Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 9 - Ray Optics And Optical Instruments

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Page No 54:

Question 1:

A ray of light incident at an angle θ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of refractive index 1.5, the angle of incidence is

(a) 7.5°.
(b) 5°.
(c) 15°.
(d) 2.5°.

Answer:

Given:
Angle of prism, A = 5°
Refractive index, μ = 1.5
Angle of emergence, e = 0°

In the above shown diagram, angle of incidence is i and angle of refraction is r.
Angle of prism, $A = r + 0 = r$
So, $r = 5^{o}$
Using Snell's law,
$1 \sin i = μ \sin r \Rightarrow θ = i = \sin^{- 1} (μ \sin r) = \sin^{- 1} (1.5 \times \sin 5^{ο}) = \sin^{- 1} (1.5 \times \sin 5^{ο}) = \sin^{- 1} (1.5 \times 0.087) = \sin^{- 1} (0.13) = 7 . 5^{ο}$

Hence, the correct answer is option (a).

Page No 54:

Question 2:

A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

(a) blue.
(b) green.
(c) violet.
(d) red.

Answer:

If the wavelength of the light is $λ$ and the velocity is $v$ , then $v \propto λ$ .
Since, the light of red colour is of highest wavelength and therefore of highest speed. Therefore, after travelling through the slab, the red colour emerge first.

Hence, the correct answer is option (d).â€‹

Page No 55:

Question 3:

An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image

(a) moves away from the lens with an uniform speed 5 m/s.
(b) moves away from the lens with an uniform acceleration.
(c) moves away from the lens with a non-uniform acceleration.
(d) moves towards the lens with a non-uniform acceleration.

Answer:

When object approaches towards the convex lens, the image keeps moving away from the lens.
Hence, when an object approaches a convergent lens from the left of the lens with a uniform speed of 5 m/s, the image away from the lens with a non-uniform acceleration.
Hence, the correct answer is option (c).

Page No 55:

Question 4:

A passenger in an aeroplane shall

(a) never see a rainbow.
(b) may see a primary and a secondary rainbow as concentric circles.
(c) may see a primary and a secondary rainbow as concentric arcs.
(d) shall never see a secondary rainbow.

Answer:

A passenger in an aeroplane may see both primary and secondary rainbow. Since there is no ground ahead of him, the rainbow appears as a full circle.

Hence, the correct answer is option (b).

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Question 5:

You are given four sources of light each one providing a light of a single colour – red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?

(a) The beam of red light would undergo total internal reflection.
(b) The beam of red light would bend towards normal while it gets refracted through the second medium.
(c) The beam of blue light would undergo total internal reflection.
(d) The beam of green light would bend away from the normal as it gets refracted through the second medium.

Answer:

According to Cauchy relationship, $μ \propto \frac{1}{λ}$
But, $μ \propto \frac{1}{\sin i_{c}}$
So, $λ \propto \sin i_{c}$
If the angle of refraction for yellow colour is $90^{o}$ .
So, corresponding to blue colour, the critical angle is least which facilitates total internal reflection for the beam of blue light. The beam of green light would also undergo total internal reflection.

Hence, the correct answer is option (c).

Page No 55:

Question 6:

The radius of curvature of the curved surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will

(a) act as a convex lens only for the objects that lie on its curved side.
(b) act as a concave lens for the objects that lie on its curved side.
(c) act as a convex lens irrespective of the side on which the object lies.
(d) act as a concave lens irrespective of side on which the object lies.

Answer:

Given:
R₁ = 0 (for plane surface)
R₂= −20 cm (for curved surface)
Refractive index, $μ$ = 1.5
From lens maker's formula:

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (1.5 - 1) (\frac{1}{\infty} - \frac{1}{- 20}) \Rightarrow f = + 40 cm$
Here, f > 0.
It signifies that the lens is converging in nature.
Therefore, lens acts as a convex lens irrespective of the side on which the object lies.
Hence, the correct answer is option (c).

Page No 55:

Question 7:

The phenomena involved in the reflection of radiowaves by ionosphere is similar to

(a) reflection of light by a plane mirror.
(b) total internal reflection of light in air during a mirage.
(c) dispersion of light by water molecules during the formation of a rainbow.
(d) scattering of light by the particles of air.

Answer:

The phenomenon involved in the reflection of radiowaves by ionosphere is similar to total internal reflection of light in air during a mirage i.e., angle of incidence is greater than critical angle.

â€‹Hence, the correct answer is option (b).

Page No 56:

Question 8:

The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (Fig 9.1). Which of the four rays correctly shows the direction of reflected ray?

(a) 1
(b) 2
(c) 3
(d) 4

Answer:

From the figure, it is clear that the incident ray passes through focus of the concave mirror before reflection.
Therefore, after reflection the ray shall become parallel to the principal axis.
Ray 2 represent the reflected ray parallel to the principle axis.

Hence, the correct answer is option (b).

Page No 56:

Question 9:

The optical density of turpentine is higher than that of water while its mass density is lower. Fig 9.2. shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in Fig 9.2, the path shown is correct?

(a) 1
(b) 2
(c) 3
(d) 4

Answer:

For the first interface(air - turpentine) the ray is travelling from rarer to denser medium. So, the ray shall bend towards the normal. That is true for ray 1 and 2 only.

For the second interface (turpentine - water), the ray is travelling from denser medium to rarer medium. Therefore the ray will bend away from the normal. Of ray 1 and ray 2, this condition is only true for ray 2.

Hence, the correct answer is option (b).

Page No 56:

Question 10:

A car is moving with at a constant speed of 60 km h^–1 on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of 5 km h ^–1. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct?

(a) The speed of the car in the rear is 65 km h^–1.
(b) In the side mirror the car in the rear would appear to approach with a speed of 5 km h^–1 to the driver of the leading car.
(c) In the rear view mirror the speed of the approaching car would appear to decrease as the distance between the cars decreases.
(d) In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases.

Answer:

When the speed of the approaching car would appear to increase as the distance between the cars decreases in the side mirror.
Hence, the correct answer is option (d).

Page No 57:

Question 11:

There are certain material developed in laboratories which have a negative refractive index (Fig. 9.3). A ray incident from air (medium 1) into such a medium (medium 2) shall follow a path given by
\

Answer:

The materials having negative refractive index are those in which incident ray from air (Medium 1) to them refract or bend differently to that of positive refractive index medium.

Hence, the correct answer is option (a).

Page No 57:

Question 12:

Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because

(a) the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.
(b) the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air.
(c) some of the points of the object far away from the edge may not be visible because of total internal reflection.
(d) water in a trough acts as a lens and magnifies the object.

Answer:

When immersed object is seen from close to the edge of the trough the object looks distorted because the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.

The angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air and some of the points of the object far away from the edge may not be visible because of total internal reflection.

Hence, the correct options are (a), (b) and (c).

Page No 57:

Question 13:

A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB (Fig. 9.4). When observed from the face AD, the pin shall

(a) appear to be near A.
(b) appear to be near D.
(c) appear to be at the centre of AD.
(d) not be seen at all.

Answer:

Consider the pin is placed at point 'L' between A and B.
When seen from face AD (so long as $i < i_{c}$ ) image of L appears to be at L', closer to A.
$\sin i_{c} = \frac{1}{μ} = \frac{1}{1.6}$
$\Rightarrow i_{c} = 38 . 7^{o}$
If $i > i_{c}$ , the ray coming from point L suffers total internal reflection. Now it is not visible at all.

Hence, the correct answers are option A and D.

Page No 57:

Question 14:

Between the primary and secondary rainbows, there is a dark band known as Alexandar’s dark band. This is because

(a) light scattered into this region interfere destructively.
(b) there is no light scattered into this region.
(c) light is absorbed in this region.
(d) angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately 42° and 50°.

Answer:

The dark bands are result of destructive interference between two light waves.
So, between the primary and secondary rainbows, there is a dark band known as Alexandar's dark band. This is because light scattered into this region interfere destructively.
Also, primary rainbow subtends an angle of 41^o− 43^o at the eye of the observer with respect to the incident light, and secondary rainbow subtends an angle of about 51^o − 54^o at the eye of the observer with respect to the incident light. Therefore, the region between the angles of 42° to 50° is dark.
Hence, the correct options are (a) and (d).

Page No 58:

Question 15:

A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in

(a) a larger angle to be subtended by the object at the eye and hence viewed in greater detail.
(b) the formation of a virtual erect image.
(c) increase in the field of view.
(d) infinite magnification at the near point.

Answer:

Magnifying glass literally means a convex lens is being used to get a magnified image of an object by placing the object between focus and optical centre of the lens.
When object is placed between optical centre and focus, the image formed is virtual and erect.

When the object's distance is increased, it subtends a large angle at the eye and hence the image viewed in greater detail.

Hence, the correct options are (a) and (b).

Page No 58:

Question 16:

An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length 2 cm.

(a) The length of the telescope tube is 20.02m.
(b) The magnification is 1000.
(c) The image formed is inverted.
(d) An objective of a larger aperture will increase the brightness and reduce chromatic aberration of the image.

Answer:

Focal length of objective lens, f_o = 20 m
Focal length of eyepiece, f_e = 2 cm = 0.02 m
The length of the telescope tube is:
f_o + f_e = 20 + (0.02) = 20.02 m
Also, m = $\frac{f_{o}}{f_{e}} = \frac{20}{0.02}$ = 1000
Also, the image formed is inverted.

Hence, the correct options are (a), (b) and (c).

Page No 58:

Question 17:

Will the focal length of a lens for red light be more, same or less than that for blue light?

Answer:

According to Cauchy's equation, the refractive index is inversely proportional to the wavelength of incident light.
Here, wavelength of red light $(λ_{R})$ > wavelength of blue light $(λ_{B})$
So, $μ_{B} > μ_{R}$
By lens maker's formula, $\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Therefore, focal length for red colour is greater than focal length of blue colour.

Page No 58:

Question 18:

The near vision of an average person is 25 cm. To view an object with an angular magnification of 10, what should be the power of the microscope?

Answer:

The least distance of distinct vision of an average person (i.e., D ) is 25 cm.
Here, v = D = 25 cm and u = f
But the magnification, m = $\frac{v}{u} = \frac{D}{f}$
$\Rightarrow f = \frac{D}{m} = \frac{0.25}{10} = 0.025 m$
Power of the lens, $P = \frac{1}{f} = 40 D$
â€‹Hence, the required power of the lens is 40 D.

Page No 58:

Question 19:

An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

Answer:

The position of the image will not change. It can be explained by lens maker's formula.
For normal conditions, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
When the lens is reversed, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} = - (μ - 1) (\frac{1}{R_{2}} - \frac{1}{R_{1}})$

On reversing the lens, the value of $R_{1}$ and $R_{2}$ reverse and also the signs reverses.
Hence, for a given position of object the image remains unaffected on reversing the lens.

Page No 58:

Question 20:

Three immiscible liquids of densities d₁ > d₂ > d₃and refractive indices μ₁ > μ₂ > μ₃ are put in a beaker. The height of each liquid column is $\frac{h}{3}$ . A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

Answer:

Let the apparent depth be O₁ for the object seen from m₂ , then $O_{1} = \frac{μ_{2}}{μ_{1}} . \frac{h}{3}$

Since, apparent depth = $\frac{real depth}{refractive index (μ)}$ .

Since, the image formed by Medium 1, O₂ act as an object for Medium 2.
If seen from µ₃ , the apparent depth is O₂ . Similarly, the image formed by Medium 2 , O₂ act as an object for Medium 3.
$\begin{array}{rcl} O_{2} & = & \frac{μ_{3}}{μ_{2}} (\frac{h}{3} + O_{1}) \\ = & \frac{μ_{3}}{μ_{2}} (\frac{h}{3} + \frac{μ_{2}}{μ_{1}} \frac{h}{3}) = \frac{h}{3} (\frac{μ_{3}}{μ_{2}} + \frac{μ_{2}}{μ_{1}}) \end{array}$
Seen from outside, the apparent height is
$\begin{array}{rcl} O_{3} & = & \frac{1}{μ_{3}} (\frac{h}{3} + O_{2}) = \frac{1}{μ_{3}} (\frac{h}{3} + \frac{h}{3} (\frac{μ_{3}}{μ_{2}} + \frac{μ_{3}}{μ_{1}})) \\ = & \frac{h}{3} (\frac{1}{μ_{1}} + \frac{1}{μ_{2}} + \frac{1}{μ_{3}}) \end{array}$
This is the required expression of apparent depth.

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Question 21:

For a glass prism (μ = $\sqrt{3}$ ) the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.

Answer:

Refractive index, $μ = \frac{\sin (\frac{A + δ_{m i n}}{2})}{\sin \frac{A}{2}}$
According to the question, $δ_{m i n} = A$
$\Rightarrow μ = \frac{\sin (\frac{A + A}{2})}{\sin \frac{A}{2}} = \frac{\sin A}{\sin \frac{A}{2}} \Rightarrow μ = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$
$\Rightarrow \frac{\sqrt{3}}{2} = \cos \frac{A}{2} \Rightarrow \cos \frac{π}{6} = \cos \frac{A}{2}$
Hence, the angle of prism is $\frac{π}{3} = 60^{o}$ .

Page No 59:

Question 22:

A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take L <<|v − f|.

Answer:

The object is at distance u.

Let us consider the two ends of the object be at distance $u_{1} = u - \frac{L}{2}$ and $u_{2} = u + \frac{L}{2}$ respectively.
So that $u_{1} - u_{2} = L$

Let the image of the two ends be formed at v₁and v₂ , respectively so that the image length would be
$L' = |v_{1} - v_{2}|$

Applying mirror formula, we have
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ or $v = \frac{f u}{u - f}$

For $v_{1} = f (\frac{u - \frac{L}{2}}{u - f - \frac{L}{2}}) or v_{2} = f (\frac{u + \frac{L}{2}}{u - f + \frac{L}{2}})$
But, $L' = |v_{1} - v_{2}| = \frac{f^{2} L}{{(u - f)}^{2} \times \frac{L^{2}}{4}}$
Since, the object is short and kept away from focus, we have
$\frac{L^{2}}{4} < < {(u - f)}^{2}$
Hence, the required expression of length of the image is $L' = \frac{f^{2}}{{(u - f)}^{2}} L$

Page No 59:

Question 23:

A circular disc of radius ‘R’ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius ‘a’ (in given figure). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index μ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

Answer:

Referring to the figure, AM is the direction of incidence ray before liquid is filled. After liquid is filled in , BM is the direction of the incident ray. Refracted ray in both cases is same as that along AM

Let the disc is separated by O at a distance d as shown in figure. Also, considering angle

N = 90°, $O M = a, C B = N B = a - R, A N = a + R$
From the figure,
$\sin t = \frac{a - R}{\sqrt{d^{2} + {(a - R)}^{2}}}$
$\Rightarrow \sin α = \cos (90 - α) = \frac{a + R}{\sqrt{d^{2} + {(a + R)}^{2}}}$
But, on applying Snell’s law,
$\frac{1}{μ} = \frac{\sin t}{\sin r} = \frac{\sin t}{\sin α}$
On substituting the values, we have the separation
$d = \frac{μ (a^{2} - R^{2})}{\sqrt{{(a + R)}^{2} - μ {(a - R)}^{2}}}$

Page No 59:

Question 24:

A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0,0) and an object placed at (–50 cm, 0). Find the coordinates of the image.

Answer:

If there was no cut, then the object would have been at a height of 0.5 cm from the principal axis.
Applying lens formula , we have
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow v = \frac{u f}{u + f} = \frac{- 50 \times 25}{- 50 + 25} = 50 cm$
Magnification, $m = \frac{v}{u} = \frac{50}{- 50} = - 1$

Thus, the inverted image would have been formed at 50 cm from the pole below the axis .

Hence, with respect to the axis passing through the edge of the cut lens, the coordinates of the image are (50 cm, –1 cm ).

Page No 59:

Question 25:

In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

Answer:

According to the principal of reversibility, the position of object and image are interchangeable.
So, by the reversibility of u and v, as seen from the formula for lens.

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

It is clear that there are two positions for which there shall be an image. On the screen, let the first position be when the lens is at O.
Finding u and v and substituting in lens formula.

Given, $- u + v = D$
$\Rightarrow u = - (D - v)$

Placing it in the lens formula
$\frac{1}{D - v} + \frac{1}{v} = \frac{1}{f}$

On solving, we have
$\Rightarrow \frac{v + D - v}{(D - v) v} = \frac{1}{f} \Rightarrow v^{2} - D v + D f = 0 \Rightarrow v = \frac{D}{2} \pm \frac{\sqrt{D^{2} - 4 D f}}{2}$

Hence, finding u
$u = - (D - v) = - (\frac{D}{2} \pm \frac{\sqrt{D^{2} - 4 D f}}{2})$

$Now if, the object distance is \frac{D}{2} + \frac{\sqrt{D^{2} - 4 D f}}{2} the image forms at \frac{D}{2} - \frac{\sqrt{D^{2} - 4 D f}}{2} Simillarly, if, the object distance is \frac{D}{2} - \frac{\sqrt{D^{2} - 4 D f}}{2} the image forms at \frac{D}{2} + \frac{\sqrt{D^{2} - 4 D f}}{2}$
The distance between the poles for these two object distances is:

$\frac{D}{2} + \frac{\sqrt{D^{2} - 4 D f}}{2} - (\frac{D}{2} - \frac{\sqrt{D^{2} - 4 D f}}{2}) = \sqrt{D^{2} - 4 D f} Let, d = \sqrt{D^{2} - 4 D f} If u = \frac{D}{2} + \frac{d}{2}, then the image is at v = \frac{D}{2} - \frac{d}{2} The magnification m_{1} = \frac{D - d}{D + d} If u = \frac{D - d}{2}, then v = \frac{D + d}{2} The magnification m_{2} = \frac{D + d}{D - d} Thus, \frac{m_{2}}{m_{1}} = {(\frac{D + d}{D - d})}^{2}$

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Question 26:

A jar of height h is filled with a transparent liquid of refractive index μ (in given figure). At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the centre, the dot is invisible.

Answer:

The diameter of the disc is d.

The spot shall be invisible if the incident rays from the dot at O to the surface at $\frac{d}{2}$ at the critical angle.
Let i be the angle of incidence.

Using relationship between refractive index and critical angle:
then, $\sin t = \frac{1}{μ}$

Using geometry and trigonometry.
Now, $\frac{(\frac{d}{2})}{h} = \tan i$
$\Rightarrow \frac{d}{2} = h \tan i = \frac{h}{[\sqrt{μ^{2} - 1}]} \Rightarrow d = \frac{2 h}{[\sqrt{μ^{2} - 1}]}$

Page No 59:

Question 27:

A myopic adult has a far point at 0.1 m. His power of accommodation is 4 diopters. (i) What power lenses are required to see distant objects? (ii) What is his near point without glasses? (iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)

Answer:

(i) Let P_f be the the power at the far point for the normal relaxed eye :
So, $P_{f} = \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{0.1} + \frac{1}{0.02} = 60 D$

By the corrective lens the object distance at the far point is ∞.
The power required is, $P_{f}' = \frac{1}{f'} = \frac{1}{\infty} + \frac{1}{0.02} = 50 D$

So for eye + lens system, we have the sum of the eye and that of the glasses P_g
$∴ P_{f}' = P_{f} + P_{g} ∴ P_{g} = - 10 D$

(ii) The power of accommodation of the person is 4 D for the normal eye.
Let P_n be the power of the normal eye for near vision.
So,
$4 = P_{n} - P_{f} \Rightarrow P_{n} = 60 + 4 = 64 D$

Let $x_{n}$ be the near point without glasses, then:
$\frac{1}{x_{n}} + \frac{1}{0.02} = 64 \Rightarrow \frac{1}{x_{n}} + 50 = 64 \Rightarrow \frac{1}{x_{n}} = 14 \Rightarrow x_{n} = 0.07 m$

(iii) Let $x'_{n}$ be the near point with glasses, then:
With glasses, $P_{n}' = P_{f}' + 4 = 54$
$\Rightarrow 54 = \frac{1}{x'_{n}} + \frac{1}{0.02} = \frac{1}{x'_{n}} + 50 \Rightarrow x'_{n} = 0.25 m$

Page No 60:

Question 28:

Show that for a material with refractive index μ ≥ $\sqrt{2}$ , light incident at any angle shall be guided along a length perpendicular to the incident face.

Answer:

Any ray entering at an angle i shall be guided along AC if the angle ray makes with the face AC ( $ϕ$ ) is greater than the critical angle as per the principle of total internal reflection $ϕ$ + r = 90^o , therefore sin $ϕ$ = cos r

$\Rightarrow \sin ϕ \geq \frac{1}{μ} \Rightarrow \cos r \geq \frac{1}{μ} or 1 - \cos^{2} r \leq 1 - \frac{1}{μ^{2}} i . e ., \sin^{2} r \leq 1 - \frac{1}{μ^{2}}$
Since, $\sin i = μ \sin r$
$\frac{1}{μ_{2}} \sin^{2} i \leq 1 - \frac{1}{μ^{2}} o r \sin^{2} i \leq μ^{2} - 1$

When, $i = \frac{π}{2}$

Then, we have smallest angle $ϕ$ .

If that is greater than the critical angle, then all other angles of incidence shall be more than the critical angle.

Thus, $1 \leq μ^{2} - 1$
or $μ^{2} \geq 2$
$\Rightarrow μ \geq 2$

Page No 60:

Question 29:

The mixture a pure liquid and a solution in a long vertical column (i.e, horizontal dimensions << vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d << h, the height of the column.

Answer:

Let the angle of incidence be θ.
The incident ray deviate continuously, but for small horizontal distance dx, the ray deviates to small length dy in vertical direction through angle (θ + dθ).
But, refractive index of the also varies with y.
So, applying Snell's law at point P at a particular instant, µ(y) sinθ =µ(y + dy) sin(θ + dθ)
Using approximation for dθ and dy to be very small
$\Rightarrow µ (y) \sin θ \approx [µ (y) + \frac{d µ}{d y} \times d y] (\sin θ \cos d θ + \cos θ \sin d θ) \Rightarrow µ (y) \sin θ \approx [µ (y) + \frac{d µ}{d y} \times d y] [\sin θ + \cos θ d θ] (∵ for small d θ, \cos d θ \to 1 and \sin d θ \to d θ) \Rightarrow µ (y) \sin θ \approx µ (y) \sin θ + µ (y) \cos θ d θ + \frac{d µ}{d y} d y \sin θ + \underset{Neglect}{\underset{⏟}{\frac{d µ}{d y} d y \cos θ d θ}} \Rightarrow µ (y) \sin θ \approx µ (y) \sin θ + µ (y) \cos θ d θ + \frac{d µ}{d y} d y \sin θ \Rightarrow µ (y) \cos θ d θ \approx - \frac{d µ}{d y} d y \sin θ \Rightarrow d θ \approx - \frac{1}{µ} \frac{d µ}{d y} d y \frac{\sin θ}{\cos θ} (∵ \tan θ = \frac{\sin θ}{\cos θ} = \frac{d x}{d y}) \Rightarrow d θ \approx \frac{1}{µ} \frac{d µ}{d y} \times d y \times \frac{d x}{d y} \Rightarrow d θ \approx - \frac{1}{µ} \frac{d µ}{d y} d x$
Now integrating both side with proper limit (for horizontal length d)
$\Rightarrow θ = - \frac{1}{µ} \frac{d µ}{d y} \int_{0}^{d} d x = - \frac{1}{µ} \frac{d µ}{d y} d$
This is the required deviation.

Page No 60:

Question 30:

If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refrative index of the medium given by

n(r) = 1 + 2 GM/rc²

where r is the distance of the point of consideration from the centre of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.

Answer:

Let us consider two spherical surfaces of radius r and r + dr with common centre O, as shown in the diagram.

Mass of the body is M.
Let the radius of the spherical body be R.
Let angle of incidence of the ray on the surface of radius r be θ and angle of emergence from surface of radius (r + dr) be (θ + dθ).
It is also given that the refractive index of the medium varies as
$n (r) = 1 + \frac{2 G M}{r c^{2}}$
So, applying Snell's law:
n(r) sinθ = n(r + dr) sin(θ + dθ)
Now using the approximation that dθ and dr be very small, we get
$\Rightarrow n (r) \sin θ \approx [n (r) + \frac{d n}{d r} \times d r] (\sin θ \cos d θ + \cos θ \cdot \sin d θ) (for d θ \to 0, \sin d θ \to d θ and \cos d θ \to 1)$

$\Rightarrow n (r) \sin θ \approx [n (r) + \frac{d n}{d r} \times d r] (\sin θ + \cos θ d θ) \Rightarrow n (r) \sin θ \approx n (r) \sin θ + n (r) \cos θ d θ + \frac{d n}{d r} d r \sin θ + \underset{(Neglect)}{\underset{⏟}{\frac{d n}{d r} d r \cos θ d θ}}$
$\Rightarrow n (r) \sin θ \approx n (r) \sin θ + n (r) \cos θ d θ + \frac{d n}{d r} d r \sin θ \Rightarrow n (r) \cos θ d θ \approx - \frac{d n}{d r} d r \tan θ \Rightarrow n (r) d θ \approx - \frac{d n}{d r} d r \tan θ \Rightarrow n (r) \frac{d θ}{d r} \approx - \frac{d n}{d r} \tan θ \Rightarrow (1 + \frac{G M}{r c^{2}}) \frac{d θ}{d r} \approx - \frac{d}{d r} (1 + \frac{G M}{r c^{2}}) \tan θ \Rightarrow (1 + \frac{G M}{r c^{2}}) \frac{d θ}{d r} \approx \frac{2 G M}{r^{2} c^{2}} \tan θ \Rightarrow \frac{d θ}{d r} \approx \frac{2 G M}{r^{2} c^{2}} \tan θ (Neglecting smaller values) \Rightarrow \int d θ = \frac{2 G M}{c^{2}} \int \frac{\tan θ}{r^{2}} d r From figure, r^{2} = x^{2} + R^{2} and \tan θ = \frac{R}{x} \Rightarrow 2 r d r = 2 x d x \Rightarrow d r = \frac{x}{r} d x$
So, $\int_{0}^{θ_{\circ}} d θ = \frac{2 G M}{c^{2}} \int_{- \infty}^{\infty} \frac{R}{x} \frac{x d x}{{(x^{2} + R^{2})}^{\frac{3}{2}}}$
Now let, x = R tanÏ•
⇒ dx = R sec²Ï• dÏ•
So, we can rewrite the integration with proper limit as:
$θ_{\circ} - 0 = \frac{2 G M R}{c^{2}} \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{R \sec^{2} ϕ d ϕ}{R^{3} \sec^{3} ϕ} \Rightarrow θ_{\circ} = \frac{4 G M}{R c^{2}}$
This is the required deviation.

Page No 60:

Question 31:

An infinitely long cylinder of radius R is made of an unusual exotic material with refractive index –1 (in given figure). The cylinder is placed between two planes whose normals are along the y direction. The center of the cylinder O lies along the y-axis. A narrow laser beam is directed along the y direction from the lower plate. The laser source is at a horizontal distance x from the diameter in the y direction. Find the range of x such that light emitted from the lower plane does not reach the upper plane.

Answer:

Refractive index of the material of the cylinder is –1.

Let the angle of incidence be θ_i.
Angle of refraction from the first surface be θ_r.
Angle of refraction from the second surface be _{$θ_{r}^{'}$ .}
Since µÎ„ = –1
_{$So, |θ_{i}| = |θ_{r}| = |θ_{r}^{'}|$}
So, the total deviation of the emergent ray will be:
δ = 4θ_i (As shown in the diagram)
Now, the ray will never reach the receiving plate only if
$\frac{π}{2} \leq 4 θ_{i} \leq \frac{3 π}{2} \Rightarrow \frac{π}{8} \leq θ_{i} \leq \frac{3 π}{8}$
But from figure, $\sin θ_{i} = \frac{x}{R}$
$\Rightarrow θ_{i} = \sin^{- 1} (\frac{x}{R})$
So,
$\Rightarrow \frac{π}{8} \leq \sin^{- 1} \frac{x}{R} \leq \frac{3 π}{8}$
Now for smaller θ_i,
$\Rightarrow \frac{π}{8} \leq \frac{x}{R} \leq \frac{3 π}{8} \Rightarrow \frac{π R}{8} \leq x \leq \frac{3 π R}{8}$
This is the required condition, such that light emitted from the source shall not reach the receiving plate.

Page No 60:

Question 32:

(i) Consider a thin lens placed between a source (S) and an observer (O) (Fig. 9.8). Let the thickness of the lens vary as $ω (b) = ω_{0} - \frac{b^{2}}{α}$ ,where b is the verticle distance from the pole. w₀ is a constant. Using Fermat’s principle i.e. the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.

(ii) A gravitational lens may be assumed to have a varying width of the form

$w (b) = k_{1} \ln (\frac{k_{2}}{b}) b_{\min} < b < b_{\max} = k_{1} \ln (\frac{k_{2}}{b_{\min}}) b < b_{\min}$
Show that an observer will see an image of a point object as a ring about the center of the lens with an angular radius
$β = \sqrt{\frac{(n - 1) k_{1} \frac{u}{v}}{u + v}}$

Answer:

(i) Let speed of light be c.
SP = u
OP = v
PP₁= b
${SP}_{1} = \sqrt{u^{2} + b^{2}} P_{1} O = \sqrt{v^{2} + b^{2}}$

Time taken by the light ray to travel from S to P₁,
$t_{1} = \frac{{SP}_{1}}{c} = \frac{\sqrt{u^{2} + b^{2}}}{c} = \frac{u}{c} {(1 + \frac{b^{2}}{u^{2}})}^{\frac{1}{2}}$
If b <<< u
Then, $t_{1} = \frac{u}{c} (1 + \frac{b^{2}}{2 c}) . . . . . (1)$
Now, time taken by the light ray to travel from P₁ to O,
$t_{2} = \frac{P_{1} O}{c} = \frac{\sqrt{v^{2} + b^{2}}}{c} = \frac{v}{c} {(1 + \frac{b^{2}}{v^{2}})}^{\frac{1}{2}}$
If b <<< v
$\Rightarrow t_{2} = \frac{v}{c} (1 + \frac{b^{2}}{2 v^{2}}) . . . . . (2)$
Time required to travel the light ray through the thickness of the lens [ω(b)],
$t_{3} = \frac{(µ - 1) ω (b)}{c} . . . . . (3)$
So, the total time taken by the light ray to travel from S to O via the lens,
$t = t_{1} + t_{2} + t_{3} = \frac{u}{c} (1 + \frac{b^{2}}{2 u^{2}}) + \frac{v}{c} (1 + \frac{b^{2}}{2 v^{2}}) + \frac{(µ - 1) ω (b)}{c} = \frac{1}{c} [u + v + \frac{b^{2}}{2} (\frac{1}{u} + \frac{1}{v}) + (µ - 1) ω (b)]$
$Let, \frac{1}{D} = \frac{1}{u} + \frac{1}{v} So, t = \frac{1}{c} [u + v + \frac{b^{2}}{2 D} + (µ - 1) ω (b)] But, ω (b) = (ω_{\circ} - \frac{b^{2}}{α}) (∵ as given in question) So, t = \frac{1}{c} [u + v + \frac{b^{2}}{2 D} + (µ - 1) (ω_{\circ} - \frac{b^{2}}{α})] . . . . . (4)$
Now, by Fermat's principle:
$\frac{d t}{d b} = \frac{1}{c} [0 + 0 + \frac{b}{D} + (µ - 1) (0 - \frac{2 b}{α})] \frac{d t}{d b} = \frac{b}{c D} - \frac{2 (µ - 1) b}{c α} \Rightarrow \frac{d t}{d b} = 0 = \frac{b}{c D} - \frac{2 (µ - 1) b}{c α} (∵ time remains constant, i . e ., \frac{d t}{d b} = 0)$
$\Rightarrow \frac{b}{c D} = \frac{2 (µ - 1) b}{2 α} \Rightarrow α = 2 (µ - 1) D$
This expression is independent of b.
So, we can say that all paraxial rays from S will converge at O passing from the convergent lens.
We have assumed that $\frac{1}{D} = \frac{1}{u} + \frac{1}{v}$
So, here D is equivalent to focal length (f) of the convergent lens.
$Thus, \frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
(ii) It is given that,
$ω (b) = k_{1} \ln (\frac{k_{2}}{b}) b_{\min} < b < b_{\max} = k_{1} \ln (\frac{k_{2}}{b_{\min}}) b < b_{\min}$
Now, from equation (4),
$t = \frac{1}{c} [u + v + \frac{b^{2}}{2 D} + (µ - 1) k_{1} \ln (\frac{k_{2}}{b})]$
So, differentiating both side with respect to b, we get:
$But, \frac{d t}{d b} = 0 = \frac{1}{c} [\frac{b}{D} - (µ - 1) \frac{k_{1}}{b}] (∵ \frac{d t}{d b} = 0, because time to travel from S to O is constant .)$
$\Rightarrow \frac{b}{D} = (µ - 1) \frac{k_{1}}{b} \Rightarrow b = \sqrt{(µ - 1) k_{1} D}$
So, the angle made by the path of the ray with the principle axis,
$\begin{array}{rcl} θ & \approx & \frac{b}{v} = \frac{\sqrt{(µ - 1) k_{1} D}}{v} = \sqrt{\frac{(µ - 1) k_{1} D}{v^{2}}} \\ = & \sqrt{\frac{(µ - 1) k_{1} (\frac{u v}{u + v})}{v^{2}}} [∵ \frac{1}{D} = \frac{1}{u} + \frac{1}{v} \Rightarrow D = \frac{u v}{u + v}] \\ = & \sqrt{\frac{(µ - 1) k_{1} u}{(u + v) v}} \end{array}$
If, µ = n
Then, $θ = \sqrt{\frac{(n - 1) k_{1} u}{(u + v) v}}$

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