Thinking Process

The ray refractive by first surface falls normally on second surface, in order to emerges from the other face normally.

Answer

(a) Since, deviation $\delta=(\mu-1) A=(1.5-1) \times 5^{\circ}=2.5^{\circ}$

By geometry, angle of refraction by first surface is $5^{\circ}$.

But $\delta=\theta-r$, so, we have, $2.5^{\circ}=\theta-5^{\circ}$ on solving $\theta=7.5^{\circ}$.

• Option (b) $5^{\circ}$: This option is incorrect because if the angle of incidence were $5^{\circ}$, the deviation would be zero, which contradicts the given deviation of $2.5^{\circ}$.

• Option (c) $15^{\circ}$: This option is incorrect because if the angle of incidence were $15^{\circ}$, the deviation would be much larger than $2.5^{\circ}$, given the refractive index and the angle of the prism.

• Option (d) $2.5^{\circ}$: This option is incorrect because if the angle of incidence were $2.5^{\circ}$, the deviation would be negative, which is not possible in this context.

Thinking Process

When light ray goes from one medium to other medium, the frequency of light remains unchanged.

Answer

(d) Since $v \propto \lambda$, 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.

• (a) blue: Blue light has a shorter wavelength compared to red light, which means it travels slower through the glass slab. Therefore, it will not be the first to emerge.

• (b) green: Green light has a shorter wavelength than red light but longer than blue and violet. It travels slower than red light through the glass slab, so it will not be the first to emerge.

• (c) violet: Violet light has the shortest wavelength among the given options, which means it travels the slowest through the glass slab. Therefore, it will not be the first to emerge.

Thinking Process

This problem has link with the formation of image when object is at different positions.

Answer

(c) 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.

• Option (a): The image does not move away from the lens with a uniform speed of $5 m / s$ because the relationship between the object distance and the image distance in a converging lens is not linear. As the object approaches the focal point, the image distance changes non-linearly, leading to non-uniform motion.

• Option (b): The image does not move away from the lens with a uniform acceleration because the rate of change of the image distance is not constant. The image distance changes according to the lens formula, which results in a non-uniform acceleration as the object approaches the focal point.

• Option (d): The image does not move towards the lens with a non-uniform acceleration because, for a converging lens, as the object moves closer to the lens, the image moves further away from the lens. Therefore, the direction of the image movement is away from the lens, not towards it.

Answer

(b) A passenger in an aeroplane may see a primary and a secondary rainbow like concentric circles.

• (a) never see a rainbow: This is incorrect because a passenger in an aeroplane can see a rainbow if the conditions are right, such as when there are water droplets in the atmosphere and the sun is shining from behind the observer.

• (c) may see a primary and a secondary rainbow as concentric arcs: This is incorrect because from an elevated position, such as in an aeroplane, rainbows appear as full circles rather than arcs. The ground usually obstructs the lower part of the circle when viewed from the ground, making them appear as arcs.

• (d) shall never see a secondary rainbow: This is incorrect because secondary rainbows can occur under the right conditions. They are formed by two reflections of sunlight inside the water droplets and can be seen from an aeroplane just like from the ground, though they are fainter and have reversed colors compared to the primary rainbow.

Thinking Process

This problem is based on the critical angle of total internal reflection.

Answer

(c) According to VIBGYOR, among all given sources of light, the blue light have smallest wavelength.According to Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle.

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.

• (a) The beam of red light would undergo total internal reflection: This statement is incorrect because red light has a longer wavelength compared to yellow light, which means it has a lower refractive index and a larger critical angle. Since the angle of incidence is the same, the red light would not reach the critical angle required for total internal reflection.

• (b) The beam of red light would bend towards normal while it gets refracted through the second medium: This statement is incorrect because if the angle of refraction for yellow light is $90^{\circ}$, it means the angle of incidence is at the critical angle for yellow light. For red light, which has a larger critical angle, the angle of incidence would be less than its critical angle, causing it to bend away from the normal, not towards it.

• (d) The beam of green light would bend away from the normal as it gets refracted through the second medium: This statement is incorrect because green light has a shorter wavelength than yellow light, resulting in a higher refractive index and a smaller critical angle. Since the angle of incidence is the same, the green light would undergo total internal reflection rather than bending away from the normal.

Thinking Process

By lens maker’s formula for plano-convex lens, focal length is given by $f=\frac{R}{\mu-1}$. This is always positive for $\mu>1$ or optically denser medium of material of lens placed in air.

Answer

(c) Here, $R=20 cm, \mu=1.5$, on substituting the values in $f=\frac{R}{\mu-1}=\frac{20}{1.5-1}=40 cm$ of converging nature as $f>0$. Therefore, lens act as a convex lens irrespective of the side on which the object lies.

• (a) act as a convex lens only for the objects that lie on its curved side: This option is incorrect because the focal length of the lens is positive, indicating that it acts as a converging lens (convex lens) regardless of the side on which the object is placed. The nature of the lens does not change based on the side of the object.

• (b) act as a concave lens for the objects that lie on its curved side: This option is incorrect because the lens has a positive focal length, which means it is a converging lens (convex lens). A concave lens would have a negative focal length, which is not the case here.

• (d) act as a concave lens irrespective of side on which the object lies: This option is incorrect because the lens has a positive focal length, indicating that it is a converging lens (convex lens). A concave lens would have a negative focal length, which is not applicable in this scenario.

Answer

(b) 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.

• (a) Reflection of light by a plane mirror: This is a simple reflection where the angle of incidence equals the angle of reflection. It does not involve the concept of a critical angle or total internal reflection, which is essential in the reflection of radiowaves by the ionosphere.

• (c) Dispersion of light by water molecules during the formation of a rainbow: Dispersion involves the splitting of light into its constituent colors due to different refractive indices for different wavelengths. This is a different phenomenon from the total internal reflection that occurs in the ionosphere.

• (d) Scattering of light by the particles of air: Scattering involves the redirection of light in various directions due to interaction with particles. This is not related to the concept of total internal reflection or the critical angle, which are key in the reflection of radiowaves by the ionosphere.

Answer

(b) The $P Q$ ray of light passes through focus $F$ and incident on the concave mirror, after reflection, should become parallel to the principal axis and shown by ray-2 in the figure.

• Option (a): The ray shown by 1 does not follow the law of reflection for a concave mirror. If the incident ray passes through the focus, the reflected ray should be parallel to the principal axis, not in the direction shown by 1.

• Option (c): The ray shown by 3 indicates a direction that does not align with the behavior of light rays in concave mirrors. If the incident ray passes through the focus, the reflected ray should be parallel to the principal axis, not in the direction shown by 3.

• Option (d): The ray shown by 4 also does not follow the correct reflection pattern for a concave mirror. If the incident ray passes through the focus, the reflected ray should be parallel to the principal axis, not in the direction shown by 4.

Thinking Process

When a light ray goes from (optically) rarer medium to (optically) denser medium, then it bends towards the normal i.e., i>r and vice-versa.

Answer

(b) Here, light ray goes from (optically) rarer medium air to optically denser terpentine, then it bends towards the normal i.e., $i>r$ whereas when it goes from to optically denser medium terpentine to rarer medium water. then it bends away the normal i.e., $i<r$.

• Option (a): The path shown in option (a) is incorrect because when light travels from air (optically rarer medium) to turpentine (optically denser medium), it should bend towards the normal. However, in the diagram, the light ray does not bend towards the normal, which violates the principle of refraction.

• Option (c): The path shown in option (c) is incorrect because when light travels from turpentine (optically denser medium) to water (optically rarer medium), it should bend away from the normal. In the diagram, the light ray does not bend away from the normal, which violates the principle of refraction.

• Option (d): The path shown in option (d) is incorrect because when light travels from air (optically rarer medium) to turpentine (optically denser medium), it should bend towards the normal. In the diagram, the light ray does not bend towards the normal, which violates the principle of refraction.

Thinking Process

The image formed by convex mirror does not depend on the relative position of object w.r.t. mirror.

Answer

(d) The speed of the image of the car would appear to increase as the distance between the cars decreases.

• (a) The speed of the car in the rear is $65 km h^{-1}$: This statement is incorrect because the speed of the car in the rear is actually $60 km h^{-1}$ + $5 km h^{-1}$ relative to the leading car. The car in the rear is approaching the leading car at a relative speed of $5 km h^{-1}$, not $65 km h^{-1}$ in absolute terms.

• (b) In the side mirror, the car in the rear would appear to approach with a speed of $5 kmh^{-1}$ to the driver of the leading car: This statement is incorrect because the side mirror does not affect the relative speed of the approaching car. The relative speed of $5 km h^{-1}$ is the same regardless of whether the driver looks in the rear view mirror or the side mirror.

• (c) In the rear view mirror, the speed of the approaching car would appear to decrease as the distance between the cars decreases: This statement is incorrect because the relative speed of the approaching car remains constant at $5 km h^{-1}$ as long as both cars maintain their speeds. The distance between the cars decreases at a constant rate, so the speed does not appear to decrease.

Answer

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

• Option (b): This option shows the ray bending in the same direction as it would in a medium with a positive refractive index. In a negative refractive index material, the ray should bend in the opposite direction.

• Option (c): This option depicts the ray continuing straight without any refraction. This is incorrect because a negative refractive index material would still cause the ray to bend, just in the opposite direction compared to a positive refractive index material.

• Option (d): This option shows the ray bending in the same direction as it would in a medium with a positive refractive index. Similar to option (b), this is incorrect because a negative refractive index material would cause the ray to bend in the opposite direction.

Thinking Process

This problem is based on the phenomenon of reflection when it goes from denser to rare medium

Answer

$(a, b, c)$

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.

• Option (d) is incorrect because water in a trough does not act as a lens that magnifies the object. The distortion observed is due to the refraction of light at the water-air interface, not due to any lensing effect.

Thinking Process

This problem is associated with the phenomenon of total internal reflection.

Answer

(d) For $\mu=1.6$, the critical angle, $\mu=1 / \sin C$, we have $C=38.7^{\circ}$, when viewed from $A D$, as long as angle of incidence on $A D$ of the ray emanating from pin is greater than the critical angle, the light suffers from total internal reflection and cannot be seen through $A D$.

• (a) appear to be near $A$: This option is incorrect because if the pin were to appear near $A$, it would imply that the light rays are bending towards $A$ when passing through the glass. However, due to the high refractive index and the critical angle being exceeded, the light rays undergo total internal reflection and do not exit through face $AD$.

• (b) appear to be near $D$: This option is incorrect because if the pin were to appear near $D$, it would imply that the light rays are bending towards $D$ when passing through the glass. Similar to option (a), the light rays undergo total internal reflection due to the critical angle being exceeded and do not exit through face $AD$.

• (c) appear to be at the centre of $A D$: This option is incorrect because if the pin were to appear at the centre of $AD$, it would imply that the light rays are passing straight through the glass without significant deviation. However, due to the refractive index and the critical angle, the light rays undergo total internal reflection and do not exit through face $AD$.

Answer

$(a, d)$

Alexandar’s dark band lies between the primary and secondary rainbows, forms due to light scattered into this region interfere destructively.

Since, primary rainbows subtends an angle nearly $41^{\circ}$ to $42^{\circ}$ at observer’s eye, whereas, secondary rainbows subtends an angle nearly $51^{\circ}$ to $54^{\circ}$ at observer’s eye w.r.t. incident light ray.

So, the scattered rays with respect to the incident light of the sun lies between approximately $42^{\circ}$ and $50^{\circ}$.

• (b) there is no light scattered into this region: This option is incorrect because there is indeed light scattered into the region between the primary and secondary rainbows. The dark band, known as Alexander’s dark band, is a result of destructive interference of the scattered light, not the absence of scattered light.

• (c) light is absorbed in this region: This option is incorrect because the dark band is not due to the absorption of light. Instead, it is caused by the destructive interference of light waves scattered into this region. Light absorption would imply that the light energy is being taken up by a medium, which is not the case here.

Answer

$(a, b)$

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 larger angle to be subtended by the object at the eye and hence, viewed in greater detail. Morever, the formation of a virtual erect and enlarged image, takes place.

• Option (c): Increase in the field of view is incorrect because a magnifying glass typically reduces the field of view. When an object is magnified, only a smaller portion of the object can be seen at a time, thus decreasing the overall field of view.

• Option (d): Infinite magnification at the near point is incorrect because no optical device, including a magnifying glass, can provide infinite magnification. Magnification is limited by the optical properties of the lens and the distance at which the object is placed.

Thinking Process

The magnifying power $m$ is the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\boldsymbol{\alpha}$ which the object subtends at the lens or the eye. Hence, in normal adjustment

$$ m \approx \frac{\beta}{\alpha} \approx \frac{h}{f_{e}} \frac{f_{o}}{h}=\frac{f_{o}}{f_{e}} $$

In this case, the length of the telescope tube is $f_{o}+f_{e}$.

Answer

$(a, b, c)$

The length of the telescope tube is $f_{o}+f_{e}=20+(0.02)=20.02 m$

Also,

$m=20 / 0.02=1000$

Also, the image formed is inverted.

• An objective of a larger aperture will increase the brightness but will not reduce chromatic aberration of the image. Chromatic aberration is dependent on the material and design of the lens, not its aperture size.

Answer

As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.

In other words, $\mu_{b}>\mu_{r}$

By lens maker’s formula

$$ \frac{1}{f}=(n_{21}-1) \frac{1}{R_1}-\frac{1}{R_2} $$

Therefore, $f_{b}<f_{r}$.

Thus, the focal length for blue light will be smaller than that for red.

Answer

The least distance of distinct vision of an average person (i.e., $D$ ) is $25 cm$, in order to view an object with magnification 10 ,

Here, $v=D=25 cm$ and $u=f$

But the magnification $m=v / u=D / f$

$$ \begin{aligned} & m=\frac{D}{f} \\ & f=\frac{D}{m}=\frac{25}{10}=2.5=0.025 m \\ & P=\frac{1}{0.025}=40 D \end{aligned} $$

This is the required power of lens

Thinking Process

One lens have unique focal length irrespective of its face or geometry taken for use.

Answer

No, the reversibility of the lens maker’s equation.

Thinking Process

The image formed by first medium act as an object for second medium .

Answer

Let the apparent depth be $O_1$ for the object seen from $m_2$, then

$$ O_1=\frac{\mu_2}{\mu_1} \frac{h}{3} $$

Ray Optics and Optical Instruments

Since, apparent depth $=$ real depth $/$ refractive index $\mu$.

Since, the image formed by Medium $1, O_2$ act as an object for Medium 2.

If seen from $\mu_3$, the apparent depth is $O_2$.

Similarly, the image formed by Medium 2, $O_2$ act as an object for Medium 3

$$ \begin{aligned} O_2 & =\frac{\mu_3}{\mu_2} \frac{h}{3}+O_1 \\ & =\frac{\mu_3}{\mu_2} \frac{h}{3}+\frac{\mu_2 h}{\mu_1 3}=\frac{h}{3} \frac{\mu_3}{\mu_2}+\frac{\mu_2}{\mu_1} \end{aligned} $$

Seen from outside, the apparent height is

$$ \begin{aligned} O_3 & =\frac{1}{\mu_3} \frac{h}{3}+O_2=\frac{1}{\mu_3} \frac{h}{3}+\frac{h}{3} \frac{\mu_3}{\mu_2}+\frac{\mu_3}{\mu_1} \\ & =\frac{h}{3} \frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3} \end{aligned} $$

This is the required expression of apparent depth.

Answer

The relationship between refractive index, prism angle $A$ and angle of minimum deviation is given by

$$ \mu=\frac{\sin \frac{(A+D_{m})}{2}}{\sin \frac{A}{2}} $$

Here,

$\therefore$ Given,

$$ D_{m}=A $$

Substituting the value, we have

$\therefore \quad \mu=\frac{\sin A}{\sin \frac{A}{2}}$

On solving, we have

$$ \begin{aligned} & =\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2} \\ \mu & =\frac{\sin A}{\sin \frac{A}{2}}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2} \end{aligned} $$

For the given value of refractive index,

we have

$$ \therefore \cos \frac{A}{2} =\frac{\sqrt{3}}{2} $$ $$ \text{or} \quad \quad s\frac{A}{2} =30^{\circ}$$ $$\therefore A =60^{\circ}$$

This is the required value of prism angle.

Thinking Process

The length of image is the separation between the images formed by mirror of the extremities of object.

Answer

Since, the object distance is $u$. Let us consider the two ends of the object be at distance $u_1=u-L / 2$ and $u_2=u+L / 2$, respectively so that $|u_1-u_2|=L$. Let the image of the two ends be formed at $v_1$ and $v_2$, respectively so that the image length would be

$$ L^{\prime}=|v_1-v_2| $$

Applying mirror formula, we have

$$ \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \text { or } v=\frac{f u}{u-f} $$

On solving, the positions of two images are given by

$$ v_1=\frac{f(u-L / 2)}{u-f-L / 2}, v_2=\frac{f(u+L / 2)}{u-f+L / 2} $$

For length, substituting the value in (i), we have

$$ L^{\prime}=|v_1-v_2|=\frac{f^{2} L}{(u-f)^{2} \times L^{2} / 4} $$

Since, the object is short and kept away from focus, we have

$$ L^{2} / 4«(u-f)^2 \\ \text {Hence, finally} L^{\prime}=\frac{f^2}{(u-f)^2} $$

This is the required expression of length of image.

Thinking Process

This problem involves the wide application of geometry and Snell’s law.

Answer

Refering to the figure, $A M$ is the direction of incidence ray before liquid is filled. After liquid is filled in , $B M$ is the direction of the incident ray. Refracted ray in both cases is same as that along $A M$.

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

Here, in figure

$$ N=90^{\circ}, O M=a, C B=N B=a-R, A N=a+R $$

and

$$ \begin{aligned} \sin t & =\frac{a-R}{\sqrt{d^{2}+(a-R)^{2}}} \\ \sin \alpha & =\cos (90-\alpha)=\frac{a+R}{\sqrt{d^{2}+(a+R)^{2}}} \end{aligned} $$

But on applying Snell’s law,

$$ \frac{1}{\mu}=\frac{\sin t}{\sin r}=\frac{\sin t}{\sin \alpha} $$

On substituting the values, we have the separation

$$ d=\frac{\mu(a^{2}-b^{2})}{\sqrt{(a+r)^{2}-\mu(a-r)^{2}}} $$

This is the required expression.

Thinking Process

There is no effect on the focal length of the lens if it is cut as given in the question.

Answer

If there was no cut, then the object would have been at a height of $0.5 cm$ from the principal axis $OO^{\prime}$.

Applying lens formula, we have

$$ \begin{aligned} & & \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ & \therefore & \frac{1}{v} & =\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50} \\ & & \text { Mangnification is } m & =\frac{v}{u}=-\frac{50}{50}=-1 \end{aligned} $$

Thus, the image would have been formed at $50 cm$ from the pole and $0.5 cm$ below the principal axis. Hence, with respect to the $X$-axis passing through the edge of the cut lens, the coordinates of the image are $(50 cm,-1 cm)$.

Thinking Process

This is also one of the methods for finding the focal length of the lens in laboratory and known as displacement method.

Answer

Principal of reversibility is states that the position of object and image are interchangeable. So, by the versibility 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,

$\Rightarrow$

Placing it in the lens formula

$$ \frac{1}{D-v}+\frac{1}{v}=\frac{1}{f} $$

On solving, we have

$$ \begin{aligned} \Rightarrow \frac{v+D-v}{(D-v) v} =\frac{1}{f} \newline \newline\\ \Rightarrow v^{2}-D v+D f =0 \newline \newline\\ \Rightarrow v =\frac{D}{2} \pm \frac{\sqrt{D^{2}-4 D f}}{2} \end{aligned} $$

Hence, finding $u$

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

When, the object distance is $\quad \frac{D}{2}+\frac{\sqrt{D^{2}-4 D f}}{2}$

the image forms at

$$ \frac{D}{2}-\frac{\sqrt{D^{2}-4 D f}}{2} $$

Similarly, when the object distance is

The image forms at

$$ \begin{aligned} & \frac{D}{2}-\frac{\sqrt{D^{2}-4 D f}}{2} \\ & \frac{D}{2}+\frac{\sqrt{D^{2}-4 D f}}{2} \end{aligned} $$

The distance between the poles for these two object distance is

$$ \frac{D}{2}+\frac{\sqrt{D^{2}-4 D f}}{2}-\frac{D}{2}-\frac{\sqrt{D^{2}-4 D f}}{2}=\sqrt{D^{2}-4 D f} $$

$$ \begin{aligned} & \text { Let } \\ & \text { If } u=\frac{D}{2}+\frac{d}{2} \text {, then the image is at } v=\frac{\sqrt{D^{2}-4 D f}}{2}-\frac{d}{2} . \\ & \therefore \quad \text { The magnification } m_1=\frac{D-d}{D+d} \\ & \text { If } u=\frac{D-d}{2} \text {, then } v=\frac{D+d}{2} \\ & \therefore \text { The magnification } m_2=\frac{D+d}{D-d} \\ & \text { Thus, } \quad \frac{m_2}{m_1}=\frac{D+d}{D-d} . \end{aligned} $$

This is the required expression of magnification.

Thinking Process

The problem is based on the principle of total internal reflection and area of visibility.

Answer

Let $d$ be the diameter of the disc. The spot shall be invisible if the incident rays from the dot at $O$ to the surface at $d / 2$ at the critical angle.

Let $i$ be the angle of incidence.

Using relationship between refractive index and critical angle,

then,

$$ \sin t=\frac{1}{\mu} $$

Using geometry and trigonometry.

Now,

$$ \begin{aligned} \frac{d / 2}{h} & =\tan i \\ \frac{d}{2} & =h \tan i=h[\sqrt{\mu^{2}-1}]^{-1} \\ d & =\frac{2 h}{\sqrt{\mu^{2}-1}} \end{aligned} $$

$$ \Rightarrow $$

This is the required expression of $d$.

Thinking Process

If two thin lenses of focal length $f_1$ and $f_2$ are in contact, the effective focal length of the combination is given by,

in terms of power $P=P_1+P_2$

$$ \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} $$

Answer

(i) Let the power at the far point be $P_{f}$ for the normal relaxed eye of an average person. The required power

$$ P_{f}=\frac{1}{f}=\frac{1}{0.1}+\frac{1}{0.02}=60 D $$

By the corrective lens the object distance at the far point is $\infty$.

The power required is

$$ P_f^{\prime}=\frac{1}{f^{\prime}}=\frac{1}{\infty}+\frac{1}{0.02}=50 D $$

So for eye + lens system,

we have the sum of the eye and that of the glasses $P_{g}$

$$ \begin{matrix} \therefore & P_f^{\prime}=P_{f}+P_{g} \\ \therefore & P_{g}=-10 D \end{matrix} $$

(ii) His power of accomodatlon is $4 D$ for the normal eye. Let the power of the normal eye for near vision be $P_{n}$.

$$ \text { Then, } \quad 4=P_{n}-P_{f} \text { or } P_{n}=64 D $$

Let his near point be $x_{n}$, then

$$ \begin{aligned} \frac{1}{x_{n}}+\frac{1}{0.02} & =64 \text { or } \frac{1}{x_{n}}+50=64 \\ \frac{1}{x_{n}} & =14, \\ \therefore \quad x_{n} & =\frac{1}{14} ; 0.07 m \end{aligned} $$

(iii) With glasses $P_n^{\prime}=P_f^{\prime}+4=54$

$$ \begin{aligned} 54 & =\frac{1}{x_n^{\prime}}+\frac{1}{0.02}=\frac{1}{x_n^{\prime}}+50 \\ \frac{1}{x_n^{\prime}} & =4 \\ \therefore \quad x_n^{\prime} & =\frac{1}{4}=0.25 m \end{aligned} $$

Answer

Any ray entering at an angle $i$ shall be guided along $A C$ if the angle ray makes with the face $A C(\varphi)$ is greater than the critical angle as per the principle of total internal reflection $\varphi+r=90^{\circ}$, therefore $\sin \varphi=\cos r$

$\begin{aligned} \Rightarrow \qquad \sin \varphi & \geq \frac{1}{\mu} \\ \Rightarrow \qquad \cos r & \geq \frac{1}{\mu} \\ \text{or} \qquad 1-\cos ^2 r & \leq 1-\frac{1}{\mu^2} \\ \text{i.e.,} \qquad \sin ^2 r & \leq \frac{1}{\mu^2} \end{aligned}$

$\begin{aligned} & \text { i.e., } \qquad \sin ^2 r \leq 1-\frac{1}{\mu^2} \\ \qquad & \text { since, } \qquad \sin i=\mu \sin r \\ & \frac{1}{\mu_2} \sin ^2 i \leq 1-\frac{1}{\mu^2} \\ & \text { when } \qquad i=\frac{\pi}{2} \\ & \end{aligned}$

Then,we have smallest angle $\varphi$.

If that is greater than the critical angle,then all other the critical angle.

$\begin{array}{ll} \text { Thus, }\qquad & 1 \leq \mu^2-1 \\ \text { or } & \mu^2 \geq 2 \\ \Rightarrow & \mu \geq \sqrt{2} \end{array}$

This is the required result.

Answer

Let us consider a portion of a ray between $x$ and $x+d x$ inside the liquid. Let the angle of incidence at $x$ be $\theta$ and let it enter the thin column at height $y$. Because of the bending it shall emerge at $x+d x$ with an angle $\theta+d \theta$ and at a height $y+d y$. From Snell’s law,

$\mu(y) \sin \theta=\mu(y+d y) \sin (\theta+d \theta)$

$\mu(y) \sin \theta ; \mu(y)+\frac{d \mu}{d y} d y \quad(\sin \theta \cos d \theta+\cos \theta \sin d \theta) \newline$

$\mu(y) \sin \theta+\mu(y) \cos \theta d \theta+\frac{d \mu}{d y} d y \sin \theta \newline$

$\mu(y) \cos \theta d \theta ; \frac{-d \mu}{d y} d y \sin \theta$ $d \theta ; \frac{-d \mu}{\mu d y} d y \tan \theta$

$$ \tan \theta=\frac{d x}{d y} $$

(from the figure)

On solving, we have

$$ \therefore \quad d \theta=\frac{-1 d \mu}{\mu d y} d x $$

Solving this variable separable form of differential equation.

$$ \therefore \quad \theta=\frac{-1 d \mu}{\mu d y} \int_0^{d} d x=\frac{-1 d \mu}{\mu d y} d $$

Answer

Let us consider two planes at $r$ and $r+d r$. Let the light be incident at an angle $\theta$ at the plane at $r$ and leave $r+d r$ at an angle $\theta+d \theta$. Then from Snell’s law,

$\Rightarrow \quad n(r) \sin \theta ; n(r)+\frac{d n}{d r} d r \quad(\sin \theta \cos d \theta+\cos \theta \sin d \theta)$

$$ n(r)+\frac{d n}{d r} d r \quad(\sin \theta+\cos \theta d \theta) $$

Ignoring the product of differentials

or we have,

$$ \begin{aligned} n(r) \sin \theta ; n(r) \sin \theta & +\frac{d n}{d r} d r \sin \theta+n(r) \cos \theta d \theta \\ -\frac{d n}{d r} \tan \theta & =n(r) \frac{d \theta}{d r} \\ \frac{2 G M}{r^{2} c^{2}} \tan \theta & =1+\frac{2 G M}{r^{2}} \frac{d \theta}{d r} \approx \frac{d \theta}{d r} \\ \int_0^{\theta_0} d \theta & =\frac{2 G M}{c^{2}} \int_{-\infty}^{\infty} \frac{\tan \theta d r}{r^{2}} \end{aligned} $$

Now substitution for integrals, we have

$$ \text { Now, } \quad \begin{aligned} r^{2} & =x^{2}+R^{2} \text { and } \tan \theta=\frac{R}{x} \\ 2 r d r & =2 x d x \\ \text { Put } \quad \int_0^{\theta_0} d \theta & =\frac{2 G M}{c^{2}} \int_{-\infty}^{\infty} \frac{R}{x} \frac{x d x}{(x^{2}+R^{2})^{\frac{3}{2}}} \\ x & =R \tan \varphi \\ d x & =R \sec ^{2} \varphi d \varphi \\ \theta_0 & =\frac{2 G M R}{c^{2}} \int_{-\pi / 2}^{\pi / 2} \frac{R \sec ^{2} \varphi d \varphi}{R^{3} \sec ^{3} \varphi} \\ & =\frac{2 G M}{R c^{2}} \int_{-\pi / 2}^{\pi / 2} \cos \varphi d \varphi=\frac{4 G M}{R c^{2}} \end{aligned} $$

This is the required proof.

Answer

Since, the material is of refractive index $-1, \theta_{r}$, is negative and $\theta_r^{\prime}$ positive.

Now,

$$ |\theta_{t}|=|\theta_{r}|=|\theta_r^{\prime}| $$

The total deviation of the outcoming ray from the incoming ray is $4 \theta$. Rays shall not reach the recieving plate if

$$ .\frac{\pi}{2} \leq 4 \theta_{t} \leq \frac{3 \pi}{2} \quad \text { [angles measured clockwise from the } y \text {-axis }] $$

On solving,

$$ \frac{\pi}{8} \leq \theta_{t} \leq \frac{3 \pi}{8} $$

Now,

or

$$ \begin{aligned} \sin \theta_{t} & =\frac{x}{R} \\ \frac{\pi}{8} & \leq \sin ^{-1} \frac{x}{R} \leq \frac{3 \pi}{8} \\ \frac{\pi}{8} & \leq \frac{x}{R} \leq \frac{3 \pi}{8} \end{aligned} $$

Thus, for light emitted from the source shall not reach the receiving plate. If $\frac{R \pi}{8} \leq x \leq \frac{R 3 \pi}{8}$.

Answer

(i) The time elapsed to travel from $S$ to $P_1$ is

$$ t_1=\frac{S P_1}{c}=\frac{\sqrt{u^{2}+b^{2}}}{c} $$

or

$$ \frac{u}{c} 1+\frac{1}{2} \frac{b^{2}}{u^{2}} \text { assuming } b«u_0 \text {. } $$

The time required to travel from $P_1$ to $O$ is

$$ t_2=\frac{P_1 O}{c}=\frac{\sqrt{v^{2}+b^{2}}}{c} ; \frac{v}{c} 1+\frac{1}{2} \frac{b^{2}}{v^{2}} $$

The time required to travel through the lens is

$$ t_1=\frac{(n-1) w(b)}{c} $$

where $n$ is the refractive index.

Thus, the total time is

$$ t=\frac{1}{c} u+v+\frac{1}{2} b^{2} \frac{1}{u}+\frac{1}{v}+(n-1) w(b) $$

Put

$$ \frac{1}{D}=\frac{1}{u}+\frac{1}{v} $$

Then,

$$ t=\frac{1}{c} u+v+\frac{1}{2} \frac{b^{2}}{D}+(n-1) w_0+\frac{b^{2}}{\alpha} $$

Fermet’s principle gives the time taken should be minimum.

For that first derivative should be zero.

$$ \begin{aligned} \frac{d t}{d b} & =0=\frac{b}{C D}-\frac{2(n-1) b}{c \alpha} \\ \alpha & =2(n-1) D \end{aligned} $$

Thus, a convergent lens is formed if $\alpha=2(n-1) D$. This is independant of and hence, all paraxial rays from $S$ will converge at $O$ i.e., for rays

and

$$ (b«v .) $$

Since, $\frac{1}{D}=\frac{1}{u}+\frac{1}{V}$, the focal length is $D$.

(ii) In this case, differentiating expression of time taken $t$ w.r.t. $b$

$$ t =\frac{1}{c} u+v+ \frac{1}{2} \frac{b^2}{D}+(n-1) k_1 \ln \frac{k_2}{b} \newline $$ $$ \Rightarrow \frac{d t}{d b} =0=\frac{b}{D}-(n-1) \frac{k_1}{b} \newline \therefore b^{2} =(n-1) k_1 D \newline \therefore b =\sqrt{(n-1) k_1 D} \newline $$

Thus, all rays passing at a height $b$ shall contribute to the image. The ray paths make an angle.

$$ \beta ; \frac{b}{v}=\frac{\sqrt{(n-1) k_1 D}}{v^{2}}=\sqrt{\frac{(n-1) k_1 u v}{v^{2}(u+v)}}=\sqrt{\frac{(n-1) k_1 u}{(u+v) v}} $$

This is the required expression.