is spent to rise linear kinetic energy, then the system gets increasing in temperature.
• Due to differences in sharing different materials have different specific heats.

10th Class Physics Textbook Page No. 8

Question 19.
Take 200 ml of water in two beakers and heat them to same temperature and pour the water of two beakers into a larger beaker.
What do you observe? What could be the reason for the fact you observed?

• The temperature of mixture remains the same.
• The reason is that the masses rise in temperature and the materials are same.

Question 20.
Heat the water in one beaker to 90°C and the other to 60°C. Mix the water from these beakers in large beaker. What will be the temperature of the mixture? What did you notice? Can you give reason for the change in temperature?

• The temperature of mixture is 75°C.
• The reason is for a given material the temperature of mixture,
• Hot water gives heat to the cold water until thermal equilibrium takes place.
• So, the thermal equilibrium attains at 75°C.

Question 21.
ake 100 ml of water at 90°C and 200 ml of water at 60°C and mix the two. What is the temperature of the mixture? What difference do you notice in change of temperature?

• The temperature of mixture is 70°C.
• The reason is here m1 = 100 gm ; m2 = 200 gm
• Final temperature of the mixture is less than the above case.
• If the quantity increases, the quantity of heat to transfer is also rises to attain thermal equilibrium.
Here hotter body quantity is less and colder body quantity is high. So, the temperature at thermal equilibrium decreases and stands at 70° C.

10th Class Physics Textbook Page No. 10

Question 22.
When floor of room is washed with water, the water on the floor disappears within minutes. Why does water on the floor disappear after some time?
Due to evaporation water disappears from the floor.

Question 23.
Pour a few drops of spirit on your palm. Why does your skin become colder? (1 Mark)
Spirit absorbs heat energy from our palm and evaporates. So our palm becomes colder.

10th Class Physics Textbook Page No. 10 & 11

Question 24.
Take a few drops of spirit in two petri dishes separately. Keep one of the dishes under a ceiling fan and switch on the fan. Keep another dish with its lid closed. What do you notice? What could be the reason for this change?
1) The spirit in the dish which is kept under the ceiling fan disappears.
2) Whereas we will find some spirit left in the dish that is kept in the lidded dish.
3) The molecules which are escaping from the surface is high and they can’t reach back to liquid due to wind blow. So, evaporation is high under fan.
4) At the same time evaporation is less in the dish which is closed by lid.

10th Class Physics Textbook Page No. 12

Question 25.
Does the reverse process of evaporation take place? When and how does it take place?
1) Yes, the reverse process of evaporation takes place.
2) When the vapour molecules lose their kinetic energy which leads to lower the temperature, they convert into droplets.
3) This process is called condensation.

10th Class Physics Textbook Page No. 13

Question 26.
In early morning, during winter, you might have noticed that water droplets form on window panes, flowers, grass, etc. How are these water droplets formed?
(OR)
Why do water drops (dew) form on flowers and grass during morning hours of winter season?

• During winter season, in the night times, atmospheric temperature goes down.
• The surfaces of window panes, flowers, grass, etc. become colder.
• The water vapour molecules touch the surfaces, gets cooled and lost its energy.
• Then water vapour condenses on the surface and water drops formed.
• The water droplets condensed on such surfaces are known as dew.

10th Class Physics Textbook Page No. 14

Question 27.
Are the process of evaporation and boiling the same? Explain.

• No, they are different.
• Evaporation takes place at any temperature.
• But boiling occurs at particular temperature called the boiling point.

10th Class Physics Textbook Page No. 16

Question 28.
You might have observed coconut oil and ghee getting converted from liquid state to solid state during winter season. What could be the reason for this change? What happens to water kept in a refrigerator? How does it get converted from liquid phase to solid phase?
1) If temperature of a substance decreases kinetic energy also decreases.
2) Kinetic energy decreases from water to ice. That means solid state to liquid state.
3) In winter season coconut oil in the form of liquid get down its temperature, hence its kinetic energy also decreases. So, it-freezes.
4) Water, which is kept in refrigerator loses the kinetic energy along with decreasing temperature and freezes.
5) In this way water converted liquid phase to solid phase.

Question 29.
Are the volumes of water and ice formed with same amount of water equal? Why?
1) No, the volume of ice is greater than volume of water.
2) Water expands on freezing.
3) That means density of ice is less than density of water.

10th Class Physics 1st Lesson Heat Activities

Activity – 1

1. Explain the term temperature with example.
(OR)
What is the name given to degree of hotness or coldness? Explain the quantity with an example.
Procedure: Take a piece of wood and a piece of metal and keep them in fridge or ice box.
Observation : When we touch both of them we feel that metal piece is colder than the wooden piece.

Explanation :

• This is due to more energy flow out of our body when we touch the metal piece as compared with wooden piece.
• The degree of coldness of metal is greater than that of the wooden piece.
• The degree of hotness or coldness is called temperature.
• From this example, we say metal piece is at a lower temperature compared to wooden piece.

Activity – 2

2. What is the measure of thermal equilibrium? How do you prove?
(OR)
How do you prove temperature is the measure of thermal equilibrium?
(OR)
Explain thermal equilibrium with an activity.
Procedure :
Take two glass tumblers and fill one of them with hot water and another with cold water.
Explanation & Observation :
1) When we place a thermometer inside the hot water the mercury level of thermometer rises from initial position due to heat transferred from hotter body (hot water) to colder body (mercury in thermometer).
2) When we place the thermometer inside the cold water the mercury level comes down from its initial position due to transfer of heat from mercury (hotter body) to water (colder body).

Conclusion :

• Heat is a form of energy that flows from a body at higher temperature to a body at lower temperature until the temperature remains same for two bodies that is called thermal equilibrium.
• In the above case, the steadiness of mercury column shows that thermal equilibrium is achieved. That reading of mercury column gives temperature.
• Thus temperature is a measure of thermal equilibrium.

Activity – 3

3. Establish the relationship between temperature and average kinetic energy.
(OR)
Suggest an activity to prove that the average kinetic energy of the molecules is directly proportional to the absolute temperature of the substances.
(OR)
How do you prove that temperature of a body is an indicator of average kinetic energy?
Procedure :

• Take two bowls one with hot water and second with cold water.
• Gently sprinkle food colour on the surface of the water in both bowls.

Observation :
We will observe the jiggling of grains of food colour in hot water is more when compared to jiggling in cold water.

Explanation :

• We know kinetic energy depends on speed motion of particles.
• So the kinetic energy of hotter body is greater than that of colder body.
• Thus the temperature of a body is an indicator of average kinetic energy of molecules of that body.

Conclusion :
Therefore average kinetic energy of molecules is directly proportional to absolute temperature.

Activity – 4

4. Write an activity which tells how heat transmits.
(OR)
In which direction does heat tend to flow? Prove it with an activity.

Procedure :

• Take water in a container and heat it to 60° C.
• Take a cylindrical transparent glass jar and fill half of it with the hot water.
• Pour coconut oil over the surface of water.
• Put a lid with two holes on the top of the glass jar.
• Insert two thermometers through the lid in such a way that one inside coconut oil and other in water.

Observation & Explanation :

• Now we can observe that the reading of thermometer kept in water decreases while the reading of thermometer kept in oil increases.
• So temperature of water decreases whereas temperature of oil increases.

Conclusion :

• Heat transmits from hotter body to colder body.
• So temperature determines direction of heat flow.

Activity – 5 Specific Heat

5. Write an activity which gives the relation between rise in temperature and nature of material.
(OR)
“The rate of rise in temperature depends on the nature of substance.” Prove it with an activity.
(OR)
Draw a diagram and label the parts to prove that the rate of increase in temperature depends on the nature of substance.
(OR)
We can observe severe burns with hot oil when compared with hot water.
Which factor will decide this aspect? Explain this process with an example.
Procedure :

• Take a large jar with water and heat it up to 80°C.
• Take two identical boiling test tubes with single-holed corks.
• Fill them, one of the boiling tubes with 50 gm of water and other with 50 gm of oil.
• Insert two thermometers in each of tubes and clamp them to retort stand and place them in a jar of hot water.

Observation :

• Observe the readings of thermometers every three minutes.
• We can observe that the rise in temperature of oil is higher than that of water.

Explanation :

• Since both the boiling tubes kept in hot water for the same interval of time, the heat supplied to oil and water is same but rise in temperature of oil is more.
• So we conclude that rise in temperature depends on the nature of substance (specific heat).

Activity – 6

6. Derive Q = mSAT.
(OR)
Establish relationship between heat energy, mass of the substance and rise in temperature.
(OR)
Derive an expression for heat energy.
(OR)
Derive an expression for factors affecting amount of heat energy absorbed.
Procedure :

• Take two beakers of equal volume and take 250 grams of water in one beaker and 1 kg of water in another beaker.
• Note their initial temperatures.
• Now heat the two beakers up to 60° C.
• Note down the heating times.

Observation:

• We observe that-the water in large beaker takes more time.
• That means we need to supply more heat enepgy to water in larger beaker (greater quantity of water).

Conclusion :
From this we conclude that for some change in temperature the amount of heat (Q) absorbed by a substance is directly proportional to its mass (m).
Q ∝ m (when ∆T is constant) ………….. (1)

Procedure :
Now take 1 litre of water in a beaker and heat it and note the temperature changes (∆T) for every two minutes and observe the rise internals.

Conclusion:
We will notice that for the same mass (m) of water the change in temperature is proportional to amount of heat (Q) absorbed by it.
Q ∝ ∆T (when m is constant) ………….. (2)
From (1) and (2) Q m∆T (or) Q = mS∆T,
where ‘S’ is called specific heat of substance.

Activity – 7

7. a) How are you able to find the final temperature of the mixture of sample?
(OR)
What is the “Principle of method of mixtures”? Verify it with an activity.
Situation – 1 :

• Take two beakers of the same size and pour 200 ml of water in each of them.
• Now heat the water in both beakers to same temperature.
• Now pour water from these beakers into a larger beaker and measure the temperature of the mixture.

Observation :
We can observe that there is no change in temperature.

Situation – 2 :

• Now heat the water in first beaker to 90° C and the other to 60° C.
• Mix the water from these beakers in a large beaker.

Observation :
We can find that the temperature of mixture is 75° C.

Situation – 3 :
Now take 100 ml of water at 90° C and 200 ml of water at 60° C and mix the two. Observation :
We can find that the temperature of mixture is 75° C.

7. b) Derive, a formula for final temperature of mixture of samples.
(OR)
Maveen added hotter water of mass m1 kept at temperature T1 to cold water of mass m2 kept at temperature T2. Find the expression to find temperature of mixture of samples.
Procedure:
1) Let the initial temperatures of the hotter and colder samples of masses m1 and m2 be T1 and T2.
2) Let T be the final temperature of mixture.

Observation :
The temperature of the mixture is lower than hotter sample and higher than colder sample. Explanation :
So hot sample has lost heat, and the cold sample has gained heat.
The heat lost by the hot sample Q1 = m1S (T1 – T)
The heat gained by the cold sample Q2 = m2S (T – T2)
We know that heat lost = heat gained
Q1 = Q2
m1 S(T1 – T) = m2 S(T – T2)
$$\mathrm{T}=\frac{\mathrm{m}_{1} \mathrm{~T}_{1}+\mathrm{m}_{2} \mathrm{~T}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}$$

Activity -8

8. Explain the process of evaporation phenomenon with an example.
(OR)
Srinu observed that spirit taken in a petri dish disappears after some time. Explain the process involved in it with an example.
Procedure :

• Take a few drops of spirit in two petri dishes separately.
• Keep one of the dishes under a ceiling fan and keep another dish with its lid closed.
• Observe the quantity of spirit in both dishes after 5 minutes.

Observation :
We will notice that spirit in the dish kept under the ceiling fan disappears whereas we will find some spirit left in the dish that is kept in the lidded dish.

Explanation :

• The reason is that the molecules of spirit in dish continuously move with random speeds and collide with other molecules.
• During the collision they transfer energy to other molecules.
• Due to this collision the molecules at the surface acquire energy and fly off from the surface.
• Some molecules come back to liquid.
• If the number of escaping molecules is greater than returned number, then the number of molecules in the liquid decreases.

Conclusion :
When a liquid is exposed to air, the molecules at the surface keep on escaping from the surface till the entire liquid disappears into air. This process is called evaporation.

Activity – 9

9. Explain the process of condensation with example.
(OR)
Explain the process of condensation with an activity.
(OR)
Karan told his friend that he observed that there are some water droplets outside a cold soft drink bottle. Explain the phenomenon involved in the formation of these droplets.
Procedure :
Place a glass tumbler on the table. Pour cold water up to half of its height.

Observation :
There are droplets formed outside of the glass.

Explanation :

• The reason is that the surrounding air contains water molecules in the form of water vapour. When the water molecules strike the surface of the glass tumbler which is cool, they lose their kinetic energy.
• This energy lowers the temperature of vapour and it turns into droplets.
• The energy lost by water molecules in air is gained by the molecules of the glass tumbler.
• Hence the average kinetic energy of glass molecules increases. In turn the energy is transferred to water molecules in die glass.

Conclusion :

• So the average kinetic energy and temperature of water in glass increases. This is called condensation.
• Condensation is the phase change from gas to liquid.

Activity – 10

10. Explain the process of boiling with an example.
(OR)
Why do we observe bubbles on the surface of water which has been heated ? What is the phenomenon involved in it? Explain.
Procedure :
Take a beaker of water, keep it on the burner and note the readings of thermometer for every two minutes.

Observation :

• We will notice that the temperature of the water rises continuously till it reaches 100° C.
• Once it reaches 100° C the temperature remains same and a lot of bubbling on the surface takes place. This is called boiling of water.

Explanation :

• It happens due to when water is heated the solubility of gases it contains reduces.
• As a result, bubbles of gas are formed in the liquid.
• Evaporation of water molecules from the surrounding liquid occurs into these bubbles and they become filled with saturated vapour.
• At a certain temperature, the pressure of the saturated vapour inside the bubbles becomes equal to the pressure exerted on the bubbles from the outside.

Conclusion :

• As a result, these bubbles rise rapidly to the surface and collapse at the surface releasing vapour present in bubbles into air at the surface. This process is called “boiling”.
• This temperature is called ‘boiling temperature”.

Activity – 11

11. Explain the process of melting and latent heat of fusion.
(OR)
When ice is heated to 0°C it starts to turn into water. But temperature remains ‘ constant for some time. What is the process involved in this? Explain.
Procedure :

• Take small ice cubes in a beaker. Insert the thermometer in the beaker.
• Now start heating the beaker and note down readings of thermometer every one minute till the ice completely melts and gets converted into water.
• Before heating the temperature of ice is 0°C or less than 0°C.

Observation :

• We will observe that the temperature of ice at the beginning is equal to or below 0°C.
• If the temperature of ice is below 0°C, it goes on changing till it reaches 0°C.
• When ice starts melting, we will observe no change in temperature though you are supplying heat continuously.

Explanation:

• Given heat energy uses to break the bonds (H2O) in ice and melts.
• So, temperature is constant while melting.

Conclusion:

• This process is called melting. In this process, heat converts solid phase to liquid phase.
• The temperature of the substance does not change until all the ice melts and converts into water.
• The heat given to melting is called latent heat of fusion.
• The heat required to convert 1 gm of solid completely into liquid at constant temperature is called “latent heat of fusion”.

Activity – 12

12. Why does a glass bottle tilled with water break when it is placed in deep freezer for some time?
(OR)
Prove that density of ice is less than that of water.
(OR)
How do you prove that volume of ice is more than that of water?
Procedure :

• Take a small glass bottle with a tight lid and fill it with water, without any gap and fix the lid tightly.
• Put the bottle into the deep freezer of a refrigerator for a few hours.

Observation :
When we take it out from the deep freezer, we can observe that the glass bottle breaks.

Explanation :

• These cracks on the bottle due to expansion of the substance in the bottle.
• This means water expands on freezing.

Conclusion :

• Water expands on freezing.
• Ice has less density than water.