Geometrical Optics

${{1.34} \over V} - {1 \over \infty } = {{1.34 - 1} \over {7.8}}$

$ \therefore $  V = 30.7 mm

${1 \over {2{f_2}}} = {1 \over {{f_1}}} = \left( {{\mu _1} - 1} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right)$

${1 \over {{f_2}}} = \left( {{\mu _2} - 1} \right)\left( {{1 \over { - R}} - {1 \over \infty }} \right)$

${{\left( {{\mu _1} - 1} \right)} \over R} = {{\left( {{\mu _2} - 1} \right)} \over {2R}}$

$2{\mu _1} - {\mu _2} = 1$

Let the angle between two mirrors are $\theta $.

We know sum of angles of triangle = 180^o

$ \therefore $  3$\theta $ = 180^o

$ \Rightarrow $  $\theta $ = 60^o

Image is formed on the screen.

So, v = 10 cm

and $\mu $ = $-$ 10 cm

Using formula,

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

$ \Rightarrow $   ${1 \over {10}} - {1 \over {\left( { - 10} \right)}}$ = ${1 \over f}$

$ \Rightarrow $   f = 5 cm

Now a glass block is placed like this,



Because of this glass block source will move t$\left( {1 - {1 \over \mu }} \right)$ in the direction of incident ray.

$ \therefore $   S' = t$\left( {1 - {1 \over \mu }} \right)$

= 1.5$\left( {1 - {2 \over 3}} \right)$

= 0.5

$ \therefore $   now distance of source from the lens = 10 $-$ 0.5 = 9.5 cm

$ \therefore $   $\mu $ = $-$ 9.5 cm

$ \therefore $   ${1 \over v} - {1 \over {\left( { - 9.5} \right)}}$ = ${1 \over 5}$

$ \Rightarrow $   ${1 \over v}$ = ${1 \over 5} - {2 \over {19}}$

$ \Rightarrow $   ${1 \over v}$ = ${9 \over {95}}$

$ \Rightarrow $   v = 10.55 cm

So, to get sharp image screen should shift away (10.55 $-$ 10) = 0.55 cm from the lens.

Case - I

H = 30 cm

n = ${3 \over 2}$

H₁ = ${H \over n}$

$ \Rightarrow $ ${{30 \times 2} \over 3}$ = 20 cm

Case - II

R = 300 cm

${{{n_2}} \over v} - {{{n_1}} \over u} = {{{n_2} - {n_1}} \over R}$

${1 \over { - {H_2}}} - {3 \over { - 2 \times 30}} = {{1 - {3 \over 2}} \over { - 300}}$

${H_2} = {{600} \over {29}} = 20.684$ cm

Case - III

${{{n_2}} \over v} - {{{n_1}} \over u} = {{{n_2} - {n_1}} \over R}$;

${1 \over { - {H_3}}} - {3 \over { - 2 \times 30}} = {{1 - {3 \over 2}} \over {300}}$

${H_3} = {{600} \over {31}} = 19.354$ cm

Given, thin convex lens made of two materials.

Radius of curvature of left spherical surface $=$ radius of curvature of right spherical surface

When $n_1=n_2=n$, focal length $=f$

When $n_1=n$ and $n_2=n+\Delta n$, focal length $=f+\Delta f$

Here, $\Delta n \ll(n-1)$ and $1< n < 2$

Now, when $n_1=n_2=n$, from lens maker's formula

$ \begin{aligned} \frac{1}{f} & =(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(n-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \\\\ \Rightarrow \frac{1}{f} & =(n-1)\left(\frac{2}{R}\right) ........(1) \end{aligned} $

When $n_1=n$ and $n_2=n+\Delta n$

$ \frac{1}{f+\Delta f}=(n-1) \frac{1}{R_1}-(n+\Delta n-1) \frac{1}{+R_2} $

$ \begin{aligned} & \quad=(n-1) \frac{1}{R}-(n+\Delta n-1) \frac{1}{(-R)} \\\\ & \Rightarrow \frac{1}{f+\Delta f}=\frac{1}{R}(n-1+n+\Delta n-1)=\frac{1}{R}(2 n+\Delta n-2) \\\\ & \Rightarrow \frac{1}{f+\Delta f}=\frac{1}{R}(2 n+\Delta n-2) ........(2) \end{aligned} $

Subtracting Eq. (2) from Eq. (1), we get

$ \begin{aligned} & \frac{1}{f}-\frac{1}{f+\Delta f}=(n-1)\left(\frac{2}{R}\right)-\frac{1}{R}(2 n+\Delta n-2) \\\\ & =\frac{1}{R}(2 n-2-2 n-\Delta n+2)=-\frac{\Delta n}{R} \\\\ & \Rightarrow \frac{1}{f}-\frac{1}{f+\Delta f}=-\frac{\Delta n}{R} \Rightarrow \frac{(f+\Delta f)-f}{f(f+\Delta f)}=-\frac{\Delta n}{R} \\\\ & \Rightarrow \frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{R} \sim \frac{\Delta f}{f}=-\frac{\Delta n}{R} \\\\ & \Rightarrow \frac{\Delta f}{f}=-\frac{\Delta n}{R} \times f \end{aligned} $

Using Eq. (1), we get $f=\frac{R}{2(n-1)}$

$ \begin{aligned} & \Rightarrow \frac{\Delta f}{f}=-\frac{\Delta n}{R} \times \frac{R}{2(n-1)} \Rightarrow \frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)} \\\\ & \Rightarrow\left|\frac{\Delta f}{f}\right|=\frac{1}{2}\left|\frac{\Delta n}{n-1}\right| \\\\ & \Rightarrow\left|\frac{\Delta f}{f}\right|>\left|\frac{\Delta n}{n}\right| \end{aligned} $

Hence, option (C) is incorrect.

Now, for $n=1.5, \Delta n=10^{-3}$ and $f=20 \mathrm{~cm}$. Using relation $\frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{R}$ and Eq. (1), we get

$ \begin{aligned} & \frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{2(n-1) f} \Rightarrow \frac{\Delta f}{f+\Delta f}=-\frac{\Delta n}{2 n-2} \\\\ & \Rightarrow \Delta f(2 n-2)=-\Delta n(f+\Delta f) \\\\ & \Rightarrow \Delta f(2 n-2)=-\Delta n f-\Delta n \Delta f \\\\ & \Rightarrow \Delta f(2 n-2+\Delta n)=-\Delta n f \\\\ & \Rightarrow \Delta f=\frac{-\Delta n \times f}{2 n-2+\Delta n} \\\\ & \Rightarrow|\Delta f|=\frac{\Delta n+f}{2 n-2+\Delta n} \end{aligned} $

Substituting all values, we get

$ |\Delta f|=\frac{10^{-3} \times 20}{2 \times 1.5-2+10^{-3}}=0.02 \mathrm{~cm} $

Hence, option (B) is correct.

Since $\frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)}$, so if $\frac{\Delta n}{n}<0$ then $\frac{\Delta f}{f}>0$.

Hence, option $(\mathrm{A})$ is correct.

Since the relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ is independent of radius of curvature so the relation remains unchanged only the sign of focal length changes.

Hence, option (D) is correct.

Given : ${i_1} = 60^\circ $; $A = 30^\circ $

Then angle of deviation is given by

$\delta = 1 - A = 60^\circ - 30^\circ = 30^\circ $

We know that,

${i_1} + {i_2} = A + \delta \Rightarrow 60^\circ + {i_2} = 30^\circ + 30^\circ \Rightarrow {i_2} = 0$

It means emergent ray leaves face AC normally.

For a convergent doublet of separated lens, we have

${1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {d \over {{f_1}{f_2}}}$ ...... (1)

where d is separation between two lens, f₁ and f₂ are focal lengths of component lenses, f is resultant focal length. Therefore, Eq. (1) becomes

${1 \over {10}} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {2 \over {{f_1}{f_2}}} \Rightarrow {1 \over {10}} = \left( {{{{f_2} + {f_1} - 2} \over {{f_1}{f_2}}}} \right)$

$ \Rightarrow {f_1}{f_2} = 10{f_2} + 10{f_1} - 20$

$ \Rightarrow 10{f_1} + 10{f_2} - {f_1}{f_2} = + 20$

For f₁ = 18 cm and f₂ = 20 cm, the above equation satisfies.

When object is at 8 cm,

the image distance (v₁) = ${{F \times 4} \over {u - f}}$ = ${{5 \times 8} \over {8 - 5}}$ = $-$ ${{40} \over 3}$ cm

When object is at 12 cm,

the image distance (v₂) = ${{f \times 4} \over {u - f}} = {{5 \times 12} \over {12 - 5}} = $ $-$ ${{60} \over 7}$ cm

$\therefore\,\,\,\,$ Separation = $\left| {{\upsilon _1} - {\upsilon _2}} \right|$ = ${{40} \over 3} - {{60} \over 7}$ = ${{100} \over {21}}\,$ cm

When plane durface is silvered then focus length of lens,

f₁ = ${R \over {2\left( {\mu - 1} \right)}}$

and when curver surface is silvered then focus length of lens,

f₂ = ${R \over {2\mu }}$

$\therefore\,\,\,\,$ ${{{f_1}} \over {{f_2}}} = {R \over {2\left( {\mu - 1} \right)}} \times {{2\mu } \over R}$

${\mu \over {\left( {\mu - 1} \right)}}$

given f₁ = 28 and f₂ = 10

$\therefore\,\,\,\,$ ${{28} \over {10}} = {\mu \over {\mu - 1}}$

$ \Rightarrow $$\,\,\,\,$ 28$\mu $ $-$ 28 = 10 $\mu $

$ \Rightarrow $$\,\,\,\,$ 18 $\mu $ = 28

$ \Rightarrow $$\,\,\,\,$ $\mu $ = ${{28} \over {18}}$ = 1.55

Consider the point F, the focus of the concave mirror. All rays emanating from F will be reflected back parallel to the principal axis PF. Thus, virtual image F' of the point F is formed at $\infty$.

Consider the point L at an object distance u = $-$ | f | / 2. For convenience, we use symbol $\widehat f$ for | f | i.e., $\widehat f$ = | f |. Use mirror formula, 1/v + 1/u = 1/f, with f = $-$ $\widehat f$ to get the image distance v = $\widehat f$. Thus, the virtual image L' of the point L is formed on the left side of the mirror at a distance $\widehat f$.

The image of a point on LF will lie on the line L'F'. Now, consider the point M. Its object distance is u = $-$ $\widehat f$/2. Thus, the virtual image M' of the point M is also formed at a distance $\widehat f$ from the mirror. The distance of M' above the principal axis is given by h_M' = $-$ (v/u) (LM) = $-$ ($-$2) ($\widehat f$/2) = $\widehat f$ (magnification by a spherical mirror).

Now consider a point N on the line segment FM. Let N₁ be the projection of N on the principal axis PF such that FN₁ = x. The object distance is u = $-$ ($\widehat f$ $-$ x). The mirror formula gives the image distance v = $\widehat f$ ($\widehat f$ $-$ x) / x. The distance of the image N' above the principal axis is given by h_N' = $-$ (v/u) (N₁N) = ($\widehat f$/x) (x) = $\widehat f$ (it is independent of x). Since, N is an arbitrary point on FM, the image of all points on FM will be formed at a distance $\widehat f$ above the optic axis.

The given optical situation is depicted in the following ray diagram:

In this case, the image forms 55 cm behind the convex mirror and then the reflection takes place due to the mirror image that forms at a distance (v) behind the mirror. Now, this image acts as an object for lens and the final image forms at a distance of 20 m from the lens if the focal length of convex mirror is as follows:

${{55} \over 2} = 27.5$ cm

Refractive index,

n₁₂ = ${{{n_D}} \over {{n_R}}}$ = ${1 \over {\sin {\theta _c}}}$

$ \Rightarrow $ $\,\,\,$ sin$\theta $_c = ${{{n_R}} \over {{n_D}}}$ . . . . . (1)

From Snell's law,

n_D sinA = n_R sin r

${{{n_R}} \over {{n_D}}}$ = ${{\sin A} \over {\sin \left( {{{90}^o} - A} \right)}}$    [as    r = 90^o $-$ A]

$ \Rightarrow $$\,\,\,$ ${{{n_R}} \over {{n_D}}} = \tan A$

$\therefore\,\,\,$ From (1) we get,

tan A = sin $\theta $_c

$ \Rightarrow $$\,\,\,$ A = tan^$-$1 (sin $\theta $_c)

As parallel beam incident on diverging lens so the image will be formed at the focus of diverging lens.

$ \therefore $ v = –25 cm
The image formed by diverging lens is used as an object for converging lens.

So for converging lens

u = –25 – 15 = –40 cm,

f = 20 cm

So Final image formed by converging lens

${1 \over {20}} = {1 \over V} - {1 \over { - 40}}$

$ \Rightarrow $ V = 40 cm

V is positive so image will be real and will form at right side of converging lens at 40 cm.

We discuss the options as follows:

$\bullet$ For option (A) : For the angle of incidence i₁ = A, we have minimum deviation and the ray inside the prism is parallel to the base of the prism.

Hence, option (A) is correct.

$\bullet$ For option (C) : The angle of deviation is given as

$\delta$ = i₁ + i₂ $-$ A

At minimum deviation, i₁ = i₂ and r₁ = r₂. Therefore,

i₁ = A ....... (1)

Also, we know that

r₁ + r₂ = A

As r₁ = r₂, we get

2r₁ = A $\Rightarrow$ r₁ = ${A \over 2}$

From Eq. (1), we get

${r_1} = {{{i_1}} \over 2}$

Hence, option (C) is correct.

According to Snell's law, we have

${{\sin {i_2}} \over {\sin {r_2}}} = \mu $

For emergent ray to be tangential to the surface, we have

i₂ = 90$^\circ$

$\Rightarrow$ sin90$^\circ$ = $\mu$sinr₂ $\Rightarrow$ $\mu$sinr₂ = 1

sinr₂ = ${1 \over \mu }$ $\Rightarrow$ r₂ = sin^$-$1$\left( {{1 \over \mu }} \right)$ ....... (2)

From the relation r₁ + r₂ = A, we get

r₁ = A $-$ r₂

$\Rightarrow$ sinr₁ = sin(A $-$ r₂)

Using sin(a $-$ b) = sinacosb $-$ sinbcosa, we have

$\Rightarrow$ sinr₁ = sinAcosr₂ $-$ sinr₂cosA ......... (3)

Using Eq. (2), we have

sinr₂ = ${{1 \over \mu }}$

Squaring it, we get

sin²r₂ = ${{1 \over {{\mu ^2}}}}$

Using the relation sin²x + cos²x = 1, we get

sin²x = 1 $-$ cos²x

$\Rightarrow$ 1 $-$ cos² r₂ = ${{1 \over {{\mu ^2}}}}$

$\Rightarrow$ cos² r₂ = 1 $-$ ${{1 \over {{\mu ^2}}}}$

That is,

cosr₂ = $\sqrt {1 - {1 \over {{\mu ^2}}}} $

Substituting the values of sinr₂ and cosr₂ in Eq. (3), we get

sinr₁ = sin A$\sqrt {1 - {1 \over {{\mu ^2}}}} $ $-$ ${{1 \over \mu }}$cos A

According to Snell's law, we have

${{\sin {i_1}} \over {\sin {r_1}}} = \mu $

That is,

$\sin {i_1} = \mu \sin {r_1} = \mu \left( {\sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - {1 \over \mu }\cos A} \right)$

$\sin {i_1} = \mu \left( {\sin A{{\sqrt {{\mu ^2} - 1} } \over \mu } - {{\cos A} \over \mu }} \right)$

$\sin {i_1} = \sin A\sqrt {{\mu ^2} - 1} - \cos A$ ....... (4)

Now, at minimum deviation, we have A = i₁ and r₁ = ${A \over 2}$. Thus, from Snell's law, we get

$\mu = {{\sin {i_1}} \over {\sin {r_1}}} = {{\sin A} \over {\sin (A/2)}}$

Using $\sin x = 2\sin {x \over 2}\cos {x \over 2}$, we get

$\mu = {{2\sin A/2\cos A/2} \over {\sin A/2}} \Rightarrow \mu = 2\cos {A \over 2}$ ....... (5)

Now, Eq. (4) becomes

$\sin {i_1} = \sin A\sqrt {{{\left( {2\cos {A \over 2}} \right)}^2} - 1 - \cos A} $

$ = \sin A\sqrt {4{{\cos }^2}{A \over 2} - 1} - \cos A$

${i_1} = {\sin ^{ - 1}}\left[ {\sin A\sqrt {4{{\cos }^2}{A \over 2} - 1} - \cos A} \right]$

Hence, option (D) is correct.

$\bullet$ For option (B) : From Eq. (5), we have

$\mu = 2\cos {A \over 2}$

$\cos {A \over 2} = {\mu \over 2}$

$ \Rightarrow {A \over 2} = {\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$

$ \Rightarrow A = 2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$

Hence, option (B) is incorrect.

The refractive index is

$\mu = {{{\mathop{\rm Real}\nolimits} \,depth} \over {Apparent\,depth}} = {{{\mathop{\rm Reading}\nolimits} \,3 - Reading\,1} \over {{\mathop{\rm Reading}\nolimits} \,3 - Reading\,2}}$

Therefore, the minimum of three readings are required.

Using mirror formula.

${1 \over v} + {1 \over { - 4}} = {1 \over { - 5}}$

$ \Rightarrow \,\,\,{1 \over v} = {1 \over 4} - {1 \over 5} = {1 \over {20}}$

$ \Rightarrow \,\,\,v = 20\,$ cm

$ \therefore $   Image distance is 20 cm



I₁ acts as object for plane glass surface.

$ \therefore $   Appartent depth =  ${{R + v} \over \mu } = {{30} \over {1.5}} = 20$ cm

$ \therefore $   Position of the image of the air bubble is 20 cm below the flat surface.

For convex lens,

${1 \over {30}} = {1 \over {{V_1}}} + {1 \over {60}}$

$ \Rightarrow $  ${1 \over {{V_1}}} = {1 \over {60}}$

$ \Rightarrow $  ${V_1} = 60$ cm

For concave lens,

${1 \over { - 120}} = {1 \over {{V_2}}} - {1 \over {40}}$

$ \Rightarrow $  ${1 \over {{V_2}}} = {1 \over {60}}$

$ \Rightarrow $  ${V_2} = 60$ cm

object will be at 60 cm distance from concave lens.

As mirror distance is 50 cm from concave lens. So virtual object is 10 cm behind mirror.

So, real image is 10 cm infront of mirror.

Distance of image from concave lens $=$ 50 $-$ 10 $=$ 40 cm. Which is same for the real object.

So similar situation will happen and final image will form at the real object it self.

For lens :

u₁ = $-$ (32.2 $-$ 22.2) cm

= $-$ 10 cm

v₁ = (71.2 $-$ 32.2) cm

= 39 cm

$ \therefore $   ${1 \over {{f_1}}}$ = ${1 \over {{v_1}}} - {1 \over {{u_1}}}$

= ${1 \over {39}}$ + ${1 \over {10}}$

= ${{49} \over {390}}$

$ \therefore $   f₁ = 7.8 cm

For mirror :

R = (71.2 $-$ 45.8) cm

= 25.4 cm

$ \therefore $   f₂ = ${R \over 2}$ = ${{25.4} \over 2}$ = 12.7 cm

We know that $i + e - A = \delta $

${35^ \circ } + {79^ \circ } - A = {40^ \circ }$

$\therefore$ $A = {74^ \circ }$

But $\mu = {{\sin \left( {{{A + {\delta _m}} \over 2}} \right)} \over {\sin A/2}} = {{\sin \left( {{{74 + \delta } \over 2}} \right)} \over {\sin {{74} \over 2}}}$

$ = {5 \over 3}\sin \left( {{{37}^ \circ } + {{{\delta _m}} \over 2}} \right)$

${\mu _{\max }}\,$ can be ${5 \over 3}.$ That is ${\mu _{\max }}$ is less than ${5 \over 3} = 1.67$

But ${\delta _m}$ will be less than ${40^ \circ }$ so

$\mu < {5 \over 3}\sin \,{57^ \circ } < {5 \over 3}\,\,$ $\sin {60^ \circ } \Rightarrow \mu = 1.45$

A telescope magnifies by making the object appearing closer.

For refraction through plano - Convex lens,

$ m=-2=\frac{v}{u} $

$ \begin{array}{ll} \Rightarrow v=-2 u \\\\ \text { As, } u=-30 \mathrm{~cm} \\\\ \Rightarrow v=60 \mathrm{~cm} \end{array} $

Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$ \begin{aligned} \frac{1}{60}-\frac{1}{-30} & =\frac{1}{f} \\\\ \Rightarrow f =20 \mathrm{~cm} \end{aligned} $



By lens maker's formula,

$ \begin{aligned} \frac{1}{f} & =(n-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\\\ \Rightarrow \frac{1}{20} & =(n-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\infty}\right)=\frac{n-1}{\mathrm{R}} \\\\ \Rightarrow n & =1+\left(\frac{\mathrm{R}}{20}\right) ........(i) \end{aligned} $

for poor reflection from convex surface,

$ \begin{aligned} & u=-30 \mathrm{~cm}, \\\\ & v=10 \mathrm{~cm}, f=\mathrm{R} / 2 \end{aligned} $



Using mirror formula,

$ \begin{aligned} \frac{1}{v}+\frac{1}{u} & =\frac{1}{f}=\frac{2}{\mathrm{R}} ; \\\\ \Rightarrow \frac{1}{10}+\frac{1}{-30} & =\frac{2}{\mathrm{R}} \end{aligned} $

$ \begin{array}{llrl} \Rightarrow \frac{2}{\mathrm{R}} =\frac{2}{30} \\\\ \therefore \mathrm{R} =30 \mathrm{~cm} \end{array} $

Substitute $R=30 \mathrm{~cm}$ in equation (i) to get $n=$ 2.5 for $f=20 \mathrm{~cm}$ (inverted image) and $n=1.5$ for $f=60 \mathrm{~cm}$ (erect image).

From Snell's law,

$ \begin{aligned} n_1 \sin \theta_i=n\left(z_1\right) \sin \theta_{z_1}=n\left(z_2\right) \sin \theta_{z_2} & =\ldots \ldots \\\\ \ldots & =n(d) \sin \theta_d=n_2 \sin \theta_f \end{aligned} $

The lateral displacement of a ray by a parallel slab depends on the angle of incidence $\theta_i$, refractive index $n_1$, slab thickness $d$ and the refractive index of the slab $n$ (it is independent of $n_2$ because $l$ is measured before the start of medium 2, there is no effect of $n_2$ on $l$ ). Consider the parallel slab shown in the figure.


By Snell's law, $n_1 \sin i_1=n \sin r_1$ and $n \sin i_2=$ $n_2 \sin r_2$. By geometry, $i_2=r_1$ and the lateral displacement is given by

$ l=d \frac{\sin r_1}{\cos r_1}=\frac{d n_1 \sin i_1}{\sqrt{n^2-n_1^2 \sin ^2 i_1}} . $

Hence, $l$ is independent of $n_2$ but it is dependent of $n(z)$.

For convex lens, u = $-$50 cm, f = 30 cm

$\therefore$ ${1 \over v} = {1 \over f} + {1 \over u} = {1 \over {30}} - {1 \over {50}} = {{5 - 3} \over {150}} = {2 \over {150}}$ or v = 75 cm

Image formed by convex lens acts as a virtual object for mirror.

If we consider that the mirror is not tilted, then for mirror,

${u_1} = 75 - 50 = 25$ cm, ${f_1} = {R \over 2} = {{100\,cm} \over 2} = 50$ cm

$\therefore$ ${1 \over {{v_1}}} = {1 \over {{f_1}}} - {1 \over {{u_1}}} = {1 \over {50}} - {1 \over {25}} = {{ - 1} \over {50}}$ or ${v_1} = 50$ cm

So, the co-ordinates of final image formed (I₂) if the mirror is not tilted are (0 cm, 0 cm)

If mirror is not tilted then this ray starts from the object, refracts through the lens without deviation, incident normally on the mirror and finally comes to the image I₂ after reflection from the mirror. If the mirror is tilted by an angle $\theta$ then this ray is incident on the mirror at an angle of reflection $\theta$ and finally forms a new image at I'₂ (see figure). Thus, the axis on which image lies makes an angle 2$\theta$ = 60$^\circ$ with the principal axis of the lens. Also, the image distance will remain same upto the first order of approximation (image distance will remain exactly same in case of plane mirror). Thus, (x, y) coordinates of the image are

$x = 50 - 50\cos 60^\circ = 25$ cm,

$y = 50\sin 60^\circ = 25\sqrt 3 $ cm.

Applying Snell's law at M,

$n = {{\sin \alpha } \over {\sin {r_1}}} \Rightarrow \sqrt 2 = {{\sin 45^\circ } \over {\sin {r_1}}}$

$ \Rightarrow \sin {r_1} = {{\sin 45^\circ } \over {\sqrt 2 }} = {{1/\sqrt 2 } \over {\sqrt 2 }} = {1 \over 2}$

${r_1} = 30^\circ $

$\sin {\theta _c} = {1 \over n} = {1 \over {\sqrt 2 }} \Rightarrow {\theta _c} = 45^\circ $

Let us take ${r_2} = {\theta _c} = 45^\circ $ for just satisfying the condition of TIR.

In $\Delta PNM$,

$\theta + 90 + {r_1} + 90 - {r_2} = 180^\circ $

or $\theta = {r_2} - {r_1} = 45^\circ - 30^\circ = 15^\circ $

Note If $\alpha$ = 45$^\circ $ (the given value). Then, r₁ > 30$^\circ $ (the obtained value)

${r_2} > {\theta _c}$ (as r₂ $-$ r₁ = $\theta$ or r₂ = $\theta$ + r₁)

or TIR will take place. So, for taking TIR under all conditions $\alpha$ should be greater than 45$^\circ$ or this is the minimum value of $\alpha$.

When ${r_2} = C,\,\angle {N_2}Rc = {90^ \circ }$

Where $C = $ critical angle

As $\sin C = {1 \over v} = \sin {r_2}$


Applying snell's law at $'R'$

$\mu \sin {r_2} = 1\sin {90^ \circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

Applying snell's law at $'Q'$

$1 \times \sin \theta = \mu \sin {r_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

But ${r_1} = A - {r_2}$

So, $\sin \theta = \mu \sin \left( {A - {r_2}} \right)$

$\theta = \mu \sin A\cos {r_2} - \cos \,A\,\,\,\,\,...\left( {iii} \right)$ [using $(i)$]

From $(1)$

$\cos \,{r_2} = \sqrt {1 - {{\sin }^2}{r_2}} = \sqrt {1 - {1 \over {{\mu ^2}}}} \,\,\,\,\,\,\,...\left( {iv} \right)$

By eq.$(iii)$ and $(iv)$

$\sin \theta = \mu \sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - \cos A$

on further solving we can show for ray not to transmitted through face $AC$

$\theta = {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right)} \right.} \right]$

So, for transmission through face $AC$

$\theta > {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right.} \right)} \right]$

Let the whole structure be placed in the medium of refractive index n₀. From geometry, if the angle of incidence at Q is $\theta$ then the angle of refraction at P is 90$^\circ$ $-$ $\theta$.

The ray will undergo total internal reflection at Q if the angle of incidence at Q is greater than or equal to the critical angle i.e.,

$\theta$ $\ge$ $\theta$_c = sin^$-$1(n₂ / n₁). ........ (1)

Apply Snell's law for refraction at P to get

n₀ sin i = n₁ sin(90$^\circ$ $-$ $\theta$) = n₁ cos$\theta$

= ${n_1}\sqrt {1 - {{\sin }^2}\theta } $. ...... (2)

From equation (2), the angle of incidence is maximum (i = i_m) when $\theta$ is minimum i.e., when $\theta$ = $\theta$_c (from equation (1)). Thus, the numerical aperture is given by

$NA = \sin {i_m} = {{{n_1}\sqrt {1 - {{\sin }^2}{\theta _c}} } \over {{n_0}}} = {{\sqrt {n_1^2 - n_2^2} } \over {{n_0}}}$ ........ (3)

Substitute ${n_1} = \sqrt {45} /4$ and ${n_2} = 3/2$ in equation (3) to get the numerical aperture for the structure S₁ is

$N{A_1} = {{\sqrt {45/16 - 9/4} } \over {{n_0}}} = {3 \over {4{n_0}}}$ ...... (4)

Similarly, substitute n₁ = 8/4 and n₂ = 7/5 in equation (3) to get the numerical aperture for the structure S₂ as

$N{A_2} = {{\sqrt {64/25 - 49/25} } \over {{n_0}}} = {{\sqrt {15} } \over {5{n_0}}}$ ...... (5)

In case (A), substitute n₀ = 4/3 in equation (4) to get NA₁ = 9/16 and substitute n₀ = 16/3$\sqrt15$ in equation (5) to get NA₂ = 9/16.

In case (B), substitute n₀ = 6/$\sqrt15$ in equation (4) to get NA₁ = $\sqrt15$/8 and substitute n₀ = 4/3 in equation (5) to get NA₂ = 3$\sqrt15$/20.

In case (C), substitute n₀ = 1 in equation (4) to get NA₁ = 3/4 and substitute n₀ = 4/$\sqrt15$ equation (5) to get NA₂ = 3/4.

In case (D), substitute n₀ = 1 in equation (4) to get NA₁ = 3/4 and substitute n₀ = 4/3 equation (5) to get NA₂ = 3$\sqrt15$/20.

$ \sin i_m=n_1 \sin \left(90-\theta_c\right) $

$ \Rightarrow \sin i_m=n_1 \cos \theta_c $

$ \begin{aligned} \Rightarrow N A & =n_1 \sqrt{1-\sin ^2 \theta_c} \\\\ & =n_1 \sqrt{1-\frac{n_2^2}{n_1^2}}=\sqrt{n_1^2-n_2^2} \end{aligned} $

Substituting the values we get,

$ N A_1=\frac{3}{4} $

$ \text { and } N A_2=\frac{\sqrt{15}}{5}=\sqrt{\frac{3}{4}} $

and $ N A_2 < N A_1 $

Therefore, the numerical aperture of combined structure is equal to the lesser of the two numerical aperture, which is $N A_2$.

Apply the formula for the refraction at the spherical surface of S₁, ${{{\mu _2}} \over v} - {{{\mu _1}} \over u} = {{{\mu _2} - {\mu _1}} \over R}$, to get

${{1.0} \over v} - {{1.5} \over {( - 50)}} = {{1 - 1.5} \over {( - 10)}}$.

Solve to get the image distance v = 50 cm (point Q in the figure).

The image Q acts as an object for refraction at the spherical surface of S₂. The object distance is $u = - (d - 50)$. The image distance is v = $\infty$ (because rays are parallel in S₂). Apply the formula for refraction at the spherical surface of S₂, to get

${{1.5} \over \infty } - {1 \over { - (d - 50)}} = {{1.5 - 1} \over {(10)}}$.

Solve to get d = 70 cm.

We are tasked with finding the relative refractive index given the relation between the critical angle and the Brewster’s angle.

Step 1: Recall the definitions

• Critical angle (from denser medium $n_1$ to rarer medium $n_2$, with $n_1 > n_2$):

$ \sin \theta_c = \frac{n_2}{n_1} = \frac{1}{\mu} $

where $\mu = \frac{n_1}{n_2}$ is the relative refractive index $> 1$.

• Brewster’s angle:

$ \tan \theta_B = \frac{n_2}{n_1} = \frac{1}{\mu} $

Step 2: Given condition

$ \frac{\sin \theta_c}{\sin \theta_B} = \eta = 1.28 $

So:

$ \frac{1/\mu}{\sin \theta_B} = 1.28 $

Step 3: Relation between $\theta_B$ and $\mu$

Since $\tan \theta_B = \tfrac{1}{\mu}$,

$ \sin \theta_B = \frac{\tan \theta_B}{\sqrt{1+\tan^2 \theta_B}} = \frac{\tfrac{1}{\mu}}{\sqrt{1+\tfrac{1}{\mu^2}}} = \frac{1}{\sqrt{\mu^2+1}} $

Step 4: Substitute into ratio

$ \frac{\sin \theta_c}{\sin \theta_B} = \frac{\tfrac{1}{\mu}}{\tfrac{1}{\sqrt{\mu^2+1}}} = \frac{\sqrt{\mu^2+1}}{\mu} $

Given this equals 1.28:

$ \frac{\sqrt{\mu^2+1}}{\mu} = 1.28 $

Step 5: Solve for $\mu$

$ \sqrt{\mu^2+1} = 1.28 \mu $

Square both sides:

$ \mu^2 + 1 = 1.6384 \, \mu^2 $

$ 1 = (1.6384 - 1)\mu^2 $

$ 1 = 0.6384 \mu^2 $

$ \mu^2 = \frac{1}{0.6384} \approx 1.565 $

$ \mu \approx 1.25 $

Step 6: Find the relative index of the two media in the question’s terms

Here we defined $\mu = \frac{n_1}{n_2} \approx 1.25$.

But the relative refractive index of the two media usually means $\frac{n_2}{n_1}$ (rarer/denser), since that's how it appears in options (<1).

$ \frac{n_2}{n_1} = \frac{1}{\mu} \approx \frac{1}{1.25} = 0.8 $

✅ Final Answer:

$ \boxed{0.8} $

Correct Option: C

By Lens maker's formula for convex lens

${1 \over f} = \left( {{\mu \over {{\mu _L}}} - 1} \right)\left( {{2 \over R}} \right)$

for, $\mu {L_1} = {4 \over 3},{f_1} = 4R$

for $\mu {L_2} = {5 \over 3},{f_2} = - 5R$

$ \Rightarrow {f_2} = \left( - \right)ve$

For critical angle ${\theta _c},$

$\sin {\theta _c} = {1 \over \mu }$

For greater wavelength or lesser frequency $\mu $ is less.



So, critical angle would be more, So, they will not suffer reflection and come out at angles less then ${90^ \circ }.$

As the film has R₁ = R₂ = R (say), its focal length

${1 \over f} = ({n_1} - 1)\left( {{1 \over R} - {1 \over R}} \right)$


${1 \over f} = 0$ or $f = \infty $. It will not cause refraction.

Refraction will take place at air and glass cylinder. Given u = $\infty$

$ - {{{n_1}} \over u} + {{{n_2}} \over v} = {{{n_2} - {n_1}} \over R}$

$ - {{{n_1}} \over \infty } + {1.5 \over {{f_1}}} = {{1.5 - 1} \over R}$ or ${f_1} = 3R$

For glass-air refraction, we have

$ - {{{n_2}} \over u} + {{{n_1}} \over v} = {{{n_1} - {n_2}} \over R}$

$ - {{{n_2}} \over \infty } + {1 \over {{b_2}}} = {{1 - 1.5} \over R}$

or ${f_2} = - 2R$ or $\left| {{f_2}} \right| = 2R$.

In $\Delta$SAB

$\tan \theta = {{AB} \over {AS}} = {{11.54/2} \over {10}}$

tan$\theta$ = 0.577

$\theta$ = 30$^\circ$

Using ${{\sin {\theta _l}} \over {\sin 90}} = {{{n_l}} \over {{n_B}}}$, we get

$\sin {\theta _l} = {{{n_l}} \over {{n_B}}}$ or ${n_l} = {n_B} \times \sin 30 = 2.72 \times {1 \over 2} = 1.36$

The lens maker's formula gives focal length of the bi-convex lens as

${1 \over {{f_1}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right]$

$ = (1.5 - 1)\left[ {{1 \over r} - {1 \over { - r}}} \right] = {1 \over r}$ ..... (1)

The focal length of a plano-convex (also convex-plane) lens is given by

${1 \over {{f_2}}} = (1.5 - 1)\left[ {{1 \over \infty } - {1 \over { - r}}} \right] = {1 \over {2r}}$ ...... (2)

and that of a plano-concave (also concave-plano) lens is given by

${1 \over {{f_3}}} = (1.5 - 1)\left[ {{1 \over \infty } - {1 \over r}} \right] = - {1 \over {2r}}$ ..... (3)

Case P:

Therefore, ${1 \over {{f_{eff}}}} = {1 \over r} + {1 \over r}$

${f_{eff}} = {r \over 2}$

(P) $\to$ (2)

Case Q :

${1 \over {{f_{eff}}}} = {1 \over {2r}} + {1 \over {2r}}$

${f_{eff}} = r$

(Q) $\to$ (4)

Case R :

${1 \over {{f_{eff}}}} = - {1 \over {2r}} - {1 \over {2r}}$

${f_{eff}} = - r$

(R) $\to$ (3)

Case S : ${1 \over {{f_{eff}}}} = {1 \over r} - {1 \over {2r}} = {1 \over {2r}}$

${f_{eff}} = 2r$

(S) $\to$ (1)

For the prism as the angle of incidence $(i)$ increase, the angle of deviation $\left( \delta \right)$ first decreases goes to minimum value and then increases.

$\therefore$ $n = {{Velocity\,\,of\,\,light\,\,in\,\,vacuum} \over {Velocity\,\,of\,\,light\,\,in\,\,medium}}$

$\therefore$ $n = {3 \over 2}$

${3^2} + {\left( {R - 3mm} \right)^2} = {R^2}$

$ \Rightarrow {3^2} + {R^2} - 2R\left( {3mm} \right) + {\left( {3mm} \right)^2} = {R^2}$

$ \Rightarrow R \approx 15\,cm$

${1 \over f} = \left( {{3 \over 2} - 1} \right)\left( {{1 \over {15}}} \right) \Rightarrow f = 30\,cm$

For e $\to$ f

$\mu$₂ > $\mu$₁, as ray bends towards the normal.

$\mu$₂ > $\mu$₃, as ray bends away from the normal.

P $\to$ 2

For e $\to$ g

$\mu$₁ = $\mu$₂ as there is no deviation.

Q $\to$ 3

For e $\to$ h

$\mu$₂ < $\mu$₁, as ray bends away from the normal.

$\mu$₂ > $\mu$₃, as ray bends away from the normal.

Also, $\mu$₁ < $\sqrt2$$\mu$₂ (No total internal reflection)

R $\to$ 4

For e $\to$ i

Total internal reflection takes place

$\therefore$ sin45$^\circ$ > sinC

But $\sin C = {{{\mu _2}} \over {{\mu _1}}}$

$\therefore$ ${1 \over {\sqrt 2 }} > {{{\mu _2}} \over {{\mu _1}}} \Rightarrow {\mu _1} > \sqrt 2 {\mu _2}$

$S \to 1$

The speed of light in a medium is given by the equation :

$ \frac{C}{V}=\frac{\lambda_0}{\lambda}=\mu $

Where :

• V is the velocity of light in the medium
• λ is the wavelength of light in the medium
• λ₀ is the wavelength in free space
• μ is the refractive index of the medium

Given :

$ \lambda=\frac{2}{3} \lambda_0 $

Using the given data :

$\begin{aligned} u & = -24 \text{ m},\\\\ h_i & = \frac{1}{3}h_0\end{aligned}$

To find the magnification (m) :

$ m = \frac{h_i}{h_0} = \frac{1}{3} = \frac{v}{u} $

Hence,

$ v = \frac{u}{3} $

Given :

$ v = 8 \text{ m},\ u = -24 \text{ m} $

Applying the lens formula :

$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Since the plane surface has an infinite radius :

$ \frac{1}{\infty} = 0 $

Therefore :

$\frac{1}{8} - \frac{1}{-24} = (1.5 - 1)\left(\frac{1}{R} - 0\right)$

$\Rightarrow \frac{1}{6} = \frac{0.5}{R}$

Solving for R :

$ \Rightarrow R = 3 \text{ m} $

Let ${\widehat v_i} = {1 \over 2}\left( {\widehat i + \sqrt 3 \widehat j} \right)$ and ${\widehat v_r} = {1 \over 2}\left( {\widehat i - \sqrt 3 \widehat j} \right)$ be the unit vectors along the incident and the reflected rays. From Snell's law, angle of incidence (say $\theta$) is equal to the angle of reflection. Thus, angle between $ - {\widehat v_i}$ and ${\widehat v_r}$ is 2$\theta$.

Using, dot products of vectors, we get

$\cos (2\theta ) = - {\widehat v_i}\,.\,{\widehat v_r} = 1/2$.

Hence, $\theta = 30^\circ $.

The focal length of the lens

${1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}$

$ = {{20 + 1} \over {240}} = {{21} \over {240}}$

$f = {{240} \over {21}}cm$

Shift $ = t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \right)$

$ = 1 \times {1 \over 3}$

Now $v' = 12 - {1 \over 3} = {{35} \over 3}cm$

Now the object distancce $u.$

${1 \over u} = {3 \over {35}} - {{21} \over {240}} = {1 \over 5}\left[ {{3 \over 7} - {{21} \over {48}}} \right]$

${1 \over u} = {1 \over 5}\left[ {{{48 - 49} \over {7 \times 16}}} \right]$

$u = - 7 \times 16 \times 5 = - 560cm = - 5.6\,m$

For meta-material, the refractive index is negative. Let n₁ is refractive index of air and n₂ is refractive index of meta-material.

$\therefore$ From Snell's law, ${{\sin {\theta _1}} \over {\sin {\theta _2}}} = {{{n_2}} \over {{n_1}}}$

Since, n₂ is negative, therefore $\theta$₂ is also negative. Hence, appropriate diagram (c) is correct.

Refractive index for a medium

$n = \left( {{c \over v}} \right)$

For meta-material, $n = \,|n|$

$\therefore$ $v = {c \over {|n|}}$

According to lens maker's formula

${1 \over f} = (n - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

For the first lens

n = 1.5, R₁ = + 14 cm, R₂ = $\infty$

$\therefore$ ${1 \over {{f_1}}} = (1.5 - 1)\left( {{1 \over {14}} - {1 \over \infty }} \right) = {{0.5} \over {14}}$

For the second lens,

n = 1.2, R₁ = $\infty$, R₂ = $-$14 cm

$\therefore$ ${1 \over {{f_2}}} = (1.2 - 1)\left( {{1 \over \infty } - {1 \over { - 14}}} \right) = {{0.2} \over {14}}$

The focal length of the bi-convex lens is

${1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} = {{0.5} \over {14}} + {{0.2} \over {14}} = {{0.7} \over {14}} = {1 \over {20}}$

According to thin lens formula

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

Here, u = $-$40 cm

$\therefore$ ${1 \over v} - {1 \over { - 40}} = {1 \over {20}}$ or ${1 \over v} = {1 \over {20}} - {1 \over {40}}$ or v = 40 cm