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Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 2nd Lesson Ray Optics and Optical Instruments Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 2nd Lesson Ray Optics and Optical Instruments

Very Short Answer Questions

Question 1.
Define focal length and radius of curvature of a concave lens.
Answer:
Focal length (f) : The distance of principal focus from the optical centre of the lens is called the focal length of the lens. The focal length (f) = CF

Radius of curvature : Radius of curvature is the radius of the sphere from which the curved surface is taken a part.

Question 2.
What do you understand by the terms ‘focus’ and ‘principal focus in the context of lenses?
Answer:
Focus : The point where image of an object placed at infinity is formed is called the focus of the lens.
Principal focus: A narrow beam of light incident on a lens in a direction parallel to its principal axis, after refraction through the lens, the rays converge to a point on the principal axis. This point is called principal focus.

Question 3.
What is optical density and how is it different from mass density ? (T.S. Mar. ’16)
Answer:
Optical density: Optical density is defined as the ratio of the speed of light in media.
Mass density: Mass per unit volume is defined as mass density.
Mass density of an optically denser medium less than that of optically rarer medium.

Question 4.
What are the laws of reflection through curved mirrors ?
Answer:

1. “The angle of reflection equals to the angle of incidence”.
2. “The incident ray, reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane”.

Question 5.
Define ‘power’ of a convex lens. What is its unit ?
Answer:
Power of a lens : Power of a lens is defined as its bending ability and is measured as reciprocal of focal length in metre.
∴ Power of a lens (P) = [latex]\frac{1}{\text { f(in metres) }}[/latex] = [latex]\frac{100}{\mathrm{f}(\text { in } \mathrm{cms})}[/latex]
Unit → Dioptre (D)

Question 6.
A concave mirror of focal length 10 cm is placed at a distance 35 cm from a wall. How far from the wall should an object be placed so that its real image is formed on the wall ?
Answer:
f = 10 cm, υ = 35 cm
[latex]\frac{1}{f}[/latex] = [latex]\frac{1}{v}[/latex] + [latex]\frac{1}{-u}[/latex] (using sign convention)
[latex]\frac{1}{\mathrm{u}}[/latex] = [latex]\frac{1}{v}[/latex] – [latex]\frac{1}{f}[/latex] = [latex]\frac{1}{35}[/latex] – [latex]\frac{1}{10}[/latex]
[latex]\frac{1}{\mathrm{u}}[/latex] = [latex]\frac{10-35}{35 \times 10}[/latex] = [latex]\frac{-1}{14}[/latex]
U = – 14 cm.
Distance of the object from the wall = 35 – 14 = 21cm.

Question 7.
A concave mirror produces an image of a long vertical pin, placed 40cm from the mirror, at the position of the object. Find the focal length of the mirror.
Answer:
Give u = υ = 40cm]
[latex]\frac{1}{f}[/latex] = [latex]\frac{1}{v}[/latex] + [latex]\frac{1}{u}[/latex]
[latex]\frac{1}{f}[/latex] = [latex]\frac{1}{40}[/latex] + [latex]\frac{1}{40}[/latex]
[latex]\frac{1}{f}[/latex] = [latex]\frac{2}{40}[/latex]
f = 20 cm.

Question 8.
A small angled prism of 40 deviates a ray through 2.48°. Find the refractive index of the prism. (A.P. Mar. ’19)
Answer:
A = 4°, D_m = 2.48°
D_m = A(μ – 1)
μ – 1 = [latex]\frac{D_m}{A}[/latex] = [latex]\frac{2.48}{4}[/latex] = 0.62
μ = 1 + 0.62
μ = 1.62

Question 9.
What is dispersion ? Which colour gets relatively more dispersed?
Answer:
Dispersion : The phenomenon of splitting of white light into its constituent colours, on passing through a prism is called dispersion of light.
The deviation is maximum for violet colour.

Question 10.
The focal length of a concave lens is 30 cm. Where should an object be placed so that its image is 1/10 of its size? (T.S. Mar. ‘19)
Answer:
f = 30 cm, h₁ = h, h₂ = [latex]\frac{\mathrm{h}}{10}[/latex]

Question 11.
What is myopia ? How can it be corrected ? (T.S. Mar. ’15)
Answer:
Myopia (or) Near sightedness :
The light from a distant object arriving at the eye-lens may get converged at a point infront of the retina. This type of defect is called myopia.
To correct this, we interpose a concave lens between the eye and the object.

Question 12.
What is hypermetropia ? How can it be corrected ? (A.P. Mar. ’16)
Answer:
Hypermetropia (or) Farsightedness :
The light from a distant object arriving at the eye-lens may get converged at a point behind the retina. This type of defect is called Hypermetropia.
To correct this, we interpose a convex lens (Convergent lens) between the eye and the object.

Short Answer Questions

Question 1.
A light ray passes through a prism of angle A in a position of minimum deviation. Obtain an expression for (a) the angle of incidence in terms of the angle of the prism and the angle of minimum deviation (b) the angle of refraction in terms of the refractive index of the prišm.
Answer:
In the quadrilateral AQNR
∠A + ∠QNR = 180° ……… (1)
From Δ^le QNR, r₁ + r₂ + ∠QNR = 180° ………. (2)
r₁ + r₂ = A …….. (3)
Total deviation δ = (i – r₁) + (e – r₂)
δ = i + e – A ……. (4)

a) At minimum deviation position M
δ = D_m, i = e and r₁ = r₂ = r
∴ From eq(4), D_m = 2i – A
i = [latex]\frac{\mathrm{A}+\mathrm{D}_{\mathrm{m}}}{2}[/latex] …… (5)

b) From eq (3), r + r = A
r = A/2 ……. (6)
Refractive index of the prism μ = [latex]\frac{\sin i}{\sin r}[/latex] …… (8)

Question 2.
Define focal length of a concave mirror. Prove that the radius of curvature of a concave mirror is double its focal length. (A.P. Mar. ’19)
Answer:
Focal length of concave mirror:
The distance between the focus F and the pole P of the mirror is called the focal length of the concave mirror.

Consider a ray AB parallel to principal axis incident on a concave mirror at B and is reflected along BF. The line CB is normal to the mirror.

Let θ be the angle of incidence, ∠ABC = ∠BCP = θ

Draw BD ⊥ CP
In right angled Δ^le BCD
Tan θ = [latex]\frac{\mathrm{BD}}{\mathrm{CD}}[/latex] ….. (1)
From Δ^le BFD, Tan2θ = [latex]\frac{\mathrm{BD}}{\mathrm{FD}}[/latex] ……. (2)
Dividing eq (2) by eq (1)
…….. (3)

If θ is very small, them tan θ ≈ θ and tan 2θ ≈ 2θ since the aperture of the lens is small
∴ The point B lies very close to p.
CD ≈ CP and FD ≈ FP
From eq (3), [latex]\frac{2 \theta}{\theta}[/latex] = [latex]\frac{C P}{F P}[/latex] = [latex]\frac{R}{f}[/latex] ⇒ 2 = [latex]\frac{R}{f}[/latex]
R = 2f

Question 3.
A mobile phone lies along the principal axis of a concave mirror longitudinally. Explain why the magnification is not uniform.
Answer:
The ray diagram for the formation of image of the phone is shown in figure. The image of part which is on the plane perpendicular to principal axis will be on the same plane. It will be the same size i.e, B’C = BC.

Question 4.
Explain the cartesian sign convention for mirrors.
Answer:
According to cartesian sign convention:

1. All distances are measured from the pole of the mirror (or) the optical centre of the lens.
2. The distances measured in the same direction as incident light are taken as positive.
   
3. The distances measured in the direction opposite to the incident light are taken as negative.
4. The heights measured upwards with respect to x-axis and normal to the x-axis are taken as positive.
5. The heights measured downwards are taken as negative.

Question 5.
Define critical angle. Explain total internal reflection using a neat diagram.(T.S. Mar. ’15)
Answer:
Critical angle:
When light ray travelling from denser medium to rarer medium, then the angle of inci-dence for which angle of refraction in air is 90° is called critical angle.
C = sin^-1 [latex]\left(\frac{1}{\mu}\right)[/latex]

Total internal reflection:
When a light ray travels from denser to rarer medium, the angle of incidence is greater than the critical angle, then it reflects into the same medium is called total internal reflection.

Explanation:
Consider an object in the denser medium. A ray OA incident on XY bends away from the normal. As the angle of incidence is increased, the angle of refraction goes on increasing. For certain angle of incidence, the refracted ray parallel to XY surface (r = 90°)

When the angle of incidence is further increased, the ray is not refracted but is totally reflected back in the denser medium. This phenomenon is called total internal reflection.

Question 6.
Explain the formation of a mirage. (T.S. Mar. 19 & A.P. Mar. ’16)
Answer:
In a desert, the sand becomes very hot during the day time and it rapidly heats the layer of air which is in its contact. So density of air decreases. As a result the successive upward layers are denser than lower layers.

When a beam of light travelling from the top of a tree enters a rarer layer, it is refracted away from the normal. As a result at the surface of layers of air, each time the angle of incidence increases and ultimately a stage is reached, when the angle of incidence becomes greater than the critical angle between the two layers, the incident ray suffers total internal reflection.

So it appears as inverted image of the tree is formed and the same looks like a pool of water to the observer.

Question 7.
Explain the formation of a rainbow.
Answer:

Figure shows how sun light is broken into its segments in the process and a rainbow appears. The dispersion of the violet and the red rays after internal reflection in the drop is shown in figure.

The red rays emerge from the drops of water at one angle (43°) and the violet rays emerge at another angle (41°). The large number of water drops in the sky makes a rainbow. The rainbow appears semicircular for an observer on earth.

Question 8.
Why does the setting sun appear red ? (Mar. ’14)
Answer:
As sunlight travels through the earths atmosphere, gets scattered by the large number of molecules present. This scattering of sun light is responsible for the colour of the sky, during sunrise and sunset etc.

The light of shorter wave length is scattered much more than light of larger wavelength. Scattering ∝ [latex]\frac{1}{\lambda^4}[/latex].

Most of blue light is scattered, hence the bluish colour of sky predominates.

At sunset (or) sunrise, sun rays must pass through a larger atmospheric distance. More of the blue colour is scattered away only red colour which is least scattered appears to come from sun. Hence it appears red.

Question 9.
With a neat labelled diagram explain the formation of image in a simple microscope. (T.S. Mar. 16 & A.P. Mar. 15)
Answer:
Simple microscope : It consists a single short focus convex lens. It increases the visual angle to see an object clearly. It is also called magnifying glass (or) reading glass.

Working : The object is adjusted within the principal focus of the convex lens to form the image at the near point. The image is formed on same side of the object and it is virtual, erect and magnified as shown in fig.

Magnifying power: The ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye is called magnifying power of a simple microscope.
It is denoted by ‘m’

Question 10.
What is the position of the object for a simple microscope ? What is the maximum magnification of a simple microscope for a realistic focal length?
Answer:
When an object is placed between principal focus and optical centre of a convex lens, a virtual and erect image will be formed on the same side of the object.

Magnifying power: It is defined as the ratio of the angle subtended at the eye by the image to the angle subtended by the object at the eye.

This shows that smaller the focal length of the lens, greater will be the magnifying power of microscope.

Long Answer Questions

Question 1.
a) What is the cartesian sign convention ? Applying this convention and using a neat diagram, derive an expression for finding the image distance using the mirror equation.
b) An object of 5 cm height is placed at a distance of 15 cm from a concave mirror of radius of curvature 20cm. Find the size of the image.
Answer:
According to cartesian sign convention:

1. All distances are measured from the pole of the mirror (or) the optical centre of the lens.
2. The distances measured in the same direction as incident light are taken as positive.
3. The distances measured in the direction opposite to the incident light are taken as negative.
4. The heights measured upwards with respect to x-axis and normal to the x-axis are taken as positive.
5. The heights measured downwards are taken as negative.

Image distance using mirror equation :

Consider an object AB is placed beyond centre of curvature of a concave mirror, on its principal axis.

A ray AD parallel to principal axis incident on the mirror at point D and is reflected to pass through F. Another ray AE passing through centre of curvature C is reflected along the same path. Two rays of light intersect at point A’. Thus A’B’ is real, inverted and diminished image of AB formed between C and F.
Δ^le DPF and Δ^le A’B’ F are similar
[latex]\frac{\mathrm{B}^{\prime} \mathrm{A}^{\prime}}{\mathrm{PD}}[/latex] = [latex]\frac{\mathrm{B}^{\prime} \mathrm{F}}{\mathrm{FP}}[/latex]
(or) [latex]\frac{\mathrm{B}^{\prime} \mathrm{A}^{\prime}}{\mathrm{BA}}[/latex] = [latex]\frac{\mathrm{B}^{\prime} \mathrm{F}}{\mathrm{FP}}[/latex] …… (1) (∴ PD = AB)

Since ∠APB = ∠A’P’V’
The right angle triangles A’B’P and ABP are similar
[latex]\frac{\mathrm{B}^{\prime} \mathrm{A}^{\prime}}{\mathrm{BA}}[/latex] = [latex]\frac{B^{\prime} P}{B P}[/latex] …….. (2)
From equations (1) and (2), [latex]\frac{\mathrm{B}^{\prime} \mathrm{F}}{\mathrm{FP}}[/latex] = [latex]\frac{\mathrm{B}^{\prime} \mathrm{P}}{\mathrm{BP}}[/latex] = [latex]\frac{\mathrm{B}^{\prime} \mathrm{P}-\mathrm{FP}}{\mathrm{FP}}[/latex] ….. (3)

Now applying the sign convention
B’P = -v, FP = -f, BP = -u
[latex]\frac{-v+f}{-f}[/latex] = [latex]\frac{-\mathrm{V}}{-\mathrm{u}}[/latex] ⇒ [latex]\frac{v-f}{f}[/latex] = [latex]\frac{\mathrm{v}}{\mathrm{u}}[/latex]
[latex]\frac{v}{\mathrm{f}}[/latex] – 1 = [latex]\frac{v}{\mathrm{u}}[/latex] ⇒ [latex]\frac{1}{\mathrm{f}}[/latex] = [latex]\frac{1}{\mathrm{v}}[/latex] + [latex]\frac{v}{\mathrm{u}}[/latex]

b) Given that h₁ = 5 cm
u = -15cm
R = 20cm
f = [latex]\frac{-\mathrm{R}}{2}[/latex] = [latex]\frac{-20}{2}[/latex] = -10cm

Question 2.
a) Using a neat labelled diagram derive the mirror equation. Define linear magnification.
b) An object is placed at 5cm from a convex lens of focal length 15cm. What is the position and nature of the image ?
Answer:
a) Derivation of mirror equation: Consider an object AB is placed beyond centre of curvature of a concave mirror, on its principal axis.

A ray AD parallel to principal axis incident on the mirror at point D and is reflected to pass through F.
Another ray AE passing through centre of curvature C is reflected along the same path. Two rays of light intersect at point A’. Thus A’ B’ is real, inverted and diminished image of AB formed between C and F.

Linear magnification:
Linear magnification is the ratio of the size of the image formed by the mirror to the size of the object.

Nature of the image is virtual.

Question 3.
a) Derive an expression for a thin double convex lens. Can you apply the same to a double concave lens too?
b) An object is placed at a distance of 20cm from a thin double convex lens of focal length 15cm. Find the position and magnification of the image.
Answer:
a)

1. A convex lens is made up of two spherical refracting surfaces of radii of curvatures, R₁ and R₂ and μ is the refractive index of the lens.
2. P₁, P₂ are the poles, C₁, C₂ are the centres of curvatures of two surfaces with optical centre C.
3. Consider a point object O lying on the principal axis of the lens and I₁ is the real image of the object.
   If CI₁ ≈ P₁ I₁ = v₁
   and CC₁ ≈ PC₁ = R₁
   CO ≈ P₁ = u
4. As refraction is taking place from rarer to denser medium
   [latex]\frac{\mu_1}{-u}[/latex] + [latex]\frac{\mu_2}{v_1}[/latex] = [latex]\frac{\mu_2-\mu_1}{R_1}[/latex] ……. (1)
5. The refracted ray suffers further refraction
   Therefore I is the final real image of O.
6. For refraction at second surface, I₁ as virtual object, whose real image is formed at I.
   ∴ u ≈ CI₁ ≈ P₂I₁ = V₁
   Let CI ≈ P₂ I = V
7. Now refraction taking place from denser to rarer medium

When the object on the left of the lens is at infinity (∝), image is formed at principal focus of the lens.
∴ u = ∝, υ = f = focal length of the lens

This is the lens maker’s formula
Yes, same formula applies to double concave lens too.

b) Given that u = 20 cm, f = 15 cm
[latex]\frac{1}{f}[/latex] = [latex]\frac{1}{v}[/latex] + [latex]\frac{1}{u}[/latex]
[latex]\frac{1}{v}[/latex] = [latex]\frac{1}{f}[/latex] – [latex]\frac{1}{u}[/latex] = [latex]\frac{1}{15}[/latex] – [latex]\frac{1}{20}[/latex]
[latex]\frac{1}{v}[/latex] = [latex]\frac{20-15}{15 \times 20}[/latex] + [latex]\frac{5}{15 \times 20}[/latex]
[latex]\frac{1}{v}[/latex] = [latex]\frac{1}{60}[/latex]
v = 60 cm.
Magnification (m) = [latex]\frac{-\mathbf{v}}{\mathbf{u}}[/latex]
m = [latex]\frac{-60}{-20}[/latex] (u = -20cm)
m = 3

Question 4.
Obtain an expression for the combined focal length for two thin convex lenses kept in contact and hence obtain an expression for the combined power of the combination of the lenses.
Answer:

1. Consider two lenses A and B of focal lengths f₁ and f₂ placed in contact with each other.
2. Let the object be placed at a point O, the first lens forms the image at I₁, it is real. It serves as the virtual object of lens B, Producing final image at I.
3. For the image formed by lens A, we get

Question 5.
a) Define Snell’s Law. Using a neat labelled diagram derive an expression for the refractive index of the material of an equilateral prism.
b) A ray of light, after passing through a medium, meets the surface separating the medium from air at an angle of 45° and is just not refracted. What is the refractive index of the medium ?
Answer:
a) Snell’s law:
The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant, called the refractive index of the medium.
[latex]\frac{\sin i}{\sin r}[/latex] = μ (constant).
Let ABC be the glass prism. Its angle of prism is A. The refractive index of the material of the prism is μ. Let AB and AC be the two refracting surfaces PQ = incident ray, RS = emergent ray.
Let angle of incidence = i₁
angle of emergence = i₂
angle of refraction = r₁
angle of refraction at R = r₂
After travelling through the prism it falls on AC and emerges as RS.
The D = angle of deviation.

From the Δ QRT
r₁ + r₂ + ∠T = 180° —— (1)
From the quadrilateral AQTR
∠A + ∠T = 180°
∠T = 180° – A ……. (2)
From the equations (1) and (2)
r₁ + r₂ + ∠T = 180° we get
r₁ + r₂ + 180° – A = 180°
r₁ + r₂ = A …… (3)
from the Δ QUR
i₁ – r₁ + i₂ – r₂ + 180° – D = 180°
i₁ + i₂ – (r₁ + r₂) = D
i₁ + i₂ – A = D [∵ r₁ + r₂ = A]
i₁ + i₂ = A + D …. (4)

Minimum deviation: Experimentally it is found that as the angle of incidence increased the angle of deviation decreases till it reaches a minimum value and then it increases. This least value of deviation is called angle of minimum deviation ‘δ’ as shown in the fig.

When D decreases the two angles i₁ and i₂ become closer to each other at the angle of minimum deviation, the two angles of incidence are same i.e, i₁ = i₂
As i₁ = i₂, r₁ = r₂
∴ i₁ = i₂ = r₁ = r₂ = r

substituting this in (1) and (2) we get
2r = A ⇒ r = A/2
i + i = A + δ ⇒ i = [latex]\frac{\mathrm{A}+\delta}{2}[/latex]

Note :The minimum deviation depends on the refractive index of the prism material and the angle of the prism.

b) Given that i = C = 45°
μ = [latex]\frac{1}{\sin c}[/latex] ⇒ μ = [latex]\frac{1}{\sin 45^{\circ}}[/latex]
μ = [latex]\frac{1}{1 / \sqrt{2}}[/latex] = [latex]\sqrt{2}[/latex]
μ = 1.414

Question 6.
Draw a neat labelled diagram of a compound microscope and explain its working. Derive an expression for its magnification.
Answer:
Description: It consists of two convex lenses separated by a distance. The lens near the object is called objective and the lens near the eye is called eye piece. The objective lens has small focal length and eye piece has of Idrger focal length. The distance of the object can be adjusted by means of a rack and pinion arrangement.

Working: The object OJ is placed outside the principal focus of the objective and the real image is formed on the other side of it. The image I₁ G₁ is real, inverted and magnified.
This image acts as the object for the eyepiece. The position of the eyepiece is so adjusted that the image due to the objective is between the optic centre and principal focus to form the final image at the near point. The final image IG is virtual, inverted and magnified.

Magnifying Power : It is defined as the ratio of the angle subtended by the final image at the eye when formed at near point to the angle subtended by the object at the eye when imagined to be at near point.

Imagining that the eye is at the optic centre, the angle subtended by the final image is α. When the object is imagined to be taken at near point it is represented by IJ’ and OJ = IJ’. The angle made by I J’ at the eye is β. Then by the definition of magnifying power

Magnifying power of the objective (me) = I₁ G₁ / OJ = Height of the image due to the objective / Height of its object.
Magnifying power of the eye piece (me) = IG/I₁G₁ = Height of the final image / Height of the object for the eyepiece.
∴ m = m_o × m_e ….. (1)

To find m_o : In figure OJ O’ and I₁ G₁ O’ are similar triangles. [latex]\left(\frac{\mathrm{I}_1 \cdot \mathrm{G}_1}{\mathrm{OJ}}\right)[/latex] = [latex]\frac{O^{\prime} I_1}{O^{\prime} O}[/latex]
Using sign convention, we find that O’I₁ = + v₀ and O’O = -u where v₀ is the image distance due to the objective and u is the object distance for the objective or the compound microscope.
I₁G₁ is negative and OJ is positive.

To find m_e : The eyepiece behaves like a simple microscope. So the magnifying power of the eye piece.
∴ m_e = [latex]\left(1+\frac{D}{f_e}\right)[/latex]
Where f_e is the focal length of the eyepiece.
Substituting m₀ and m_e in equation (1),
m = [latex]+\frac{\mathrm{v}_0}{\mathrm{u}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)[/latex]
When the object is very close to the principal focus F₀ of the objective, the image due to the objective becomes very close to the eyepiece.
u ≈ f₀ and v₀ ≈ L
Where L is the length of the microscope. Then
m = [latex]-\frac{L}{f_0}\left(1+\frac{D}{f_e}\right)[/latex]

Problems

Question 1.
A light wave of frequency 4 × 10¹⁴ Hz and a wavelength of 5 × 10^-7 m passes through a medium. Estimaté the refractive index of the medium.
Answer:
υ = 4 × 10¹⁴ Hz
A = 5 × 10^-7 m
v = υλ = 4 × 10¹⁴ × 5 × 10^-7 = 20 × 10⁷
= 2 × 10⁸ m fs
we know that C = 3 × 10⁸ m/s
µ = [latex]\frac{C}{v}[/latex]
µ = [latex]\frac{3 \times 10^8}{2 \times 10^8}[/latex]
µ = [latex]\frac{3}{2}[/latex] = 1.5

Question 2.
A ray of light is incident at an angle of 60° on the face of a prism of angle 30°. The emergent ray makes an angle of 30° with the incident ray. Calculate the refractive index of the material of the prism.
Answer:
i₁ = 60°, r = 30°, i² = 30° .
µ = [latex]\frac{\sin i}{\sin r}[/latex]
µ = [latex]\frac{\sin 60^{\circ}}{\sin 30^{\circ}}[/latex]
µ = [latex]\frac{\sqrt{3}}{2 \times \frac{1}{2}}[/latex]
µ = [latex]\sqrt{3}[/latex]
µ = 1.732

Question 3.
Two lenses of power – 1.75D and +2.25D respectivetly, are placed in contact. Calculate the focal length of the combination.
Answer:
P₁ = – 1.75 D, P₂ = + 2.25 D.
P = P₁ + P₂
P = -1.75 + 2.25
P = 0.5
[latex]\frac{1}{\mathrm{~F}}[/latex] = P
F = [latex]\frac{1}{p}[/latex] = [latex]\frac{1}{0.5}[/latex] = 2m
F = 200cm

Question 4.
Some rays falling on a converging lens are focussed 20cm from the lens. When a diverging lens is placed in contad with the converging lens, the rays are focussed 30cm from the combination. What is the focal length of the diverging lens?
Answer:
u = -20 cm
υ = 30cm
[latex]\frac{1}{f}[/latex] = [latex]\frac{1}{v}[/latex] + [latex]\frac{1}{\mathrm{u}}[/latex] ⇒ [latex]\frac{1}{\mathrm{f}}[/latex] = [latex]\frac{1}{\mathrm{30}}[/latex] – [latex]\frac{1}{\mathrm{20}}[/latex]
[latex]\frac{1}{\mathrm{f}}[/latex] = [latex]\frac{20-30}{30 \times 20}[/latex] = [latex]\frac{-10}{30 \times 20}[/latex]
[latex]\frac{1}{\mathrm{f}}[/latex] = [latex]-\frac{1}{60}[/latex]
f = -60cm.

Question 5.
A double convex lens of focal length 15cm is used as a magnifying glass in order to produce an erect image which
is 3 times magnified. What is the distance between the object and the lens?
Answer:
f = 15cm
m = 3
Magnifying power (m) = [latex]\frac{-v}{\mathbf{u}}[/latex] = [latex]\frac{f}{f-u}[/latex]
3 = [latex]\frac{15}{15-u}[/latex]
45 – 3u = 15
3u = 45 – 15
3u = 30
u = [latex]\frac{30}{3}[/latex] = 10cm.

Question 6.
A compound microscope consists of an object lens of focal length 2cm and an eyepiece of focal length 5cm. When an object is placed at 2.2cm from the object lens, the final Image is at 25cm from the eye lens. What is the distance between the lenses ? What is the total linear magnification?
Answer:
Given that f₀ = 2, f_e = u₀ = 2.2, D = 25cm
[latex]\frac{1}{\mathrm{f}_0}[/latex] = [latex]\frac{1}{\mathrm{u}_0}[/latex] + [latex]\frac{1}{\mathrm{v}_0}[/latex]
[latex]\frac{1}{\mathrm{v}_0}[/latex] = [latex]\frac{1}{\mathrm{f}_0}[/latex] – [latex]\frac{1}{\mathrm{u}_0}[/latex] = [latex]\frac{1}{2}[/latex] – [latex]\frac{1}{2.2}[/latex] ⇒ [latex]\frac{1}{v_0}[/latex] = [latex]\frac{2.2-2}{2 \times 2.2}[/latex]
v₀ = 22 cm
For the eye-piece, the distance of the image V_e = 25cm
[latex]\frac{1}{f_e}[/latex] = [latex]\frac{1}{\mathrm{u}_{\mathrm{e}}}[/latex] – [latex]\frac{1}{v_e}[/latex] (For virtual image)
[latex]\frac{1}{u_e}[/latex] = [latex]\frac{1}{f_e}[/latex] + [latex]\frac{1}{v_e}[/latex] = [latex]\frac{1}{5}[/latex] + [latex]\frac{1}{25}[/latex]
⇒ [latex]\frac{1}{u_e}[/latex] = [latex]\frac{25+5}{5 \times 25}[/latex]
u_e = 4.166
i) The distance between the two lenses
L = v₀ + u_e
L = 22 + 4.166
L = 26.166

Question 7.
The distance between two point sources of light is 24cm. Where should you place a converging lens, of focal length 9 cm, so that the images of both sources are formed at the same point ?
Answer:
Distance between two point sources of light = 24cm Focal length (f) = 9cm Radius of curvature (R) = 2f
R = 2 × 9 = 18cm
∴ Converging lens is placed at 18 cm (or) Second position of converging
lens = 24 – 18 = 6cm.
∴ position of converging lens = 18 cm (or) 6cm.

Question 8.
Find two positions of an object, placed in front of a concave mirror of focal length 15cm, so that the image formed is 3 times the size of the object.
Answer:
f = 15cm
m = 3
i) m = [latex]\frac{-v}{u}[/latex] = [latex]\frac{f}{f-u}[/latex]
3 = [latex]\frac{15}{15-u}[/latex]
45 – 3u = 15
3u = 30
u = 10 cm

ii) m = [latex]\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}[/latex]
3 = [latex]\frac{15}{\mathrm{u}-15}[/latex]
3u – 45 = 15
3u = 60
u = 20

Question 9.
When using a concave mirror, the magnification is found to be 4 times as much when the object is 25cm from the mirror as it is with the object at 40cm from the mirror, the image being real in each case. What is the focal length of the mirror?
Answer:
Given that m = 4
u = 25cm
m = [latex]\frac{f}{u-f}[/latex]
4 = [latex]\frac{f}{25-f}[/latex]
100 – 4f = f
100 = 5f
f = [latex]\frac{100}{5}[/latex] = 20cm.

Question 10.
The focal length of the objective and eyepiece of a compound microscope are 4cm and 6cm respectively. If an object is placed at a distance of 6cm from the objective, what is the magnification produced by the microscope ?
Answer:
Given that f₀ = -4cm, f_e = 6cm, u₀ = 6
[latex]\frac{1}{\mathrm{f}_0}[/latex] = [latex]\frac{1}{\mathrm{v}_0}+\frac{1}{\mathrm{u}_0}[/latex]
[latex]\frac{1}{v_0}[/latex] = [latex]\frac{1}{\mathrm{f}_0}[/latex] – [latex]\frac{1}{\mathrm{u}_0}[/latex] = [latex]\frac{1}{4}-\frac{1}{6}[/latex]
[latex]\frac{1}{v_0}[/latex] = [latex]\frac{2}{24}[/latex]
υ₀ = 12 cm.
Magnifying power m = [latex]\frac{\mathrm{v}_0}{\mathrm{u}_0}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)[/latex]
= [latex]\frac{12}{6}\left[1+\frac{25}{6}\right][/latex] = [latex]2\left[\frac{31}{6}\right][/latex] = [latex]\frac{62}{6}[/latex]
m = 10.33

Textual Exercises

Question 1.
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved ?
Answer:
u = – 27cm, R = – 36cm, f = – 18cm
[latex]\frac{1}{\mathrm{u}}[/latex] + [latex]\frac{1}{v}[/latex] = [latex]\frac{1}{\mathrm{f}}[/latex] ⇒ [latex]\frac{-1}{27}+\frac{1}{v}[/latex] = [latex]\frac{-1}{18}[/latex]
[latex]\frac{1}{v}[/latex] = [latex]\frac{1}{27}-\frac{1}{18}[/latex] ⇒ [latex]\frac{1}{\mathrm{v}}[/latex] = [latex]\frac{-1}{54}[/latex]
v = – 54 cm.
The Screen should be placed 54cm from the mirror
m = [latex]\frac{\mathrm{I}}{\mathrm{O}}[/latex] = [latex]\frac{-\mathrm{V}}{\mathrm{u}}[/latex] ⇒ [latex]\frac{\mathrm{I}}{2.5}[/latex] = [latex]\frac{-54}{27}[/latex]
U = – 5cm
∴ The image is real, inverted and magnified. If the candle is moved closer, the screen would have moved farther and farther. Closer than 18cm from the mirror, the image gets virtual and cannot be collected on the screen.

Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification.
Answer:
O = 4.5 cm, u = -12cm, f = 15cm

Image is virtual and erect and is formed behind the mirror.

As the needle is moved further from the mirror, the image moves towards the focus and gets progressively diminished in size.

Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 2.4 cm. What is the refractive index of water ? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again ?
Answer:

Distance by which image becomes raised = 9.4 – 7.67 = 1.73 = 1.7cm the microscope will be moved up by 1.7cm to focus on the needle again.

Question 4.

The above 3 Figures (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water- air figures interface respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with, the normal to a water-glass interface [Fig. (c)]
Answer:
1st Case:
Angle of incidence i = 60°
Angle of refraction r = 35°

3rd Case :
Angle of incidence i = 45°
Angle of refraction = ?
w_rg = [latex]\frac{\sin i}{\sin r}[/latex] = [latex]\frac{\sin i}{\sin r}[/latex] = 1.28
Sin r = [latex]\frac{\sin 45^{\circ}}{1.28}[/latex] = 0.5525
sin r = sin 33°54′
r = 33°44′

Question 5.
A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out ? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Answer:
If r is the radius of the large circle from which light comes out, C is the critical angle for water – air interface, then
tan C = [latex]\frac{\mathrm{DB}}{\mathrm{DO}}[/latex] = [latex]\frac{\mathbf{r}}{\mathbf{d}}[/latex]
r = d tan C
Area of circle,
A = πr²
A = π(d tan C)²
A = πd².[latex]\frac{\sin ^2 C}{\cos ^2 C}[/latex]
A = πd².[latex]\frac{\sin ^2 C}{1-\sin ^2 C}[/latex]
But Sin C = [latex]\frac{1}{\mu}[/latex] = [latex]\frac{1}{1.33}[/latex] ≈ 0.75
A = [latex]\frac{\pi(0.8)^2(0.75)^2}{1-(0.75)^2}[/latex] = 2.6m²

Question 6.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism ? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Answer:

Question 7.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm ?
Answer:
[latex]\frac{\mu_2}{\mu_1}[/latex] = μ = 1.55
R₁ = R, R₂ = -R
f = 20

Question 8.
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm ?
Answer:
Here the object is virtual and the image is real u = + 12 cm (object on right and virtual)
a) f = + 20cm
Lens formula is [latex]\frac{-1}{\mathrm{u}}[/latex] + [latex]\frac{1}{v}[latex] = [/latex]\frac{1}{f}[/latex]

i. e., u = 7.5 cm (image on right and real) It is located 7.5 cm from the lens,

b) f = – 16cm

Image will be located 48cm from the lens.

Question 9.
An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens ?
Answer:
‘O’ = 3.0cm
u = – 14cm, f = -21cm
[latex]\frac{1}{v}+\frac{1}{u}[/latex] = [latex]\frac{1}{f}[/latex] ⇒ [latex]\frac{1}{v}+\frac{1}{14}[/latex] = [latex]\frac{-1}{21}[/latex]
[latex]\frac{1}{v}[/latex] = [latex]\frac{-1}{21}-\frac{1}{14}[/latex]
υ = [latex]\frac{-42}{5}[/latex] = – 8.4 cm
Image is erect, virtual and located 8.4cm from the lens on the same side as the object. Using the relation,
[latex]\frac{I}{O}[/latex] = [latex]\frac{\mathrm{v}}{\mathrm{u}}[/latex]
υ = [latex]\frac{8.4}{15}[/latex] × 5 = 1.8cm

As the object is moved away from the lens, the virtual image moves towards the focus of the lens and progressively diminishes in size. (When u = 21 cm, v = -10.5 cm and when u = ∞, v = -21 cm)

Question 10.
What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20cm ? Is the system a converging or a diverging lens ? Ignore thickness of the lenses.
Answer:
Given f₁ = 30 cm, f₂ = -20 cm, f = ?
f = [latex]\frac{\mathrm{f}_1 \mathrm{f}_2}{\mathrm{f}_1+\mathrm{f}_2}[/latex]
f = [latex]\frac{30 \times(-20)}{30-20}[/latex] = -60cm
Thus the system is a diverging lens of focal length 60cm.

Question 11.
A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.2 5cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (2 5cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Answer:
a) V_e = 25 cm
f_e = 6.25 cm.
Using lens formula,

Question 12.
A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 2.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Answer:
Angular magnification of the eye piece for image at 25 cm

u = -[latex]\frac{25}{11}[/latex] = -2.27 cm
∴ Distance between objective and eye piece
= v + |u| = 7.2 + 2.27 = 9.47cm
Magnifying power of microscope
= [latex]\frac{7.2}{0.9} \times \frac{-25}{\frac{-25}{11}}[/latex] = 88

Question 13.
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope ? What is the separation between the objective and the eyepiece?
Answer:
a) For normal adjustment
M.P of telescope = [latex]\frac{f_0}{f_e}[/latex] = [latex]\frac{144}{6}[/latex] = 24
b) The length of the telescope in normal adjustment
L = f₀ + f_e = 144 + 6
= 150 cm.

Question 14.
a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of
focal length 1.0 cm is used, what is the angular magnification of the telescope?
b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 × 10⁶m, and radius of lunar orbit is 3.8 × 10⁸m.
Answer:
a) Anglular magnification
= [latex]\frac{f_0}{f_e}[/latex] = [latex]\frac{15}{0.01}[/latex] = 1500
b) if d is the diameter of the image (in cm)
[latex]\frac{\mathrm{d}}{1500}[/latex] = [latex]\frac{3.48 \times 10^6}{3.8 \times 10^8}[/latex]
d = 13.7

Question 15.
Use the mirror equation to deduce that:
a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
b) a convex mirror always produces a virtual image independent of the location of the object.
c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

[Note : This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Answer:

Since for concave mirror, f is negative, υ becomes negative.
It means image produced is real and beyond 2f.

b) For mirror formula,
υ = [latex]\frac{\mathbf{u f}}{\mathbf{u}-\mathbf{f}}[/latex]
since for a convex mirror, fis positive and u is always negative, u will be always positive image and will always be formed behind the mirror and will be virtual.

c) For relation, m = [latex]\frac{v-f}{f}[/latex] positive for convex mirror, m will always be negative and less than one. Hence virtual image formed will always be diminished.
For relation, m = [latex]\frac{v-f}{f}[/latex] and m being negative, υ will always be less than f. Hence image will be formed between pole and focus.

d) When u > 0 < f we get m = [latex]\frac{f}{u-f}[/latex] = [latex]\frac{f}{(>0<f)-f}[/latex] = [latex]\frac{f}{(>-f<0)}[/latex] = > -1
(∵ m is negative, image is virtual and enlarged because is numerically >1).

Question 16.
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By What distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table ? Refractive index of glass = 1.5. Does the answer depend on the location of the slab ?
Answer:
μ = 1.5; real thickness = 15 cm

∴ Apparent Depth = [latex]\frac{15}{1.5}[/latex] = 10 cm
∴ Pin appears raised by 15 – 10 = 5 cm.
The result is independent of the location of the slab.

Question 17.
a) Figure shows a cross-section of a light pipe made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.

b) What is the answer if there is no outer covering of the pipe ?

Answer:
a)
μ = [latex]\frac{1.68}{1.44}[/latex] = [latex]\frac{1}{\sin C}[/latex]
Sin C = [latex]\frac{1.44}{1.68}[/latex] = 0.8571
C = 59°
Total internal reflection takes place when i > 59° or angle r may have value between 0 to 31°
r_max = 31°
Now [latex]\frac{\sin \mathrm{i}_{\max }}{\sin \mathrm{r}_{\max }}[/latex] = 1.68
[latex]\frac{\sin i_{\max }}{\sin 31^{\circ}}[/latex] = 1.68
sin i_max = 0.8562, i<sub.max = 60°
Thus all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe.

b) If there is no outer covering of the pipe
Sin C = [latex]\frac{1}{\mu}[/latex]
= [latex]\frac{1}{1.68}[/latex] = 0.5962
sin C = sin 36.5°
C = 36.5°

Question 18.
Answer the following questions :
a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce, real images under some circumstances ? Explain.
b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction ?
c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is ?
d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decrease ?
e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter ?
Answer:
a) Rays converging to a point (behind) a plane or convex mirror are reflected to a point infront of the mirror on the screen. In other words a plane or convex mirror can produce a real image if the object is virtual

b) When the reflected or refracted rays are divergent, the image is virtual. The divergent rays can be converged on to a screen by means of an appropriate converging lens. The convex lens of the eye does just that the virtual image here serves as a object for the lens to produce a real image. The screen here is not located at the position of the virtual image. There is no contradiction.

c) The rays starting from the head of the fisherman and incident on water become bent towards normal and appear to come from a higher point.
AF is real height of fisherman. Rays starting from A, bend towards normal. For diver they appear to come from A₁, A₁ F becomes apparent height of fisherman, which is more than real height.

d) The apparent depth for oblique viewing decreases from its value for near-normal viewing.

e) Refractive index of diamond is about 2.42, much larger than that of ordinary glass. The critical angle for diamond is above 24°, much less than that of glass. A skilled diamond cutter exploits the large range of angles of incidence (in the diamond) 24° to 90° to ensure that light entering the diamond is totally reflected from many faces before getting out thus producing a sparkling effect.

Question 19.
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens: What is the maximum possible focal length of the lens required for the purpose ?
Answer:
Let,
υ = + υ
∴ u = -(3 – υ)
f_max = ?
Now, [latex]\frac{1}{f}[/latex] = [latex]\frac{1}{v}-\frac{1}{u}[/latex] = [latex]\frac{1}{v}-\frac{1}{-(3-v)}[/latex]
[latex]\frac{1}{f}[/latex] = [latex]\frac{1}{v}+\frac{1}{3-v}[/latex] ⇒ [latex]\frac{1}{f}[/latex] = [latex]\frac{3-v+v}{(3-v) v}[/latex]
3υ – υ² = 3f
For f to be maximum d(f) = 0
i.e d(3υ – υ²) = 0
3 – 2 υ = 0
υ = 3/2 = 1.5 m
Hence, u = -(3 – 1.5)
= -1.5m
and
[latex]\frac{1}{f}[/latex] = [latex]\frac{1}{v}[/latex] – [latex]\frac{1}{u}[/latex] = [latex]\frac{1}{1.5}[/latex] – [latex]\frac{1}{-1.5}[/latex] = [latex]\frac{1+1}{1.5}[/latex]
= [latex]\frac{2}{1.5}[/latex] = 0.75m

Question 20.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Answer:
a) Distance between object and image D = 90 cm = u + υ
Distance between two’ positions of lens d = 20 = u = υ
u = 55 cm and υ = 35 cm.
For lens formula,
[latex]\frac{1}{\mathrm{f}}[/latex] = [latex]\frac{1}{55}[/latex] + [latex]\frac{1}{35}[/latex] = [latex]\frac{18}{385}[/latex]
f = [latex]\frac{385}{18}[/latex] = 21.4

Question 21.
a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 10, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all ?
b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40cm. Determine the magni-fication produced by the two-lens is 40cm.
Answer:
a) Here f₁ = 30cm, f₂ = -20cm,
d = 8.0cm, f = ?

i) Let a parallel beam be incident on the convex lens first. If 2nd lens were absent, then
u₁ = ∞ and f₁ = 30cm
As [latex]\frac{1}{v_1}-\frac{1}{\mathrm{u}_1}[/latex] = [latex]\frac{1}{f_1}[/latex]
[latex]\frac{1}{v_1}[/latex] = [latex]\frac{1}{\infty}[/latex] = [latex]\frac{1}{30}[/latex]
υ₁ = 30 cm
The image would now act as a virtual object for 2nd lens.
υ₂ = + (30 – 8) = + 22 cm
υ₂ = ? f₂ = – 20 cm
As [latex]\frac{1}{v_2}[/latex] = [latex]\frac{1}{\mathrm{f}_2}+\frac{1}{\mathrm{u}_2}[/latex]
[latex]\frac{1}{v_2}[/latex] = [latex]\frac{1}{-20}+\frac{1}{22}[/latex] = [latex]\frac{-11+10}{220}[/latex] = [latex]\frac{-1}{220}[/latex]
υ₂ = -220 cm
∴ Parallel incident beam would appear to diverge from a point 220 – 4 = 216 cm from the centre of the two lens system,

ii) Suppose a parallel beam of light from the left is incident first on the concave lens.

This image acts as a real object for the 2nd lens.

∴ The parallel beam appears to diverge from a point 420 – 4 = 416 cm, on the left of the centre of the two lens system From the above discussion, we observe that the answer depends on which side of the lens system the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be useful here.

b) Here, h₁ = 1.5 cm, u₁ = 40 cm, m = ?, h₂ = ? for the 1st lens, [latex]\frac{1}{v_1}-\frac{1}{u_1}[/latex] = [latex]\frac{1}{f_1}[/latex]
[latex]\frac{1}{v_1}[/latex] = [latex]\frac{1}{p_1}+\frac{1}{u_1}[/latex] = [latex]\frac{1}{30}-\frac{1}{40}[/latex] = [latex]\frac{1}{120}[/latex]
υ₁ = 120 cm
Magnitude of magnification produced by first lens,
m₁ = [latex]\frac{v_1}{u_1}=\frac{120}{40}[/latex] = 3
The image formed by 1st lens as virtual object for the 2nd lens.
υ₂ = 120 – 8 = 112 cm, f₂ = – 20cm
υ₂ = ?

Magnitude of Magnification produced by second lens
m₂ = [latex]\frac{\mathrm{v}_2}{\mathrm{u}}[/latex] = [latex]\frac{112 \times 20}{92 \times 112}[/latex] = [latex]\frac{20}{92}[/latex]
Net magnification produced by the combination
m = m₁ × m₂
= 3 × [latex]\frac{20}{92}[/latex] = [latex]\frac{60}{92}[/latex] = 0.652
∴ size of image h₂ = mh₁
= 0.652 × 1.5
= 0.98 cm

Question 22.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face ? The refractive index of the material of the prism is 1.524.
Answer:
i₁ = ?, A = 60°, μ = 1.524
μ = [latex]\frac{1}{\sin C}[/latex]
C = r₂
sin C = sin r₂ = [latex]\frac{1}{\mu}=\frac{1}{1.524}[/latex] = 0.6561
r₂ = 41°
As r₁ + r₂ = A
r₁ = A – r₂ = 60° – 41° = 19°
μ = [latex]\frac{\sin i_1}{\sin r_1}[/latex]
sin i₁ = 1.524 sin 19°
= 1.524 × 0.3256
= 0.4962
i₁ = 29°45¹

Question 23.
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
a) deviate a pencil of white light without much dispersion.
b) disperse (and displace) a pencil of white light without much deviation.
Answer:
i) For no dispersion, angular dispersion produced by two prisms should be zero
i.e. (μ_b – μ) A + (μ_b – μ’_r) A’ = 0
As (μ’_b – μ’_r) for flint glass is more than that for grown glass, therefore A’ < A i.e ., flint glass prism of smaller angle has to be suitably combined with crown glass prism of larger angle.

ii) For almost no deviation (μ_v – 1) A + (μ’_y – 1) A’ = 0
Taking crown glass prism of certain angle, we go on-increasing angle of flint glass prism till this condition is met. In the final combination however, angle of flint glass prism will be smaller than the angle of crown glass prism as μ’_y for flint glass is more than μ_y for crown glass.

Question 24.
For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 diopters, and the least converging power of the eye- lens behind the cornea is about 20 diopters. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye. ”
Answer:
To observe objects at infinity, the eye uses its least converging power = 40 + 20 = 60D
∴ Distance between cornea eye lens and retina
focal length of eye lens = [latex]\frac{100}{p}[/latex] = [latex]\frac{100}{60}[/latex] = [latex]\frac{5}{3}[/latex] cm
To focus an object at the near point u = -25cm, v = 5/3 cm f = ? –
[latex]\frac{1}{f}[/latex] = [latex]-\frac{1}{\mathrm{u}}+\frac{1}{v}[/latex] ⇒ [latex]\frac{1}{f}[/latex] = [latex]\frac{1}{25}+\frac{3}{5}[/latex] = [latex]\frac{1+15}{25}[/latex]
= 16/25
f = 25/16 cm
Power = [latex]\frac{100}{f}[/latex] = [latex]\frac{100}{25 / 16}[/latex] = 64D
Power of eye lens = 64 – 40 = 24D
Hence range of accommodation of eye lens is roughly 20 to 24 dioptre.

Question 25.
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?
Answer:
No, a person may have normal ability of accommodation and yet he may be myopic or hyper metropic. Infact, myopia arises when length of eye ball gets shortened.
However, when eye ball has normal length, but the eye lens loses partially its power of accommodation, the defect is called Presbiopia.

Question 26.
A myopic person has been using spectacles of power – 1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain whàt may have happened.
Answer:
Here u = -25cm, y = -50cm, f = ?
[latex]\frac{1}{v}-\frac{1}{u}[/latex] = [latex]\frac{1}{\mathrm{f}}[/latex] ⇒ -[latex]\frac{1}{50}[/latex] + [latex]\frac{1}{25}[/latex] = [latex]\frac{1}{f}[/latex]
[latex]\frac{-1+2}{50}[/latex] = [latex]\frac{1}{\mathrm{f}}[/latex] or f = 50 cm
As P = [latex]\frac{100}{f}[/latex] = [latex]\frac{100}{50}[/latex] = + 2 dioptres.

Question 27.
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to ? How is such a defect of vision corrected ?
Answer:
This defect is called Astigmatism. It arises because curvature of cornea plus eye lens refracting system is-not the same in diffreht planes. As vertical lines are seen distinctly, the curvature in the vertical plane is enough, but in the horizontal , plane, curvature is insufficient.
This defect is removed by using a cylindrical lens with its axis along the vertical.

Question 28.
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass ?
b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope ?
Answer:
a) Here, f = 5 cm, u = ?
For the closest distance; v = – 25cm
As [latex]\frac{1}{f}[/latex] = [latex]\frac{1}{v}-\frac{1}{u}[/latex]
[latex]\frac{1}{u}[/latex] = [latex]\frac{1}{v}-\frac{1}{f}[/latex] = [latex]\frac{1}{-25}-\frac{1}{5}[/latex] = [latex]\frac{-1-5}{25}[/latex]
u’ = [latex]\frac{25}{-6}[/latex] – 4.2 cm
This is the closest distance at which he can read the book.
For the farthest distance, v’ = ∞, u’ = ?
As [latex]\frac{1}{\mathrm{f}}[/latex] = [latex]\frac{1}{\mathrm{v}^{\prime}}-\frac{1}{\mathrm{u}^{\prime}}[/latex]
[latex]\frac{1}{v^{\prime}}[/latex] = [latex]\frac{1}{v^i}-\frac{1}{f}[/latex] = [latex]\frac{1}{\infty}-\frac{1}{5}[/latex] = [latex]\frac{-1}{5}[/latex]
u’ = -5cm
This is the farthest distance at which he can real the book.

b) Max. Angular magnification
m = [latex]\frac{\mathrm{d}}{\mathrm{u}}[/latex] = [latex]\frac{25}{25 / 6}[/latex] = 6
Min. Angular magnification
m’ = [latex]\frac{\mathrm{d}}{\mathrm{u}^{\prime}}[/latex] = [latex]\frac{25}{5}[/latex] = 5

Question 29.
A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9cm) held close to the eye.
a) What is the magnification in produced by the lens ? How much is the area of each square in the virtual image ?
b) What is the angular magnification (magnifying power) of the lens ?
c) Is the magnification in (a) equal to the magnifying power in (b) ? Explain.
Answer:
a) Here, area of each (object) square 1mm²,
u – 9cm, f = 10cm
[latex]\frac{1}{v}[/latex] = [latex]\frac{1}{f}+\frac{1}{u}[/latex] = [latex]\frac{1}{10} \frac{-1}{9}[/latex] = [latex]\frac{-1}{90}[/latex]
v = -90cm
Magnification, m = [latex]\frac{v}{|u|}=\frac{90}{9}=10[/latex]
∴ Area of each square in virtual image = (10)² × 1 = 100 sq,mm

b) Magnifying power = [latex]\frac{\mathrm{d}}{\mathrm{u}}[/latex] = 25/9 = 2.8
c) No, Magnification in (a) which is (υ/u) cannot be equal to magnifying power in (b) which is (dlu) unless v = d ie., image is located at the least distance of distinct vision.

Question 30.
a) At what distance should the lens be held from the figure in Exercise 2.29 in order to view the squares distinctly with the maximum possible magnifying power ?
b) What is the magnification in this case ?
c) Is the magnification equal to the magnifying power in this case ? Explain.
Answer:
i) Here, υ = -25cm, f = 10cm, u = ?

Yes, the magnification and magnify-ing power in this case are equal, because image is formed at the least distance of distinct vision.

Question 31.
What should be the distance between the object in Exercise 30 and the magnifying glass if the virtual image of each square in the figure is to have an , area of 6.25 mm². Would you be able to see the squares distinctly with your eyes very close to the magnifier ?
[Note : Exercises 29 to31 will help you clearly understand the difference between magnification in absolute size ‘ and the angular magnification (or magnifying power) of an instrument.]
Answer:
Here, magnification in area = 6.25
linear magnification m = [latex]\sqrt{6.25}[/latex] = 2.5
As m = [latex]\frac{v}{u}[/latex] or v = mu = 2.5u
As [latex]\frac{1}{v}[/latex] – [latex]\frac{1}{\mathrm{u}}[/latex] = [latex]\frac{1}{\mathrm{f}}[/latex]
[latex]\frac{1}{2.5 \mathrm{u}}[/latex] – [latex]\frac{1}{\mathrm{u}}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{1-2.5}{2.5 u}[/latex] = [latex]\frac{1}{10}[/latex]
u = -6cm
y = 2.5u = 2.5 (-6) = -15cm
as the virtual image is at 15cm; where as distance of distinct vision is 25cm, therefore, the image cannot be seen distinctly by the eye.

Question 32.
Answer the following questions :
a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification ?
b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back ?
c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power ?
d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths ?
e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why ? How much should be that short distance between the eye and eyepiece ?
Answer:
a) It is true that angular size of image is equal to the angular size of the object. By using magnifying glass, we keep the object far more closer to the eye than at 25cm, its normal position without use of glass. The closer object has larger angular size than the same object at 25cm. It is in this sense that angular magnification is achieved.

b) yes, the angluar magnification changes, if the eye is moved back. This is because angle subtended at the eye would be slightly less than the angle subtended at the lens. The effect is negligible when image is at much larger distance.

c) Theoretically, it is true, However, when we decrease focal length, observations both spherical and chromatic become more pronounced. Further, it is difficult to grind lenses of very small focal lengths.

d) Angular magnification of eye piece is [latex]\left(1+\frac{d}{f_e}\right)[/latex]. This increase as f_e decreases. Further, magnification if objective lens is [latex]\frac{v}{u}[/latex]. As object lies close to focus of objective lens u ≈ f₀. To increase this magnification (υ/f₀), f₀ should be smaller.

e) The image of objective lens in eye pie is called ‘eye ring’ All the rays from the object refracted by the objective go through the eye ring. Therefore, ideal position for our eyes for viewing is this eye ring only.

When eye is too close to the eye piece, field of view reduces and eyes do not collect much of the light. The precise, location of the eye ring would depend upon the separation between the objective and eye piece, and also on focal length of the eye piece.

Question 33.
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5cm. How will you set up the compound microscope ?
Answer:
In normal ajustment, image is formed at least distance of distinet vision, d = 25cm Angular Magnification of eye piece = [latex]\left(1+\frac{d}{f_e}\right)=\left(1+\frac{25}{5}\right)[/latex] = 6 As total Magnification is 30, Magnification of objective lens.

i.e. object should be held at 1.5cm in-front of objective lens

∴ Seperation between the objective lens and eye piece
= |u_e| + |v₀|
= 4.17 + 7.5.
= 11.67cm

Question 34.
A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
a) the telescope is in normal adjustment (i.e., when the final image is at infinity) ?
b) the final image is formed at the least distance of distinct vision (25 cm) ?
Answer:
Here, f₀ = 140cm, f_e = 5.0cm
Magnifying power = ?
a) In normal adjustment,
Magnifying power
= [latex]\frac{f_0}{-f_e}[/latex] = [latex]\frac{140}{-5}[/latex] = -28

b) When final image is at the least distance of distinct vision, Magnifying power
= [latex]\frac{-\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\left(1+\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{d}}\right)[/latex] = [latex]\frac{-140}{5}\left(1+\frac{5}{25}\right)[/latex] = -33.6

Question 35.
a) For the telescope described in Exercise 34 (a), what is the separation between the objective lens and the eyepiece ?
b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens ?
c) What is the height of the final image of the tower if it is formed at 25 cm ?
Answer:
a) Here, in normal adjustment, seperation between objective lens and eye lens = f₀ + f_e = 140 + 5 = 145cm

b) Angle subtended by tower 100m tall at 3km
α = [latex]\frac{100}{3 \times 1000}[/latex] = [latex]\frac{1}{30}[/latex] radian
if his the height of image formed by the objective, then
α = [latex]\frac{\mathrm{h}}{\mathrm{f}_0}[/latex] = [latex]\frac{h}{140}[/latex]
∴ [latex]\frac{\mathrm{h}}{140}[/latex] = [latex]\frac{\mathrm{1}}{30}[/latex]
h = [latex]\frac{\mathrm{140}}{30}[/latex] = 4.7 cm

c) Magnifying produced by eye piece
= (1 + [latex]\frac{d}{f e}[/latex]) = 1 + [latex]\frac{25}{5}[/latex] = 6
∴ Height of final image = 4.7 × 6 = 28.2cm

Question 36.
A cassegrain telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140mm. where will the final image of an object at infinity be ?
Answer:
Here, radius of curvature of objective mirror R₁ = 220 mm radius of curvature of secondary mirror R₂ = 140mm;
f₂ = [latex]\frac{\mathrm{R}_2}{2}=\frac{140}{2}[/latex] = 70mm
Distance between the two mirrors, d = 20mm. When object is at infinity, parallel rays falling on objective mirror, on reflection, would collect at its focus at
f₁ = [latex]\frac{\mathrm{R}_1}{2}=\frac{220}{2}[/latex] = 110mm
Instead, they fall on secondary mirror at 20mm from objective mirror.
∴ For secondary mirror, u = f₁ – d = 110 – 20 = 90 mm
From [latex]\frac{1}{v}+\frac{1}{u}[/latex] = [latex]\frac{1}{\mathrm{f}_2}[/latex]
υ = [latex]\frac{1}{\mathrm{f}_2} \frac{-1}{\mathrm{u}}[/latex] = [latex]\frac{1}{70}-\frac{1}{90}[/latex] = [latex]\frac{9-7}{630}[/latex] = [latex]\frac{2}{630}[/latex]
υ = [latex]\frac{630}{2}[/latex] = 315 mm = 31.5 cm to the right of secondary mirror

Question 37.
Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Answer:
Here, θ = 3.5°
x = 1.5m, d = ?
When the mirror turns through an angle θ, the reflected ray turns through double the angle.
2θ = 2 × 3.5° = 7° = [latex]\frac{7 \pi}{180}[/latex] rad
from figure,
tan 2θ = [latex]\frac{\text { SA }}{\mathrm{OS}}[/latex] = [latex]\frac{\mathrm{d}}{1.5} \times \mathrm{d}[/latex]
= 1.5 × [latex]\frac{7 \pi}{180} \mathrm{~m}[/latex] = 0.18m
d = 1.5 tan 2θ
≈ 1.5(2θ)
= 1.5 × [latex]\frac{7 \pi}{180} \mathrm{~m}[/latex] = 0.18m

Question 38.
Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A
small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?

Answer:
Let focal length of convex lens of glass = f₁ = 30cm focal length of plano concave lens of liquid = f₂ combined focal length, F = 45.0cm

For liquid lens

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 1st Lesson Waves Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 1st Lesson Waves

Very Short Answer Questions

Question 1.
What does a wave represent?
Answer:
A wave represents the transport of energy through a medium from one point to another without translation of the medium.

Question 2.
Distinguish between transverse and longitudinal waves.
Answer:
Transverse waves

1. The particles of the medium vibrate perpendicular to the direction of wave propagation.
2. Crests and troughs are formed alternatively.

Longitudinal waves

1. The particles of the medium vibrate parallel to the direction of wave propagation.
2. Compressions and rare fractions are formed alternatively.

Question 3.
What are the parameters used to describe a progressive harmonic wave ?
Answer:
Progressive wave equation is given y = a sin (ωt – kx)
Where ω = 2πv = [latex]\frac{2 \pi}{T}[/latex]

Parameters :

1. a = Amplitude
2. λ = Wavelength
3. T = Time period
4. v = Frequency
5. k = Propagation constant
6. ω = Angular frequency.

Question 4.
Obtain an expression for the wave velocity in terms of these parameters.
Answer:
Let ‘v’ be the velocity of a wave, ‘v’ be frequency and ‘λ’ be the wavelength. If T is the time period, then v = [latex]\frac{1}{\mathrm{~T}}[/latex]
The distance travelled by the wave in the time T = λ.
Distance travelled in one second = [latex]\frac{\lambda}{T}[/latex]
which is equal to wave velocity v = [latex]\frac{\lambda}{T}[/latex].
∴ v = vλ

Question 5.
Using dimensional analysis obtain an expression for the speed of transverse waves in a stretched string.
Answer:
Wave velocity v ∝ T^a µ^b ⇒ V = K T^a µ^b ——-> (1)
Dimensions of v = M⁰L¹ T^-1, Tension T = M¹L¹T^-2,
Linear mass µ = M¹L^-1, Constant K = M⁰L⁰T⁰
Now (1) becomes M⁰L¹T^-1 = [M¹L¹T^-2]^a [M¹L^-1]^b
M⁰L¹T¹ = M^{a + b}L^a-bT^-2a Comparing the powers of same physical quantity.
-1 = -2a ⇒ a = [latex]\frac{1}{2}[/latex]
a + b = 0 ⇒ b = -[latex]\frac{1}{2}[/latex]
⇒ v = (1)[latex]T^{\frac{1}{2}} \mu^{\frac{1}{2}}[/latex] [∵ K = 1 Practically]
∴ v = [latex]\sqrt{\frac{\mathrm{T}}{\mu}}[/latex]

Question 6.
Using dimensional analysis obtain an expression for the speed of sound waves in a medium. .
Answer:
Speed of sound v ∝ B^a ρ^b ⇒ v = K B^a ρ^b ——–> (1)
Dimensions of v = M⁰L¹T^-1,
Elasticity of medium,
B = M¹L^-1T^-2, density ρ = M¹L^-3, constant K = M⁰L⁰T⁰.
Now (1) becomes M⁰L¹T^-1 = M⁰L⁰T⁰ [M¹L^-1T^-2]^a [M¹L^-3]^b
0 = a + b
1 = -a – 3b
-1 = -2a ⇒ a = [latex]\frac{1}{2}[/latex]
b = -[latex]\frac{1}{2}[/latex]
v = K [latex]B^{\frac{1}{2}} \rho^{\frac{1}{2}}[/latex]
∴ v = [latex]\sqrt{\frac{B}{\rho}}[/latex] [∵ K = 1, practically]

Question 7.
What is the principle of superposition of waves ?
Answer:
When two or more waves, are acting simultaneously on the particle of the medium, the resultant displacement is equal to the algebraic sum of individual displacements of all the waves. This is the principle of superposition of waves.

If y₁, y₂, …… y_n be the individual displacements of the particles,then resultant displacement y = y₁ + y₂ + ……. + y_n

Question 8.
Under what conditions will a wave be reflected ?
Answer:

1. When the medium ends abruptly at any point.
2. If the density and rigidity modulus of the medium changes at any point.

Question 9.
What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary.
Answer:
π Radian or 180°.

Question 10.
What is a stationary or standing wave ?
Answer:
When two identical progressive (Transverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are superimposed then the resultant wave is called stationary waves or standing wave.

Question 11.
What do you understand by the terms ‘node ‘and’ antinode’?
Answer:
Node : The points at which the amplitude is zero, are called nodes.
Antinodes : The points at which the amplitude is maximum, are called antinodes.

Question 12.
What is the distance between a node and an antinode in a stationary wave ?
Answer:
The distance between node and antinode is [latex]\frac{\lambda}{4}[/latex]

Question 13.
What do you understand by ‘natural frequency’ or ‘normal mode of vibration’ ?
Answer:
When a body is set into vibration and then left to itself, the vibrations made by it are called natural or free vibrations. Its frequency is called natural frequency or normal mode of vibration.

Question 14.
What are harmonics ?
Answer:
The frequencies in which the standing waves can be formed are called harmonics.
(Or)
The integral multiple of fundamental frequencies are called harmonics.

Question 15.
A string is stretched between two rigid supports. What frequencies of vibration are possible in such a string ?
Answer:
The possible frequencies of vibrations in a stretched string between two rigid supports is given by
v_n = (n + [latex]\frac{1}{2}[/latex])[latex]\frac{v}{21}[/latex] where n = 0, 1, 2, 3, ……

Question 16.
The air column in a long tube, closed at one end, is set in vibration. What harmonics are possible in the vibrating air column ?
Answer:
The possible harmonics in the vibrating air column of a long closed tube is given by
v_n = [2n + 1][latex]\frac{v}{4 l}[/latex] where n = 0, 1, 2, 3, ……..

Question 17.
If the air column in a tube, open at both ends, is set in vibration; what harmonics are possible?
Answer:
The possible harmonics in vibrating air column of a long open tube is given by
V_n = [latex]\frac{n v}{21}[/latex]
where n = 1, 2, 3, ……….

Question 18.
What are ‘beats’ ?
Answer:
Beats : When two sound notes of nearly frequency travelling in the same direction and interfere to produce waxing and waning of sound at regular intervals of time is called “Beats”.

Question 19.
Write down an expression beat frequency and explain the terms there in.
Answer:
Expression of beat frequency, Δv = v₁ ~ v₂
where v₁ and v₂ are the frequencies of two waves.

Question 20.
What is ‘Doppler effect’? Give an example.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between source of sound and observer is called “Doppler effect”.

E.g.: When the whistling railway engine approaches the stationary observer on the platform, the frequency of sound appears to increase above the actual frequency. When it moves away from the observer, the apparent frequency decreases.

Question 21.
Write down an expression for the observed frequency when both source and observer are moving relative to each other in the same direction.
Answer:
Apparent frequency of sound heard by an observer,
v’ = [latex]\left[\frac{v-v_0}{v-v_s}\right] v[/latex]

where v = frequency of sound
v = velocity of sound
v₀ = velocity of observer
v_S = velocity of source

Short Answer Questions

Question 1.
What are transverse waves ? Give illustrative examples of such waves.
Answer:
Transverse waves: In a wave motion, the vibration of the particles and the direction of the propagation of the waves are perpendicular to each other, the waves are said to be transverse waves.

Illustration:

1. Waves produced in the stretched strings are transverse.
2. When a stretched string is plucked, the waves travel along the string.
3. But the particles in the string vibrate in the direction perpendicular to the propagation of the wave.
4. They can propagate only in solids and on the surface of the liquids.
5. Ex : Light waves, surface water waves.

Question 2.
What are longitudinal waves ? Give illustrative example of such waves.
Answer:
Longitudinal waves : In a wave motion, the direction of the propagation of the wave and vibrations of particles are in the same direction, the waves are said to be longitudinal waves.
Illustration:

1. Longitudinal waves may be easily illustrated by releasing a compressed spring.
2. A series of compressions and rarefactions (expansions) propagate along the spring.
   
   C = Compression; R = Rarefaction.
3. They can travel in solids, liquids and gases.
4. Ex : Sound waves.

Question 3.
Write an expression for a progressive harmonic wave and explain the various parameters used in the expression.
Answer:
The expression of a progressive harmonic wave is written as y = a sin(ωt – [latex]\frac{2 \pi}{\lambda} x[/latex])
or y = a sin(ωt – kx) where ω = 2πv, k = [latex]\frac{2 \pi}{\lambda}[/latex]

Parameters:

1. Amplitude (a) : It is the maximum displacement of a vibrating particle from its mean position.
2. Frequence (v): It is the number of complete vibrations made by a vibrating body in one second.
3. Wave length (λ) : It is defined as the distance covered by a wave while it completes one vibration, (or) It is the distance between two consecutive points in the same phase.
4. Phase of vibration (ϕ) : The phase of vibration of a vibrating particle gives its state of displacement at a given instant. It is generally given by the phase angle.

Question 4.
Explain the modes of vibration of a stretched string with examples.
Answer:
Modes of vibrations of a stretched string :

1. In sitar or Guitar, a stretched string can vibrate, in different frequencies and form stationary waves. This mode of vibrations are known as harmonics.
2. If it vibrates in one segment, which is known as fundamental harmonic. The higher harmonics are called the overtones.
   
3. It vibrates in two segments then the second harmonic is called first overtone. Similarly the pattern of vibrations are shown fig.
4. If a stretched string vibrates with P ’Seg’ ments (loop) then frequency of vibration v = [latex]\frac{\mathrm{P}}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}[/latex] where T = tension in the string, µ = linear density = [latex]\frac{\text { mass }}{\text { length }}[/latex]
5. In first mode of vibration, P = 1, then v = [latex]\frac{1}{2l} \sqrt{\frac{T}{\mu}}[/latex] (1^st hamonic (or) fundamental frequency)
6. second mode of vibration, P = 2, then v₁ = [latex]\frac{2}{2l} \sqrt{\frac{T}{\mu}}[/latex] = 2v (2^nd harmonic (or) 1^st overtone)
7. In third mode of vibration, P = 3, then v₂ = [latex]\frac{3}{2l} \sqrt{\frac{T}{\mu}}[/latex] = 3v (3^rd harmonic (or) 2^nd overtone)
   The ratio of the frequency of Harmonics are, v : v₁ : v₂ = v : 2v : 3v = 1 : 2 : 3

Question 5.
Explain the modes of vibration of an air column in an open pipe.
Answer:
Modes of vibration of an air column in an open pipe :
1) For a open pipe both the ends are open. So antinodes will be formed at both the ends. But two antinodes cannot exist without a node between them.
2) The possible harmonics in vibrating air column of a open pipe is given by .
Where n = 1, 2, 3
(1^st harmonic or fundamental frequence)

3) In first normal Mode of vibrating air column in a open pipe v₁ = [latex]\frac{v}{2l}[/latex] = v
(2nd harmonic 1^st overtone)

4) In second normal Mode of vibrating air column in a open pipe, v₂ = [latex]\frac{2 v}{2l}[/latex] = 2v

5) In third, normal Mode of vibrating air column in a open pipe, v₃ = [latex]\frac{3 v}{21}[/latex] = 3u
(3^rd harmonic 2^nd overtone)

6) In open pipe the ratio of frequencies of harmonics is
v₁ : v₂ : v₃ = v : 2v : 3v = 1 : 2 : 3

Question 6.
What do you understand by ‘resonance’ ? How would you use resonance to determine the velocity of sound in air ?
Answer:
Resonance: If the natural frequency of a vibrating body is equal to the frequency of external periodic force then the two bodies are said to be in resonance. At resonance the bodies will vibrate with increasing amplitude.

Determination of velocity of sound in air using resonance :

1) In resonance tube, an air column is made to vibrate by means of vibrating fork. At certain length of air column, the air column would have the same frequency as that of the fork. Then the air column vibrates with the maximum amplitude and the intense sound is produced.

2) The vibrating fork of known frequency (v) is placed above the open end of the tube.

3) The length of air column is gradually increased until the booming sound can be heard at two different lengths of air column.

4) In first resonance, l₁ be the length of air column, then [latex]\frac{\lambda}{4}[/latex] = l₁ + C …….. (1)
Where λ is the wavelength of sound emitted by the fork and C is the end correction of the tube.

5) In second resonance, l₂ be the length of air column, then [latex]\frac{3 \lambda}{4}[/latex] = l₂ + C …… (2)
(2) – (1) ⇒ [latex]\frac{\lambda}{2}[/latex] = l₂ – l₁

λ = 2 (l₂ – l₁)
Speed of sound is given by
υ = vλ = v[2(l₂ – l₁)]
∴ υ = 2v (l₂ – l₁)

6) Hence by knowing v, l₁, l₂ the speed of sound is calculated.

Question 7.
What are standing waves ? Explain how standing waves may be formed in a stretched string.
Answer:
Standing wave or stationary: When two identical progressive (Transverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are super-imposed then the resultant wave is called standing wave.

Formation of standing wave in a stretched string : –

1. If a string of length ‘l’ is stretched between two fixed points and set into vibration, a transverse progressive wave begins to travel along the string.
2. The wave is get reflected at the other fixed end.
3. The incident and reflected waves interfere and produce a stationary wave.
4. The stationary wave with nodes and antinodes is shown below.
   

Question 8.
Describe a procedure for measuring the velocity of sound in a stretched string.
Answer:
The velocity of a transverse wave travelling along a stretched string in fundamental mode is given by v = 2vl, where v = frequency, l = resonating length.

Measurement of velocity of sound in a stretched string using sonometer :

1. The wire is subjected to a fixed tension with suitable load.
2. A tuning fork of known frequency (v), is excited and the stem is held against the sono – meter box.
   
3. The distance between the two bridges is adjusted such that a small paper rider at the middle of B₁ B₂ vibrates vigorously and flies off due to resonance.
4. The resonating length ‘l’ can be measured between two bridges with scale.
5. By knowing v and l; we can find the velocity of a wave using υ = 2vl.

Question 9.
Explain, using suitable diagrams, the formation of standing waves in a closed pipe. How may this be used to determine the frequency of a source of sound ?
Answer:
Formation of standing waves in a closed pipe :

1. In closed pipe one end is closed and the other end is open. So antinode is formed at open end and antinode is formed at closed end.
2. The possible harmonics in vibrating air column in a closed pipe v_n = [latex]\frac{(2 n+1) v}{4 l}[/latex] where v = 0, 1, 2, 3, …….
3. In first normal mode of vibrating air column in a closed pipe, v₁ = [latex]\frac{v}{41}[/latex]
   [first harmonic (or) fundamental frequency]
   
4. In second normal mode of vibrating air column in a closed pipe,
   v₃ = [latex]\frac{3 \mathrm{v}}{4l}[/latex] [Third harmonic (or) first overtone]
5. In third normal mode of vibrating air column in a closed pipe,
   v₅ = [latex]\frac{5 \mathrm{v}}{4 \mathrm{l}}[/latex] [Fifth harmonic (or) second overtone]

Determination of frequency of a source of sound :

1. The vibrating fork of unknown frequency (v) is placed above the open end of the tube.
   
2. Reservoir is slowly lowered, until a large booming sound is heard. Measure 1st resonating, air column length l₁.
3. Further lower the reservoir, until second time a large booming sound is heard. Measure 2^nd resonating air column length l₂.
4. Velocity of a wave at 0°C is v = 331 m/s.
5. By knowing v, l₁ and l₂ we can find unknown frequency of a tuning fork using
   v = [latex]\frac{v}{2\left(l_2-l_1\right)}[/latex]

Question 10.
What are ‘beats’ ? When do they occur ? Explain their use, if any.
Answer:
Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing (maximum) and waning (minimum) in the intensity of the resultant sound waves at regular intervals of time is called beats.

It two vibrating bodies have slightly difference in frequencies, beats can occur.
No. of beats can be heard Δυ = υ₁ – υ₂

Importance :

1. It can be used to tune musical Instruments.
2. Beats are used to detect dangerous gases

Explanation-for tuning musical instruments with beats :
Musicians use the beat phenomenon in tuning their musical instruments. If an instrument is sounded against a standard frequency and tuned until the beats disappear. Then the instrument is in tune with the standard frequency.

Question 11.
What is ‘Doppler effect’? Give illustrative examples.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between the observer and the source of sound is called doppler effect.

Examples:

1. The frequency of whistling engine heard by a person standing on the platform appears to increase, when the engine is approaching the platform and it appears to decrease when the engine is moving away from the platform.
2. Due to Doppler effect the frequency of sound emitted by the siren of an approaching ambulance appears to increase. Similarly the frequency of sound appears to drop when it is moving away.

Long Answer Questions

Question 1.
Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse wave in stretched strings. (A.P. Mar. ’19)
Answer:
A string is a metal wire whose length is large when compared to its thickness. A stretched string is fixed at both ends, when it is plucked at mid point, two reflected waves of same amplitude and frequency at the ends are travelling in opposite direction and overlap along the length. Then the resultant waves are known as the standing waves (or) stationary waves.

Let two transverse progressive waves of same amplitude a, wave length λ and frequency ‘v’, travelling in opposite direction be given by
y₁ = a sin (kx – ωt) and y₂ = a sin (kx + ωt)
where ω’ = 2πv and k = [latex]\frac{2 \pi}{\lambda}[/latex]

The resultant wave is given by y = y₁ + y₂
y = a sin (kx – ωt) + a sin (kx + ωt)
y = (2a sin kx) cos ωt
2a sin kx = Amplitude of resultant wave.
It depends on ‘kx’. If x = 0, [latex]\frac{\lambda}{2}[/latex], [latex]\frac{2 \lambda}{2}[/latex], [latex]\frac{3 \lambda}{2}[/latex],……. etc, the amplitude = zero
These positions are known as “Nodes”.
If x = [latex]\frac{\lambda}{4}[/latex], [latex]\frac{2 \lambda}{2}[/latex], [latex]\frac{3 \lambda}{2}[/latex] ……… etc, the amplitude = zero

The positions are known as “Nodes”
If x = [latex]\frac{\lambda}{4}[/latex], [latex]\frac{3 \lambda}{4}[/latex], [latex]\frac{5 \lambda}{4}[/latex] ……. etc, the amplitude = maximum (2a).
These positions are called “Antinodes”.

If the string vibrates in ‘P’ segments and ‘l’ is its length then length of each segment = [latex]\frac{l}{p}[/latex]
Which is equal to [latex]\frac{\lambda}{2}[/latex]
∴ [latex]\frac{l}{\mathrm{p}}[/latex] = [latex]\frac{\lambda}{2}[/latex] ⇒ λ = [latex]\frac{2 l}{\mathrm{P}}[/latex]
Harmonic frequency v = [latex]\frac{v}{\lambda}=\frac{v \mathrm{P}}{2 l}[/latex]
v = [latex]\frac{v P}{2 l}[/latex] ——- (1)
If ‘ T’ is tension (stretching force) in the string and ‘μ’ is linear density then velocity of transverse wave (v) in the string is v = [latex]\sqrt{\frac{\mathrm{T}}{\mu}}[/latex] —– (2)
From the Eqs (1) and (2)
Harmonic frequency v = [latex]\frac{p}{2 l} \sqrt{\frac{T}{\mu}}[/latex]
If P = 1 then it is called fundamental frequency (or) first harmonic frequency
∴ Fundamental Frequency v = [latex]\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}[/latex] —— (3)

Laws of Transverse Waves Along Stretched String:

Fundamental frequency of the vibrating string v = [latex]\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}[/latex]

First Law : When the tension (T) and linear density (μ) are constant, the fundamental frequency (v) of a vibrating string is inversely proportional to its length.
∴ v ∝ [latex]\frac{1}{l}[/latex] ⇒ vl = constant, when T and ‘μ’ are constant.

Second Law: When the length (I) and its, linear density (m) are constant the fundamental frequency of a vibrating string is directly proportional to the square root of the stretching force (T).
∴ v ∝ [latex]\sqrt{\mathrm{T}}[/latex] ⇒ [latex]\frac{v}{\sqrt{T}}[/latex] = constant, when ‘l’ and ‘m’ are constant.

Third Law: WHien the length (J) and the tension (T) are constant, the fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m).
∴ v ∝ [latex]\frac{1}{\sqrt{\mu}}[/latex] ⇒ [latex]v \sqrt{\mu}[/latex] = constant, when ‘l’ and ‘T’ are constant.

Question 2.
Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced. (T.S. Mar. ’16, A.P. Mar. ’15)
Answer:
A pipe, which is opened at both ends is called open pipe. When a sound wave is sent through a open pipe, which gets reflected by the earth. Then incident and reflected waves are in same frequency, travelling in the opposite directions are super – imposed stationary waves are formed.

Harmonics in open pipe : To form the stationary wave in open pipe, which has two antinodes at two ends of the pipe with a node between them.
∴ The vibrating length (l) = half of the wavelength [latex]\left(\frac{\lambda_1}{2}\right)[/latex]
l = [latex]\frac{\lambda_1}{2}[/latex] ⇒ λ₁ = 2l
fundamental frequency v₁ = [latex]\frac{\mathrm{v}}{\lambda_1}[/latex] where v is velocity of sound in air v₁ = [latex]\frac{v}{2 l}[/latex] = v —— (1)
For second harmonic (first overtone) will have one more node and antinode than the fundamental.
If λ₂ is wavelength of second harmonic l = [latex]\frac{2 \lambda_2}{2}[/latex] ⇒ λ₂ = [latex]\frac{2l}{2}[/latex]
If ‘v₂‘ is frequency of second harmonic then v₂ = [latex]\frac{v}{\lambda_2}[/latex] = [latex]\frac{v \times 2}{2 l}[/latex] = 2v
v₂ = 2v —– (2)

Similarly for third harmonic (second overtone) will have three nodes and four antinodes as shown in above figure.
If λ₃ is wave length of third harmonic l = [latex]\frac{3 \lambda_3}{2}[/latex]
λ₃ = [latex]\frac{2l}{3}[/latex]
If ‘v₂’ is frequency of third harmonic then
v₃ = 3v —– (3)

Similarly we can find the remaining or higher harmonic frequencies i.e v₃, v₄ etc, can be determined in the same way.
Therefore the ratio of the harmonic frequencies in open pipe can be written as given below.
v : v₁ : v₂ = 1 : 2 : 3 ………

Question 3.
How are stationary waves formed in closed pipes ? Explain the various modes of vibrations and obtain relations for their frequencies. (A.P. & T.S. Mar. ’15)
Answer:
A pipe, which is closed at one end and the other is opened is called closed pipe. When a sound wave is sent through a closed pipe, which gets reflected at the closed end of the pipe. Then incident and reflected waves are in same frequency, travelling in the opposite directions are superimposed stationary waves are formed.

To form the stationary wave in closed pipe, which has atleast a node at closed end and antinode at open end of the pipe, it is known as first harmonic in closed pipe. Then length of the pipe (l) is equal to one fourth of the wave length.
∴ l = [latex]\frac{\lambda_1}{4}[/latex] ⇒ λ₁ = 4l
If ‘v₁‘ is fundamental frequency then
v₁ = [latex]\frac{v}{\lambda_1}[/latex] where ‘υ’ is velocity of sound in air
v₁ = [latex]\frac{v}{4l}[/latex] = v ——- (1)

To form the next harmonic in closed pipe, two nodes and two antinodes should be formed. So that there is possible to form third harmonic in closed pipe. Since one more node and antinode should be included.
Then length of the pipe is equal to [latex]\frac{3}{4}[/latex] of the wavelength.
∴ l = [latex]\frac{3 \lambda_3}{4}[/latex] where ‘λ₃‘ is wave length of third harmonic.
λ₃ = [latex]\frac{4l}{3}[/latex]
If ‘v₃‘ is third harmonic frequency (first overtone)
∴ v₃ = [latex]\frac{v}{\lambda_3}[/latex] = [latex]\frac{3 v}{41}[/latex]
v₃ = 3v ——- (2)

Similarly the next overtone in the close pipe is only fifth harmonic it will have three nodes and 3 antinodes between the closed end and open end.
Then length of the pipe is equal to [latex]\frac{5}{4}[/latex] of wave length (λ₅)
∴ l = [latex]\frac{5 \lambda_5}{4}[/latex] where ’λ₅‘ is wave length of fifth harmonic. .
λ₅ = [latex]\frac{4l}{5}[/latex]
If ‘V₅‘ is frequency of fifth harmonic (second overtone)
V₅ = [latex]\frac{v}{\lambda_5}=\frac{5 v}{4 I}[/latex]
v₅ = 5v —– (3)

∴ The frequencies of higher harmonics can be determined by using the same procedure. Therefore from the Eq (1), (2) and (3) only odd harmonics are formed.
Therefore the ratio of the frequencies of harmonics in closed pipe can be written as
v₁ : v₃ : v₅ = v : 3v : 5v
v₁ : v₃ : v₅ = 1 : 3 : 5

Question 4.
What are beats ? Obtain an expression for the beat frequency ? Where and how are beats made use of ?
Answer:
Beats : Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing and waning in the intensity of the resultant sound waves at regular intervals of time are called Beats.
If v₁ and v₂ are the frequencies of two sound notes superimposed along the same direction, no of beats heard per second = Δv = v₁ – v₂.
Maximum no. of beats heard per sec is 10 due to persistence of hearing.

Expression for the beat frequency :

1. Consider the two wave trains of equal amplitude but of nearly equal frequencies.
2. Let the frequencies of the waves be v₁ and v₂. Say v₁ is slightly greater than v₂.
3. Let the beat period be T seconds.
4. No.of vibrations, made by the first wave train in T seconds – v₁T
   [∵ no.of oscillations in 1 sec = v]
   [∵ no.of oscillations in T sec = vt]
5. No.of vibrations, made by the second wave train in T seconds = v₂T
6. During the time interval T, the first wave train would have completed one vibration more than the second wave train.
7. Hence, v₁T – v₂T = 1 or v₁ – v₂ = [latex]\frac{1}{\mathrm{~T}}[/latex]
8. Since, T is the beat period, no.of beats per seconds = [latex]\frac{1}{\mathrm{~T}}[/latex]
9. Hence the beat frequency = [latex]\frac{1}{\mathrm{~T}}[/latex] = v₁ – v₂ = Δv
10. That is the beat frequency is the difference between the frequencies of the two wave trains.

Practical applications of beats:

1. Determination of an unknown frequency: Out of two tuning forks, one is loaded with wax and the other is filed. The excited tuning forks are close together and no.of beats can be heard. Then after unknown frequencies of them will be found practically.
2. For tuning musical instruments : Musicians use the beat phenomenon in tuning their musical instruments.
3. For producing colourful effects in music: Sometimes, a rapid succession of beats is knowingly introduced in music. This produces an effect similar to that of human voice and is appreciated by the audience.
4. For detection of Marsh gas (dangerous gases) in mines.

Question 5.
What is Doppler effect? Obtain an expression for the apparent frequencý of sound heard when the source is in motion with respect to an observer at rest. (A.P. Mar. ’16, Mar. ’14)
Answer:
Doppler effect : The apparent change in the frequency heard by the observer due to the relative motion between the observer and the source of sound is called doppler effect.

When a whistling railway engine approaches an observer standing on the platform, the frequency of sound appears to increase. When it moves away the frequency appear to decrease.

Expression for apparent frequency when source is in motion and listener at rest:
Let S = Source of sound
O = listener

Let ‘S be the source, moving with a velocity ‘υ_s‘ towards the stationary listener.
The distance travelled by the source in time period T’ = υ_s. T
Therefore the successive compressions and rarefactions are drawn closer to listener.
∴ Apparent wavelength = λ’ = λ – υ_sT.

If ‘v’ “is apparent frequency heard by the listener
then v’ = [latex]\frac{v}{\lambda^{\prime}}[/latex] where ‘υ’ is velocity of sound in air
v’ = [latex]\frac{v . V}{v-v_s}[/latex]
Therefore the apparent frequency is greater than the actual frequency.
Similarly, if the source is away from the stationary listener then apparent frequency
v’ = [latex]\frac{v . V}{v+v_s}[/latex], which is less than the actual frequency.

Limitation : Doppler effect is applicable when the velocities of the source and listener are much less than that of sound velocity

Question 6.
What is Doppler shift? Obtain an expression for the apparent frequency of sound heard when the observer Is In motion with respect to a source át rest.
Answer:
Doppler Shift: Due to the relative motion, when the source comes closer to listener, the apparent frequency is greater than actual frequency and source away from listener; the apparent frequency is less than actual frequency So the difference in apparent and actual frequencies is known as Doppler shift.
Expression for the apparent frequency heard by a moving observer:

Case (1) : When observer Is moving towards source:
Let ‘υ₀’ be velocity of listener ‘O’, moving towards the stationary source ‘s’ as shown in figure. So observer will receive more number of waves in each second.
The distance travelled by observer in one second = υ₀

The number of extra waves received by the observer = [latex]\frac{v_0}{\lambda}[/latex]
We know v = vλ ⇒ λ = [latex]\frac{v}{v}[/latex]

Where υ = Velocity of sound
v = Frequency of sound
If ‘v’ is apparent frequency heard by him then

Therefore the apparent frequency is greater than actual frequency.

Case (2) : When observer Is moving away from rest source:
If the observer is moving away from the stationary source then he loses the number of waves [latex]\frac{v_0}{\lambda}[/latex]
∴ Apparent frequency v’ = v – [latex]\frac{v_0}{\lambda}[/latex] = v – [latex]\frac{v_0 \cdot v}{v}[/latex]
v’ = [latex]\left[\frac{v-v_0}{v}\right] \cdot v[/latex]
Hence the apparent frequency is less than actual frequency.

Problems

Question 1.
A stretched wire of length 0.6m is observed to vibrate with a frequency of 30Hz in the fundamental mode. If the string has a linear mass of 0.05 kg/m find
(a) the velocity of propagation of transverse waves in the string
(b) the tension in the string.
Solution:
v = 30 Hz; l = 0.6 m ; μ = 0.05 kg m^-1 υ = ? ; T = ?
a) υ = 2vl = 2 × 30 × 0.6 = 36 m/s
b) T = υ²μ = 36 × 36 × 0.05 = 64.8 N

Question 2.
A steel cable of diameter 3 cm is kept under a tension of 10kN. The density of steel is 7.8 g/cm³. With what speed would transverse waves propagate along the cable ?
Solution:
T = 10 kN = 10⁴ N
D = 3 cm; r = [latex]\frac{\mathrm{D}}{2}[/latex] = [latex]\frac{3}{2}[/latex]cm
= [latex]\frac{3}{2}[/latex] × 10^-2m;
A = πr² = [latex]\frac{22}{7} \times\left[\frac{3}{2} \times 10^{-2}\right]^2[/latex]

Question 3.
Two progressive transverse waves given by y₁ = 0.07 sinπ(12x-500t) and y₂ = 0.07 sinπ(12x + 500t) travelling along a stretched string from nodes and antinodes. What is the displacement at the (a) nodes (b) antinodes ? What is the wavelength of the standing wave ?
Solution:
A₁ = 0.07; A₂ = 0.07; K = 12π
a) At nodes, displacement
y = A₁ – A₂ = 0.07 – 0.07 = 0
b) At antinodes, displacement
y = A₁ + A₂ = 0.07 + 0.07 = 0.14 m
c) Wavelength λ = [latex]\frac{2 \pi}{\mathrm{K}}=\frac{2 \pi}{12 \pi}[/latex] = 0.16 m

Question 4.
A string has a length of 0.4m and a mass of 0.16g. If the tension in the string is 70N, what are the three lowest frequencies it produces when plucked ?
Solution:
I = 0.4 m; M = 0.16g = 0.16 × 10^-3 kg;

Question 5.
A metal bar when clamped at its centre resonantes in its fundamental frequency with longitudinal waves of frequency 4 kHz. If the clamp is moved to one end. What will be its fundamental resonance frequency ?
Solution:
When a metal bar of length l is clamped in the middle, it has one node in the middle and two antinodes at its free ends. In the fundamental mode. l = [latex]\frac{\lambda}{2}[/latex] ⇒ λ = 21

In fundamental mode of frequency of bar
= frequency of wave = 4 kHz.
∴ v = [latex]\frac{\mathrm{v}}{\lambda}[/latex] = [latex]\frac{\mathrm{v}}{2l}[/latex] = 4kHz —– (1)
When clamp is moved to one end,
l = [latex]\frac{\lambda^{\prime}}{4}[/latex] ⇒ λ’ = 4l
∴ V¹ = [latex]\frac{\mathrm{v}}{\lambda}[/latex] = [latex]\frac{\mathrm{v}}{4 \mathrm{l}}[/latex] = [latex]\frac{4 \mathrm{kHz}}{2}[/latex] = 2kHz
[∵ from (1)]

Question 6.
A closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m/s, what is the fundamental frequency of vibration of the air column ?
Solution:
I = 70 cm = 70 × 10^-2m; v = 331 m/s ; v = ?
v = ?
v = [latex]\frac{v}{4 l}[/latex] = [latex]\frac{331}{4 \times 70 \times 10^{-2}}[/latex] = 118.2 Hz

Question 7.
A vertical tube is made to stand in water so that the water level can be adjusted. Sound waves of frequency 320 Hz are sent into the top of the tube. If standing waves are produced at two successive water levels of 20 cm and 73 cm, what is the speed of sound waves in the air in the tube ?
Solution:
v = 320 Hz; l₁ = 20cm = 20 × 10^-2
l₂ = 73 cm = 73 × 10^-2m; v = ?
v = 2v(l₂ – l₁)
= 2 × 320 (73 × 10^-2 – 20 × 10^-2)
∴ v = 339 m/s .

Question 8.
Two organ pipes of lengths 65cm and 70cm respectively, are sounded simultaneously. How many beats per second will be produced between the fundamental frequencies of the two pipes ? (Velocity of sound = 330 m/s).
Solution:
l₁ = 65 cm = 0.65 m
₂ = 70 cm = 0.7 m
v = 330 m/s
No. of beats per second ∆υ = υ₁ – υ₂
= [latex]\frac{v}{2 h}[/latex] – [latex]\frac{\mathrm{v}}{2 l_2}[/latex] = [latex]\frac{330}{2 \times 0.65}[/latex] – [latex]\frac{330}{2 \times 0.7}[/latex]
∴ ∆v = 253.8 – 235.8 = 18Hz

Question 9.
A train sounds its whistle as it approaches and crosses a level crossing. An observer at the crossing measures a frequency of 219 Hz as the train, approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m/s, find the speed of the train and the frequency of its whistle.
Solution:
When a whistling train approaches to rest observer,

v’ = [latex]\left[\frac{v}{v-v_s}\right] v[/latex] ——– (1)
When a whistling train away from rest observer

v” = [latex]\left[\frac{v}{v+v_{\mathrm{S}}}\right] v[/latex] —— (2)
Here v’ = 219 Hz; V” = 184 Hz;
v = 340 m/s
[latex]\frac{(1)}{(2)}[/latex] ⇒ [latex]\frac{v^{\prime}}{v^{\prime \prime}}[/latex] = [latex]\frac{\left(v+v_s\right)}{\left(v-v_s\right)}[/latex]
[latex]\frac{219}{184}[/latex] = [latex]\frac{340+v_{\mathrm{s}}}{340-v_{\mathrm{s}}}[/latex]
219(340 – υ_s) = 184(340 + υ_s)
219 × 340 – 219 υ_s = 184 × 340 + 184 υ_s
403 υ_s = 35 × 340
∴ Velocity of train υ_s = 29.5 m/s
Frequency of whistle, v = v’ × [latex]\left[\frac{v-v_{\mathrm{S}}}{v}\right][/latex]
= 219 × [latex]\left[\frac{340-29.5}{340}\right][/latex]
= 199.98
∴ v = 200 Hz.

Question 10.
Two trucks heading in opposite directions with speeds of 60 kmph and 70 kmph respectively, approach each other. The driver of the first truck sounds his horn of frequency 400Hz. What frequency does the driver of the second truck hear ? (Velocity of sound = 330 m/ s). After the two trucks have passed each other, what frequency does the driver of the second truck hear ?
Solution:
v_s = 60 kmph = 60 × [latex]\frac{5}{18}[/latex] m/s = [latex]\frac{300}{18}[/latex] m/s
v₀ = 70 kmph = 70 × [latex]\frac{5}{18}[/latex] m/s = [latex]\frac{350}{18}[/latex] m/s
v = 400 Hz
When two trucks approach each other

When two trucks crossed each other,

Textual Exercises

Question 1.
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is caused at one end of the string, how long does the disturbance take to reach the other end ?
Answer:
Here M = 2.50 kg, T = 200 N, l = 20.0M
Mass per unit length; μ = [latex]\frac{\mathrm{M}}{l}[/latex] = [latex]\frac{2.5}{20.0}[/latex]
= 0.125 kg/m
Velocity V = [latex]\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{200}{0.125}}[/latex] = 40 m/s
Time taken by disturbance to reach the other end
t = [latex]\frac{l}{\mathrm{~V}}[/latex] = [latex]\frac{20}{40}[/latex] = 0.5s.

Question 2.
A stone dropped from the top of a tower of height 300 m high splashes into the pond of water near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s^-1, (g = 9.8m s^-2)
Answer:
Here, h = 300m, g = 9.8 m/s², V = 340 m/s. If t₁ = time taken by stone to strike the surface of water in the pond, then from
S = ut + [latex]\frac{1}{2}[/latex] at²
300 = 0 + [latex]\frac{1}{2}[/latex] × 9.8 [latex]\mathrm{t}_1^2[/latex]
t₁ = [latex]\sqrt{\frac{300}{4.9}}[/latex] = 7.82s.
Time taken by sound to reach the top of tower t₂ = [latex]\frac{\mathrm{h}}{\mathrm{v}}[/latex]
= [latex]\frac{300}{400}[/latex] = 0.88s
Total time after which splash of sound is heard = t₁ + t₂ = 7.82 + 0.88 = 8.70s

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20° C = 343 m s^-1.
Answer:
Here, l = 12.0M, M = 2.10kg, T = ?
V = 343 m/s
Mass per unit length μ = [latex]\frac{\mathrm{M}}{l}[/latex] = [latex]\frac{2.10}{12.0}[/latex]
= 0.175 kg/m
As V = [latex]\sqrt{\frac{\mathrm{T}}{\mu}}[/latex]
T = V² . μ = (343)² × 0.175 = 2.06 × 10⁴N.

Question 4.
Use the formula υ = [latex]\sqrt{\frac{\gamma P}{\rho}}[/latex] to explain why the speed of sound in air
a) is independent of pressure,
b) increases with temperature,
c) increases with humidity.
Answer:
a) Effect of pressure:
The speed of sound in gases, υ = [latex]\sqrt{\frac{\gamma \mathrm{P}}{\rho}}[/latex]
At constant temperature, PV = constant
P[latex]\frac{m}{\rho}[/latex] = constant ⇒ [latex]\frac{\mathrm{P}}{\rho}[/latex] = constant
If P increases, ρ also increases. Hence speed of sound in air is independent of pressure.

b) Effect of temperature:
As PV = nRT, P[latex]\frac{\mathrm{m}}{\rho}[/latex] = [latex]\frac{m}{M} R T[/latex]
⇒ [latex]\frac{P}{\rho}[/latex] = [latex]\frac{\mathrm{RT}}{\mathrm{M}}[/latex]
∴ υ = [latex]\sqrt{\frac{R T}{M}}[/latex]
Since R, M are constants υ ∝ [latex]\sqrt{\mathrm{T}}[/latex]
∴ Velocity of sound in air depends on temperature.

c) Effect of humidity:
As υ = [latex]\sqrt{\frac{\gamma \mathrm{P}}{\rho}}[/latex] ∴ υ ∝ [latex]\frac{1}{\sqrt{\rho}}[/latex]
As the density of water vapour is less than density of dry air at STP. So the presence of moisture in air decreases the
density of air. Since the speed of sound is inversely proportional to the square root of density. So sound travels faster in moist air than dry air. Hence velocity of sound
V ∝ humidity

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f(x, t) where x and t must appear in the combination x – υt or x + υt, i.e. y = f(x ± υ t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:
Answer:
No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that for all values of x & t, wave function must have a finite value.

Out of the given functions y, no one satisfies this condition therefore, none can represent a travelling wave.

Question 6.
A bat emits ultrasonic sound a frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of
(a) the reflected sound,
(b) the transmitted sound?
Speed of sound in air is 340 m s^-1 and in water 1486 m s^-1.
Answer:
Here V = 100 KHz = 10⁵Hz, V_a = 340m/s, V_w = 1486 m/s^-1
Wavelength of reflected sound, λ_a = [latex]\frac{\mathrm{V}^{\mathrm{a}}}{\mathrm{V}}[/latex]
= [latex]\frac{340}{10^5}[/latex] = 3.4 × 10^-3 m
Wavelength of transmitted sound,
λ_w = [latex]\frac{V_w}{V}[/latex] = [latex]\frac{1486}{10^5}[/latex] = 1.486 × 10^-2 m

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1? The operating frequency of the scanner is 4.2 MHz.
Answer:
λ = ? υ = 1.7 Km/s = 1700 ms^-1
y = 4.2 MHz = 4.2 × 10⁶Hz
λ = [latex]\frac{v}{v}[/latex] = [latex]\frac{1700}{4.2 \times 10^6} \mathrm{~m}[/latex] = 0.405 × 10^-3 m
= 0.405 mm

Question 8.
A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right.
a) Is this a travelling wave or a stationary wave?
If it is travelling, what are the speed and direction of its propagation?
b) What are Its amplitude and frequency?
C) What is the initial phase at the origin?
d) What is the least distance between two successive crests in the wave?
Answer:
Compare the given equation with that of plane progressive wave of amplitude r, travelling with a velocity V from right to left.
y(x, t) = rsin[latex]\left[\frac{2 \pi}{\lambda}(v t+x)+\phi_0\right][/latex] ……… (1)
We find that
a) The given equation represents a transvërse harmonic wave travelling from right to left. It is ñot a stationary wave.

b) The given equation can be written as
Y(x, t) = 3.0sin[0.018([latex]\frac{36}{0.018}[/latex] + x) + [latex]\frac{\pi}{4}[/latex]] ……… (2)
equating coefficient of t in the two
(1) & (2) we get. :
V = [latex]\frac{.36}{0.018}[/latex] = 2000 cm/sec.
Obviously, r = 3.0 cm
Also, [latex]\frac{2 \pi}{\lambda}[/latex] = 0.018
λ = [latex]\frac{2 \pi}{0.018} \mathrm{~cm}[/latex]
Frequency, v = [latex]\frac{v}{\lambda}[/latex] = [latex]\frac{2000}{2 \pi}[/latex] × 0.018
= 5.7351.

c) Intial phase, φ₀ = [latex]\frac{\pi}{4}[/latex] radian.

d) Least distance between two successive crests of the wave =
Wave length, λ = [latex]\frac{2 \pi}{0.018 \mathrm{~cm}}[/latex] = 349 cm,

Question 9.
For the wave described in the last problem plot the displacement (y) versus (t) graphs for x = 0.2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Answer:
The transverse harmonic wave is
y(x,t) = 3.0 sin[36t + 0.018x + [latex]\frac{\pi}{4}[/latex]]
For x = 0
y(0, t) = 3.0 sin(36t + [latex]\frac{\pi}{4}[/latex]) —— (i)
Here w = [latex]\frac{2 \pi}{T}[/latex] = 36, T = [latex]\frac{2 \pi}{36}[/latex] = [latex]\frac{\pi}{18}[/latex]-sec.

For different values of t, we calculate y using eq(i). These values are tabulated below.

On plotting y versus t graph, we obtain a sinusoidal curve as shown in fig.

Similar graphs are obtained for x = 2cm & x = 4cm. The oscillary motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three areas.

Question 10.
For the travelling harmonic wave
y(x, t) = 2.0 cos 2 π (10t – 0.0080 x + 0.35)
where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
a) 4 m,
b) 0.5 m,
c) λ/2,
d) 3λ/4,
Answer:
The given equation be written as
y = 2.0 cos[2π(10t – 0.0080x) + 2π × 0.35]
y = 2.0 cos[2π × (0.0080([latex]\frac{10 \mathrm{t}}{0.0080}[/latex] – x) + 0.7π]
Compare it with the standard equation of a travelling harmonic, we have
y = r.cos[[latex]\frac{2 \pi}{\lambda}(v t-x)+\phi_0[/latex]
We get, [latex]\frac{2 \pi}{\lambda}[/latex] = 2π × 0.0080
Further we know that phase diff. φ = [latex]\frac{2 \pi}{\lambda} \mathrm{x}[/latex]
a) When x = 4m = 400 cm
φ = [latex]\frac{2 \pi}{\lambda}[/latex] . x = 2π × 0.0080 × 400
= 6.4 π rad.

b) When x = 0.5 = 50 cm
φ = [latex]\frac{2 \pi}{\lambda}[/latex] . x = 2π × 0.0080 × 50
= 0.8π rad.

c) When x = [latex]\frac{\lambda}{2}[/latex]
φ = [latex]\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}[/latex] = π rad.

d) When x = [latex]\frac{3 \lambda}{4}[/latex]
φ = [latex]\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4}[/latex] = [latex]\frac{3 \lambda}{2} \mathrm{rad}[/latex]

Question 11.
The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin[latex]\left(\frac{2 \pi}{3} x\right)[/latex]cos (120 πt)
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10^-2 kg
Answer the following:
a) Does the function represent a travelling wave or a stationary wave?
b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
c) Determine the tension in the string.
Answer:
The given equation is
y(x, t) = 0.06 sin [latex]\frac{2 \pi}{3} \mathrm{x} \cos 120 \pi \mathrm{t}[/latex]

a) As the equation involves harmonic functions of x and t seperately, it represents a stationary wave.

b) We know that when a wave pulse
y₁ = r sin [latex]\frac{2 \pi}{\lambda}(v t+x)[/latex] —– (i)
travelling along + direction of x-axis is super imposed by the reflected wave
y = y₁ + y₂ = -2rsin[latex]\frac{2 \pi}{\lambda}[/latex] xcos [latex]\frac{2 \pi}{\lambda}[/latex] vt is formed. ——- (ii)
Comparing (i) & (ii) we find that
[latex]\frac{2 \pi}{\lambda}[/latex] = [latex]\frac{2 \pi}{3}[/latex] ⇒ λ = 3m.
Also [latex]\frac{2 \pi}{\lambda} v[/latex] = 120π (Or)
V = 60λ = 60 × 3 = 180m/s.
frequency, v = [latex]\frac{v}{\lambda}[/latex] = [latex]\frac{180}{3}[/latex] = 60 hertz.
Note that both the waves have same wave length, same frequency and same speed.

c) Velocity of transverse waves is
υ = [latex]\sqrt{\frac{T}{\mu}}[/latex] (or) υ² = T/μ
T = V² × μ where μ = [latex]\frac{3 \times 10^{-2}}{1.5}[/latex]
= 2 × 10^-2 kg/m
T = (180)² × 2 × 10^-2 = 648 N.

Question 12.
i) For the wave on a string described in previous problem do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end?
Answer:
i) All the points on the string
a) have the same frequency except at the nodes (where frequency is cos θ),
b) have the same phase every where in one loop except at the nodes,
c) however, the amplitude of vibration at different points is different.

ii) From y(x, t) = 0.06 sin[latex]\left(\frac{2 \pi}{3} x\right)[/latex] cos (120 πt)
The amplitude at x = 0.375 m is 0.06
sin [latex]\frac{2 \pi}{3} x \times 1[/latex] = 0.06 × sin [latex]\frac{2 \pi}{3} \times 0.375[/latex]
= 0.06sin[latex]\frac{\pi}{4}[/latex] = [latex]\frac{0.06}{\sqrt{2}}[/latex] = 0.042 M

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave State which of these represent
(i) a travelling wave,
(ii) a stationary wave or
(iii) none at all:
a) y = 2 cos (3x) sin (10t)
b) y = [latex]2 \sqrt{x-v t}[/latex]
c) y = 3 sin(5x – 0.5t) + 4 cos(5x – 0.5t)
d) y = cos x sin t + cos 2x sin 2t
Answer:
a) It represents a stationary wave as harmonic functions of x & t are contained separetely in the equation.
b) It cannot represent any type of wave.
c) It represents a progressive / travelling harmonic wave.
d) This equation is sum of two functions each representing a stationary wave. Therefore it represents superposition of two stationary waves.

Question 14.
A wire stretched between two right supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10^-2 kg and its linear mass density is 4.0 × 10^-2 kg m^-1. What is (a) the speed of transverse wave on the string, and (b) the tension in the string?
Answer:
Here, v = 45Hz, μ = 3.5 × 10^-2 kg
Mass/length = μ = 4.0 × 10^-2 kg/m
l = [latex]\frac{\mu}{\mu}[/latex] = [latex]\frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}}[/latex] = [latex]\frac{7}{8}[/latex]
As [latex]\frac{\lambda}{2}[/latex] = l = [latex]\frac{7}{8}[/latex] ∴ λ = [latex]\frac{7}{4}[/latex]m = 1.75m
a) The speed of transverse wave
υ = vλ = 45 × 1.75 = 78.75 m/s.

b) As υ = [latex]\sqrt{\frac{\mathrm{T}}{\mu}}[/latex]
∴ T = υ² × μ = (78.75)² × 4.0 × 10^-2
= 248.06 N.

Question 15.
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length Is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.
Answer:
As there is a piston at one end of the tube, it behaves as a closed organ pipe, which produces odd harmonics only. Therefore the pipe is in resonance with the fundamental note at the third harmonic (79.3 cm is about 3 times 25.5 cm)
In the fundamental note = [latex]\frac{\lambda}{4}[/latex] = l₁ = 25.5
λ = 4 × 25.5 = 102 cm = 1.02m
Speed of sound in air.
υ = vλ = 340 × 1.02
= 346.0 m/s

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
Answer:
Here, l = 100 cm = 1 m, y = 2.53 KHz
= 2.53 × 10³ Hz
When the rod is clamped at the middle, then in the fundamental mode of vibration of the rod, anode is formed at the middle and antinode is formed at each end.
Therefore, it is clear from fig

l = [latex]\frac{\lambda}{4}[/latex] + [latex]\frac{\lambda}{4}[/latex] + [latex]\frac{\lambda}{2}[/latex]
λ = 2l = 2m
As v = λl
v = 2.53 × 10³ × 2
= 5.06 × 10³ ms^-1

Question 17.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s^-1).
Answer:
Here l = 20 cm = 0.2m, v_n = 430 Hz,
υ = 340 m/s
The frequency of n^th normal mode of vibration of closed pipe is
v_n = (2n – 1)[latex]\frac{v}{4l}[/latex]
∴ 430 = (2n – 1)[latex]\frac{340}{4 \times 0.2}[/latex]
2n – 1 = [latex]\frac{430 \times 4 \times 0.2}{340}[/latex] = 1.02
2n = 2.02, n = 1.01
Hence it will be the 1^st normal mode of vibration. In a pipe, open at both ends we have
v_n = n × [latex]\frac{\mathrm{v}}{2l}[/latex] = [latex]\frac{\mathrm{n} \times 340}{2 \times 0.2}[/latex] = 430.
∴ n = [latex]\frac{430 \times 2 \times 0.2}{340}[/latex] = 0.5
As n has to be an integer, therefore open organ pipe cannot be in resonance with the source.

Question 18.
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324Hz. What is the frequency of B?
Answer:
Let original frequency of sitar string A be n_a & original frequeny of sitar string B be n_b.
As number of beats / sec = 6
∴ n_b = n_a ± 6 = 330 (or) 318Hz.
When tension in A is reduced, its frequency reduces (∴ n ∝ [latex]\sqrt{T}[/latex])
As number of beats/sec decreases to 3 therefore, frequency of B = 324 – 6
= 318Hz.

Question 19.
Explain why (or how):
a) in a sound wave, a displacement node is a pressure antinode and vice versa,
b) bats can ascertain distances, directions, nature and sizes of the obstacles without any ‘eyes”.
c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
d) Soils can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases and
e) The shape of a pulse gets distorted during propagation In a dispersive medium.
Answer:
a) Node (N) is a point where the amplitude of oscillation is 0. (and pressure is maximum)
Antinode (A) is a point where the amplitude of oscillation is maximum (and pressure is min).
These nodes & antinodes do not coincide with pressure nodes & antinodes.
Infact, N coincides with pressure antinode and A coincides with pressure node, as is clear from the definitions.

b) Bats emit ultrasonic wave of large frequencies, when these waveš are reflected from the obstacles in their path,
they give them the idea about the distance, direction, size & nature of the obstacle.

c) Though the violin note and sitar note have the same frequency, yet the over tones produced and their reactive strengths are different in the two flotes that is why we can distinguish between the two notes.

d) This is because solids have both, the elasticity of volume and elasticity of shape where as gases have only the volume elasticity.

e) A sound pulse is a combination of waves of different wavelengths. As waves of different wavelengths travel in a disperse medium with different velocities, therefore the shape of the pulse gets distorted.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.
i) What is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10ms^-1,
(b) recedes from the platform with a speed of 10 m s^-1 ?
ii) What is the speed of sound in each case ? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
i) Here, y = 400 Hz, υ = 340 m/s
a) Train approaches the platform
υ_s = 10m/s
v’ = [latex]\frac{v}{v-v_s}[/latex] = [latex]\frac{340 \times 400}{340-10}[/latex] = 412.12 Hz.

b) Train recedes from the platform
υ_s = 10m/s
v’ = [latex]\frac{v \times v}{v \times v_s}[/latex] = [latex]\frac{340 \times 400}{340+10}[/latex]
= 388.6Hz

ii) The speed of sound in each case is the same = 340 m/s

Question 21.
A train, stañdingin a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10m s^-1. What are the frequency, wavelength and speed of sound for an observer standing on the station’s platform ? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s^-1 ? The speed of sound in still air can be taken as 340 m^-1
Answer:
Here y = 400 Hz, υ_m = 10ms^-1, υ = 340m/s
As the wind is blowing in the direction of sound, therefore effective speed of sound
= υ + υ_m = 340 + 10 = 350m/s
As the source & Iistner both are at rest, therefore, frequency remains unchanged
i.e. v = 400 Hz.
Wavelength, λ = [latex]\frac{v+v_m}{v}[/latex] = [latex]\frac{350}{400}[/latex]
= 0.875 M.
When air is still, υ_m = 0
Speed of observer υ₁ = 10m/s υ_s = 0
As observer moves toward the source
υ’ = [latex]\frac{\left(v+v_l\right)}{v} \times v[/latex] = [latex]\frac{(340+10)}{340} \times 400[/latex]
= 411.76 Hz.
As source is at rest, wavelength does not change
i.e, λ’ = λ = 0.875M.
Also, speed of sound = υ + υ_m = 340 + 0
= 340 m/s
The situations in the two cases are entirely different.

Additional Exercises

Question 1.
A travelling harmonic wave on a string is described by
y(x, t) = 7.5 sin (0.0050x + 12t + π/4)
a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1cm
point at t = 2 s, 5 s and 11 s.
Answer:
a) The travelling harmonic wave is y(x, t)

From (1), y(1, 1) = 7.5 sin(732.55°)
= 7.5 sin (720 + 12.55°)
= 7.5 sin 12.55° = 7.5 × 0.2173 = 1.63 cm
velocity of oscillation, v = [latex]\frac{d y}{d t} y(1,1)[/latex]
= [latex]\frac{\mathrm{d}}{\mathrm{dt}}[/latex][7.5 sin (0.005x + 12t + [latex]\frac{\pi}{4}[/latex])
= 7.5 × 12 cos (0.005x + 12t + [latex]\frac{\pi}{4}[/latex])
At x = 1 cm, t = 1 sec
= 7.5 × 12 cos (o.oo5 + 12 + [latex]\frac{\pi}{4}[/latex])
= 90 cos (732.55°).
= 90 cos(720 + 12.55°)
= 90 cos (12.55°)
= 90 × 0.9765
= 87.89 cm/s.
Comparing the given equation with the standard form
y(x, t) = r sin[latex]\left[\frac{2 \pi}{\lambda}(v t+x)+\phi_0\right][/latex]
We get r = 7.5 cm, [latex]\frac{2 \pi v}{\lambda}[/latex] = 12 (or) 2πV = 12
V = [latex]\frac{6}{\pi}[/latex]
2[latex]\frac{\pi}{\lambda}[/latex] = 0.005
∴ λ = [latex]\frac{2 \pi}{0.005}[/latex] = [latex]\frac{2 \times 3.14}{0.005}[/latex] = 1256 cm
= 12.56 m.
Velocity of wave propagation, υ = Vλ
= [latex]\frac{6}{\pi}[/latex] × 12.56 .
= 24 m/s.
We find that velocity at x = 1 cm t = 1 sec is not equal to velocity of wave propagation.

b) Now, all points which are at a distance of ±λ, ± 2λ, ± 3λ from x = 1 cm will have same transverse displacement and velocity. As λ = 12.56 m, therefore, all points at distances ± 12.6m, ± 25.2 m displacement and velocity As λ = 12.56m, therefore all points at distances ± 12.6m, ± 25.2m, ± 37.8m from x = 11m will have same displacement & velocity at x = 1 cm point at t = 25.55 & 115s.

Question 2.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (I) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s (that is the whistle blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:
a) A short pip by a whistle has neither a definite wavelength nor a definite frequency. However its speed of propagation is fixed, being equal to speed of sound in air.

b) No, frequency of the note produced by whistle is not 1/20 = 0.05 Hz. Rather 0.05 Hz is the frequency of repetition of the short pipe of the whistle.

Question 3.
One end of a long string of linear mass density 8.0 × 10^-3 kg m^-1 is connected to an electrically driven tuniúg fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the Incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
Answer:
Here, μ = 8.0 × 10^-3 kg/m, y = 256 Hz, T = 90kg = 90 × 9.8 = 882N.
Amplitude of wave, r = 5.0 1m = 0.05m.
As the wave propagation along the string is a transverse travelling wave, the velocity of the wave is given by

As the wave is propagating along x direction, the equation of the wave is
y(x, t) = r sin (ωt – kx)
= 0.05 sin (1.61 × 10³t – 4.84x)
Here x, y are in mt & t in sec

Question 4.
A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine ? Take the speed of sound in water to be 1450 m s^-1.
Answer:
Here, frequency of SONAR,
v = 40 KHz = 40 × 10³ Hz.
Speed of observer / enemy’s submarine
υ₁ = 360km/h .
= 360 × [latex]\frac{5}{18}[/latex]m/s = 100m/s.
Speed of sound wave in water; υ = 1450 m/s.
As the source is at rest & observer is moving towards the source, therefore, apparent frequency received by enemy submarine
v’ = [latex]\frac{\left(v+v_1\right) v}{v}[/latex]
= [latex]\frac{(1450+100) 40 \times 10^3 \mathrm{~Hz}}{1450}[/latex]
= 4.27 × 10⁴ Hz.
This frequency is reflected by enemy submarine (source) and is observed by SONAR. Therefore in this case,
υ_s = 360 km/s = 100 m/s, υ₁ = 0
∴ Apparent frequency, v¹¹ = [latex]\frac{v \times v}{v_i-v_s}[/latex]
= [latex]\frac{1450 \times 4.276 \times 10^4}{1450-10}[/latex]
= 4.59 × 10⁴ Hz = 45.9 Hz.

Question 5.
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s^-1 and that of P wave is 8.0 km s^-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
Answer:
Let υ₁, υ₂ be the velocity of S waves & P waves & t₁, t₂ be the time taken by these waves to travel to the position of seismograph. If l is the distance of occurrence of earth quake from the seismograph, then
l = υ₁t₁ = υ₂t₂ ——- (i)
now υ₁ = 4 km/s & υ₂ = 8 km/s .
∴ 4t₁ = 8t₂ (or) t₁ = 2t₂ ——- (ii)
Also t₁ – t₂ = 4min = 240s.
using (iii), 2t₂ – t₂ = 240s, t₂ = 240s
t₁ = 2t₂ = 2 × 240 = 480s.
Now from (i) l = υ₁t₁ = 4 × 480 = 1920 km.
Hence earthquake occurs 1920 km away from the seismograph.

Question 6.
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Here, the frequency of sound emitted by the bat, v = 40 kHz.
velocity of bat, υ_s = 0.03υ, where υ is velocity of sound.
Apparent frequency of sound striking the wall.
v’ = [latex]\frac{v \times v}{v-v_s}[/latex] = [latex]\frac{v}{v-0.03 v}[/latex] × 40 kHxz
= [latex]\frac{40}{0.97}[/latex] kHZ.
This frequency is reflected by the wall & is received by the bat moving towards the wall, So υ_s = 0.
υ₁ = 0.03 υ
v’ = [latex]\frac{\left(v+v_1\right) v^{\prime}}{v}[/latex] = [latex]\frac{(v+0.03 v)}{v}\left(\frac{40}{0.97}\right)[/latex]
= [latex]\frac{1.03}{0.97} \times 40 \mathrm{kHz}[/latex]
= 42.47 kHz