Geometrical Optics

${n_1} = {n_2} = n = {3 \over 2}$ for minimum deviation.

${r_m} = {A \over 2} = {{60} \over 2} = 30^\circ $ and $i = e = \theta $

$n = {{\sin \left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}} $

$\Rightarrow {3 \over 2} = {{\sin \theta } \over {\sin \left( {{{60} \over 2}} \right)}}$

$ \Rightarrow \sin \theta = {3 \over 4} \Rightarrow \cos \theta = {{\sqrt 7 } \over 4}$

If ${n_1} = n = {3 \over 2}$ and ${n_2} = n + \Delta n$

$e = \theta + \Delta \theta $

$(1)\sin \theta = n\sin 30^\circ $

$(n + \Delta n)\sin 30^\circ = (1)\sin (\theta + \Delta \theta )$

Solving ${1 \over 2}(\Delta n) = \sin (\theta + \Delta \theta ) - \sin \theta $

${{\Delta n} \over 2} = {{\sin (\theta + \Delta \theta ) - sin\theta } \over {\Delta \theta }} \times \Delta \theta $

${{d(\sin \theta )} \over {d\theta }} = \cos \theta $

${{\Delta n} \over 2} = (\cos \theta )(\Delta \theta ) \Rightarrow {{\Delta n} \over 2} = {{\sqrt 7 } \over 4}(\Delta \theta )$

$ \Rightarrow {{\Delta n} \over {\Delta \theta }} = {{\sqrt 7 } \over 2} = {{2.64} \over 2}$

${{\Delta n} \over {\Delta \theta }} = 1.34 > 1$

$\Rightarrow$ $\Delta$n > $\Delta$$\theta$, so option (a) is incorrect.

$\Delta n = (1.34)\Delta \theta $

$ \Rightarrow \Delta \theta \propto \Delta n \Rightarrow $ option (b) is correct.

${{\Delta n} \over {\Delta \theta }} = {{\sqrt 7 } \over 2} \Rightarrow {{2.8 \times {{10}^{ - 3}}} \over {\Delta \theta }} = {{\sqrt 7 } \over 2}$

$\Delta \theta = {{5.6 \times {{10}^{ - 3}}} \over {\sqrt 7 }} = \sqrt 7 \times 0.8 \times {10^{ - 3}}$

$\Delta \theta = (2.64 \times 0.8) \times {10^{ - 3}} = 2.11 \times {10^{ - 3}}$ rad

So, option (c) is correct.

$\sin \theta > {1 \over {{n_1}}}$ (Given)

i.e., $\sin {\theta _1} > {1 \over {{n_1}}}$

${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$

$\sin {\theta _2} = {{{n_1}\sin {\theta _1}} \over {{n_2}}}$

if ${n_1} = {n_2}$ then ${\theta _2} = {\theta _1}$

${n_2}\sin {\theta _2} = (1)\sin {\theta _3}$

$\sin {\theta _3} = {n_2}\sin {\theta _2}$

$\sin {\theta _3} = {n_1}\sin {\theta _1}$

$\sin {\theta _1} = {{\sin {\theta _3}} \over {{n_1}}} > {1 \over {{n_1}}}$

$\sin {\theta _3} > 1$

${\theta _3} > 90^\circ $

This means ray cannot enter air.

For ${n_1} > {n_2};\sin {\theta _1} = {{{n_2}} \over {{n_1}}}\sin {\theta _2} > {1 \over {{n_1}}}$

$\sin {\theta _2} > {1 \over {{n_2}}}$

For surface 2 - air interface

${n_2}\sin {\theta _2} = \sin {\theta _3}$

$\sin {\theta _2} = {{\sin {\theta _3}} \over {{n_2}}} > {1 \over {{n_2}}}$

${\theta _2} > 90^\circ $

It means ray is reflected back in medium-2


For surface 1 and surface 2 interface

${n_2}\sin {\theta _2} = {n_1}\sin {\theta _1}$

$\sin {\theta _{2C}} = {{{n_1}} \over {{n_2}}}$

${\theta _{2C}}$ : critical angle

For ray to enter medium - 1

${\theta _2} < {\theta _{2C}}$

$\sin {\theta _2} < \sin 2{\theta _C}$

${{{n_1}} \over {{n_2}}}\sin {\theta _1} < {{{n_1}} \over {{n_2}}}$

$\sin {\theta _1} < 1$

${\theta _1} < 90^\circ $, which is true.

Hence, ray enters medium - 1

For ${n_2} > {n_1}$

${{{n_2}} \over {{n_1}}}\sin {\theta _2} > {{{n_2}} \over {{n_1}}}$

$\sin {\theta _2} > {1 \over {{n_2}}}$

For surface 2 - air interface

${n_2}\sin {\theta _2} = \sin {\theta _3}$

$\sin {\theta _2} = {{\sin {\theta _3}} \over {{n_2}}} > {1 \over {{n_2}}}$

${\theta _2} > 90$

It means ray reflected back in medium - 2


${n_2}\sin {\theta _2} = {n_1}\sin {\theta _1}$

$\sin {\theta _1} = {{{n_2}} \over {{n_1}}}\sin {\theta _2}$

$\sin {\theta _{2C}} = {{{n_1}} \over {{n_2}}};{\theta _{2C}}$ $\to$ critical angle

For ray to enter medium - 1

${\theta _2} < {\theta _{2C}}$

$\sin {\theta _2} < \sin {\theta _{2C}}$

${{{n_1}} \over {{n_2}}}\sin {\theta _1} < {{{n_1}} \over {{n_2}}}$

$\sin {\theta _1} < 1$

${\theta _1} = 90^\circ $, which is true.

Hence, ray enters medium - 1

Let, n₂ = 1


${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2} \Rightarrow {n_2} = 1$

${n_1}\sin {\theta _1} = \sin {\theta _2}$

$\sin {\theta _1} = {{\sin {\theta _2}} \over {{n_1}}} > {1 \over {{n_1}}}$

$\sin {\theta _2} > 1 \Rightarrow {\theta _2} = 90^\circ $

$\therefore$ Ray is reflected back in medium.

Focal length of convex lens f₁,

Therefore ${1 \over {{f_1}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right] = (1.5 - 1)\left[ {{1 \over {20}} - \left( {{1 \over { - 20}}} \right)} \right]$

$\Rightarrow$ f₁ = + 20 cm

Focal length of concave lens f₂,

$\therefore$ ${1 \over {{f_2}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right]$

${1 \over {{f_2}}} = (1.5 - 1)\left[ { - {1 \over {20}} - {1 \over {20}}} \right] = {1 \over { - 20}}$

$\Rightarrow$ f₂ = $-$ 20 cm

For lens 1


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

$\Rightarrow$ v = $-$ 20 cm

${m_1} = {v \over u} = {{ - 20} \over { - 10}} = 2$

For lens 2

u = $-$ 30n cm, f = $-$ 20 cm,

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

v = $-$ 12 cm

${m_2} = {v \over u} = {{ - 12} \over { - 30}} = {2 \over 5}$

Net magnification,

$m = {m_1}{m_2} = 2 \times {2 \over 5} = {4 \over 5} = 0.8$

Assume refractive index = $\mu $_$l$

P = $\left( {{\mu _l} - 1} \right)\left( {{2 \over R}} \right)$ .....(1)

${3 \over 2}P = \left( {{\mu _l} - 1} \right)\left( {{1 \over {{R_1}}}} \right)$ ......(2)

from (1)/(2)

${P \over {{3 \over 2}P}} = {{\left( {{2 \over R}} \right)} \over {\left( {{1 \over {{R_1}}}} \right)}}$

$ \Rightarrow $ R₁ = ${R \over 3}$

Object is at 2f. So image will also be at 2f. (I₁).

Image(I₂) of I₁ will be 1 m behind mirror.

Now I₂ will be object for lens.

$ \therefore $ u = -3 m

f = + 0.5 m

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

= ${1 \over {0.5}} + {1 \over { - 3}}$

$ \Rightarrow $ v = ${3 \over 5}$ = 0.6 m

So total distance from mirror = 2 + 0.6 = 2.6 m and real image.

$v = {{uf} \over {u + f}}$

Case-I

If v = u

$ \Rightarrow $ f + u = f

$ \Rightarrow $ u = 0

Case-II

If u = $\infty $

then v = f

So you can see both of them are equal only when v = 0 and u = 0.

f = $ - {8 \over 2}$ = -4 cm

u = –10 cm

v = ?

Using mirror formula

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

$ \Rightarrow $ ${1 \over v} + {1 \over { - 10}} = {1 \over { - 4}}$

$ \Rightarrow $ v = ${{ - 20} \over 3}$

m = ${{ - \left( { - {{20} \over 3}} \right)} \over { - 10}}$ = $ - {2 \over 3}$

So image would be real, inverted and smaller than object.

${4 \over 3}$ sin $\theta $ = 1sin90^o

sin $\theta $ = ${3 \over 4}$

cos $\theta $ = ${{\sqrt 7 } \over 4}$

Surface area in solid angle d$\Omega $ = 2$\pi $R²(1 - cos $\theta $)

= 2$\pi $R²(1 - ${{\sqrt 7 } \over 4}$)

Percentage of light = ${{2\pi {R^2}\left( {1 - {{\sqrt 7 } \over 4}} \right)} \over {4\pi {R^2}}}$ $ \times $ 100%

= ${{{4 - \sqrt 7 } \over 8}}$ $ \times $ 100%

= 17%

$\theta $ = 1.22${\lambda \over a}$

Distance O₁O₂ = ($\theta $)d

= (1.22${\lambda \over a}$)d

= ${{\left( {1.22} \right)\left( {5500 \times {{10}^{ - 10}}} \right)\left( {4 \times {{10}^5}} \right) \times {{10}^3}} \over 5}$

= 5368 × 10^–2 m

= 53.68 m

D = ${{{t_1}} \over {{\mu _1}}} + {{{t_2}} \over {{\mu _2}}}$

= ${h \over {\sqrt 2 }} + {h \over {2\sqrt 2 }}$

= ${3 \over 4}h\sqrt 2 $

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

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

$ \Rightarrow $ v = ${{uf} \over {u - f}}$

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

as u approaches f, m approaches infinity

And at u = 2f, m = –1 or |m| = 1

For telescope

Tube length (L) = f₀ + f_e

and magnification (m) = ${{{f_0}} \over {{f_e}}}$ = 5

where f_o and f_e are focal length of objective and eyepiece.

$ \therefore $ f₀ + f_e = 60

$ \therefore $ f_o = 50 cm

f_e = 10 cm

n = $\sqrt {{\varepsilon _r}{\mu _r}} = \sqrt {3 \times {4 \over 3}} $ = 2

n sin c = 1 sin 90^o

sin c = ${1 \over 2}$

$ \Rightarrow $ c = 30^o

Using formula
${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

${1 \over {{f}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ ...(1)

and ${1 \over {{f_1}}} = \left( {{{1.5} \over {1.42}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ ...(2)

Dividing (1) by (2), we get

${{{f_1}} \over f} = {{0.5} \over {0.056}}$ = 8.93 $ \approx $ 9

Case 1 : Near – point adjustment

M.P = ${L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)$

$ \Rightarrow $ 375 = ${{150} \over 5}\left( {1 + {{250} \over {{f_e}}}} \right)$

$ \Rightarrow $ f_e = 21.7 mm $ \approx $ 22 mm

Case-2 : If final image is at inifinity

M.P = ${L \over {{f_0}}}\left( {{D \over {{f_e}}}} \right)$

$ \Rightarrow $375 = ${{150} \over 5}\left( {{{250} \over {{f_e}}}} \right)$

$ \Rightarrow $ f_e = 20 mm

${(R - h)^2} + {r^2} = {R^2}$

${R^2} + {h^2} - 2Rh + {r^2} = {R^2}$

${h^2} + {r^2} = 2Rh \Rightarrow {{{h^2} + {r^2}} \over {2h}} = R$ .....(i)

When r > > > h, then

$R = {{{r^2}} \over {2h}} \Rightarrow h = {{{\omega ^2}{r^2}} \over {2g}}$

$R = {{{r^2}2g} \over {2{\omega ^2}{r^2}}} \Rightarrow R = {g \over {{\omega ^2}}}$ ..... (ii)

${{{\mu _1}} \over v} - {{{\mu _2}} \over u} = {{{\mu _1} - {\mu _2}} \over R} $

$\Rightarrow {1 \over v} + {4 \over {3u}} = {{1 - {4 \over 3}} \over R}$

${1 \over v} = - \left( {{1 \over {3R}} + {4 \over {3(H - h)}}} \right)$

${1 \over v} = - \left( {{1 \over {3R}} + {4 \over {3H}}} \right)$

Putting the value of Eq. (ii) h < < H

${1 \over v} = - \left[ {{4 \over {3H}}\left( {1 + {{3H} \over 4}{{{\omega ^2}} \over {39}}} \right)} \right]$

$v = - \left[ {{{3H} \over 4}{{\left( {1 + {{{\omega ^2}H} \over 4}} \right)}^{ - 1}}} \right]$

By Snell’s law at AB

$\mu $₁ sin $\theta $ = $\mu $₂ sin $\alpha $

sin $\alpha $ = ${{{\mu _1}} \over {{\mu _2}}}\sin \theta $ ....... (1)

Critical angle at BC,

$\mu $₂ sin i_c = $\mu $₁ sin 90^o

$ \Rightarrow $ sin i_c = ${{{\mu _1}} \over {{\mu _2}}}$ ..... (2)

For Total internal reflection to occur,

90^o - $\alpha $ > i_c

Taking sin both side,

sin(90^o - $\alpha $) > sin i_c

$ \Rightarrow $ cos $\alpha $ > ${{{\mu _1}} \over {{\mu _2}}}$ (from equation 2)

$ \Rightarrow $ $\sqrt {1 - {{\mu _1^2} \over {\mu _2^2}}{{\sin }^2}\theta } $ > ${{{\mu _1}} \over {{\mu _2}}}$ (from equation 1)

Squaring both sides, we get

${1 - {{\mu _1^2} \over {\mu _2^2}}{{\sin }^2}\theta }$ > ${{{\mu _1^2} \over {\mu _2^2}}}$

$ \Rightarrow $ ${1 - {{\mu _1^2} \over {\mu _2^2}}}$ > ${{{\mu _1^2} \over {\mu _2^2}}{{\sin }^2}\theta }$

$ \Rightarrow $ ${{\mu _2^2} \over {\mu _1^2}} - 1$ > ${{{\sin }^2}\theta }$

$ \Rightarrow $ sin $\theta $ < $\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $

$ \Rightarrow $ $\theta < {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1} $

Light incident from particle P will be reflected at mirror.

u = 5cm, f = $m - {R \over 2} = - 20cm$

${1 \over v} + {1 \over u} = {1 \over f};$   ${v_1} = + {{20} \over 3}cm$

This image will act as object for light getting refracted at water surface.

So, object distance $d = 5 + {{20} \over 3} = {{35} \over 3}cm$

Below water surface.
After refraction, final image is at

$d' = d\left( {{{{\mu _2}} \over {{\mu _1}}}} \right) = \left( {{{35} \over 3}} \right)\left( {{1 \over {4/3}}} \right)$

= ${{35} \over 4} = 8.75\,cm$

$ \approx $ 8.8 cm

As the graph between magnification (m) and image distance (v) varies linearly, then
m = k₁v + k₂

$ \Rightarrow {v \over u} = {k_1}v + {k_2}$

$ \Rightarrow {1 \over u} = {k_1} + {{{k_2}} \over v}$

$ \Rightarrow {{{k_2}} \over v} - {1 \over u} = {k_1}$

Clearly, ${k_1} = {1 \over f}$ and k₂ = 1 here

$ \therefore f = {1 \over {slope\,of\,m - y\,graph}} = {b \over c}$

For 1^st lens ${1 \over {{f_1}}} = \left( {{{{\mu _1} - 1} \over 1}} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right) = {{{\mu _1} - 1} \over R}$

For 2^nd lens ${1 \over {{f_2}}} = \left( {{{{\mu _2} - 1} \over 1}} \right)\left( {{1 \over { - R}}} \right) - 0 = {{{\mu _2} - 1} \over R}$

${1 \over {{f_{eq}}}} = {1 \over {{f_1}}} + {1 \over {{f_2}}}$

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

$ \Rightarrow {f_{eq}} = {R \over {{\mu _1} - {\mu _2}}}$

From Snell’s law
1 sin 60° = $\mu $ sin 30°

$ \Rightarrow $ $\mu $ = $\sqrt 3 $

Optical path = AO + $\mu $(OB)

= ${a \over {\cos {{60}^o}}} + \sqrt 3 {b \over {\cos {{30}^o}}}$

= 2a + 2b

Magnification is 2

If image is real, ${x_1} = {{3f} \over 2}$

If image is virtual, ${x_2} = {f \over 2}$

${{{x_1}} \over {{x_2}}} = 3:1$

For image to form at object itself, says must retrace their path back to object. Hence must incident on mirror normally.

Case 1: Object will be at focus of lens
${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over { - R}}} \right) = {1 \over { - 18}}$ $ \Rightarrow $ R = 18 cm

Case 2: Retraction at 1^st surface:
${1 \over { - 27}} - {{1.5} \over {{V_1}}} = {{1 - 1.5} \over R}$    ... (i)

2^nd retraction:
${{1.5} \over {{V_1}}} - {\mu \over \infty } = {{15 - u} \over { - R}}$   ...(ii)

From (i) and (ii)
$\mu = {4 \over 3}$

$m = {f \over {f - u}}$

$5 = {{ - 40} \over { - 40 - u}};\,u = - 32cm$

Image formed by lens
${1 \over v} - {1 \over u} = {1 \over f};{1 \over v} + {1 \over {30}} = {1 \over {20}}$

v = +60 cm
If image position does not change even when mirror is removed it means image formed by lens is formed at centre of curvature of spherical mirror.
Radius of curvature of mirror = 80 – 60 = 20 cm.

$ \Rightarrow $ Focal length of mirror f = 10 cm for virtual image, object is to be kept between focus and pole.

$ \Rightarrow $ Maximum distance of object from spherical mirror for which virtual image is formed, is 10 cm.

If we approximate the angle $\theta _2$ as 30° initially then answer will be closer to 57000. but if we solve thoroughly, answer will be close to 55000.
So both the answers must be awarded. Detailed solution as following.

EXACT SOLUTION
By Snell’s law 1.sin 40° = (1.31) sin $\theta _2$

$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$

Now tan$\theta _2$
= ${{64} \over {\sqrt {{{\left( {131} \right)}^2} - {{\left( {64} \right)}^2}} }} = {{64} \over {\sqrt {13065} }} \approx {{64} \over {114.3}} = {d \over x}$

Now number of reflections

$ = {{2 \times 64} \over {114.3 \times 20 \times {{10}^{ - 6}}}} = {{64 \times {{10}^5}} \over {114.3}}$

$ \approx 55991 \approx 55000$

APPROXIMATE SOLUTION
By Snell’s law 1.sin 40° = (1.31)sin $\theta _2$

$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$

If assume $ \Rightarrow $ $\theta _2$ $\approx $ 30°

tan 30° = ${d \over x} \Rightarrow x = {\sqrt3} d$

Now number of reflections
$ = {t \over {\sqrt 3 d}} = {2 \over {\sqrt 3 \times 20 \times {{10}^{ - 6}}}} = {{{{10}^5}} \over {\sqrt 3 }}$

$ \approx 57735 \approx 57000$

In given system of lens and mirror, position of object O in front of lens is at a distance 2f. i.e. u = 2f = 40 cm

So, image (I₁) formed is real, inverted and at a distance, v = 2f = 2 $\times$ 20 = 40 cm, (behind lens) magnification, ${m_1} = {v \over u} = {{40} \over {40}} = 1$

Thus, size of image is same as that of that of object. This image (I₁) acts like a real object for mirror.

As object distance for mirror is

u = C = 2f = $-$ 20 cm

where, C = centre of curvature.

So, image (I₂) formed by mirror is at 2f.

$\therefore$ For mirror v = 2f = 2($-$ 10) = $-$ 20 cm

Magnification, ${m_2} = - {v \over u} = - {{( - 20)} \over {( - 20)}} = - 1$

Thus, image size is same as that of object.

The image I₂ formed by the mirror will act like an object for lens.

As the object is at 2f distance from lens, so image (I₃) will be formed at a distance 2f or 40 cm. Thus, magnification,

${m_3} = {v \over u} = {{40} \over {40}} = 1$

So, final magnification, $m = {m_1} \times {m_2} \times {m_3} = - 1$

Hence, final image (I₃) is real, inverted of same size as that of object and coinciding with object.

From ${1 \over F} = \left( {{\mu _{rel}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

Focal length of lens will change hence image disappears from the screen.

${1 \over F} = {1 \over {{f_1}}} + {1 \over {{f_2}}} = {{1 - {\mu _1}} \over R} + {{{\mu _2} - 1} \over R}$

3d

For first lens

${1 \over V} - {1 \over { - 20}} = {1 \over 5}$

V $=$ ${{20} \over 3}$

For second lens

V = ${{20} \over 3}$ $-$ 2 $=$ ${{14} \over 3}$

${1 \over V} - {1 \over {{{14} \over 3}}} = {1 \over { - 5}}$

V $=$ 70cm

i = e

r₁ = r₂ = ${A \over 2}$ = 30^o

by Snell's law

1 $ \times $ sin i = $\sqrt 3 \times {1 \over 2} = {{\sqrt 3 } \over 2}$

i = 60

From lens equation

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

${1 \over v} - {1 \over {\left( { - 20} \right)}} = {1 \over {\left( {.3} \right)}} = {{10} \over 3}$

${1 \over v} = {{10} \over 3} - {1 \over {20}}$

${1 \over v} = {{197} \over {60}};v = {{60} \over {197}}$

m = $\left( {{v \over u}} \right)$ = ${{\left( {{{60} \over {197}}} \right)} \over {20}}$

velocity of image wrt. to lens is given by

v_I/L = m²v_O/L

direction of velocity of image is same as that of object

v_O/L = 5 m/s

Since D_m = ($\mu $ $-$ 1) A & on increasing the wavelength, $\mu $ decreases & hence D_m decreases.