Geometrical Optics

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$

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

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 $u.$

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

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

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

From mirror formula

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

so, $\,\,\,{{dv} \over {dt}} = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$

$ \Rightarrow {{dv} \over {dt}} = - {\left( {{f \over {u - f}}} \right)^2}{{du} \over {dt}}$

$ \Rightarrow {{dv} \over {dt}} = {1 \over {15}}m/s$

Angle of incidence is given by

$\cos \left( {\pi - i} \right) = {{\left( {6\sqrt 3 \widehat i + 8\sqrt 3 \widehat j - 10\widehat k} \right).\widehat k} \over {20}}$

$ - \cos \,i = - {1 \over 2}$

$\angle i = {60^ \circ }$

From Snell's law, $\sqrt 2 \sin i = \sqrt 3 \sin r$

$ \Rightarrow $ $\angle r = {45^ \circ }$

In the medium, the refractive index will decreases from the axis forwards the periphery of the beam.

Therefore, the beam will move as one move from the axis to the periphery and hence the beam will converge.

The speed of light $(c)$ in a medium of refractive index $\left( \mu \right)$ is given by

$\mu = {{{c_0}} \over c},$ where ${c_0}$ is the speed of light in vacuum

$\therefore$ $c = {{{c_0}} \over \mu } = {{{c_0}} \over {{\mu _0} + {\mu _2}\left( I \right)}}$

As $I$ is decreasing with increasing radius, it is maximum

on the axis of the beam. Therefore, $c$ is minimum on the axis of the beam.

Here $u = - 2f,v = 2f$

As $|u|$ increases, $v$ decreases for $|u| > f.$ The graph between $|v|$ and $|u|$ is shown in the figure. A straight line passing through the origin and making an angle of ${45^ \circ }$ with the $x$-axis meets the experimental curve at $P\left( {2f,2f} \right).$

Applying Snell's law at $Q$

$n = {{\sin {{90}^ \circ }} \over {\sin C}} = {1 \over {\sin C}}$

$\therefore$ $\sin C = {1 \over n} = {{\sqrt 3 } \over 2}$

$\therefore$ $C = {60^ \circ }$

Applying Snell's Law at $P$

$n = {{\sin \theta } \over {\sin \left( {90 - C} \right)}}$

$ \Rightarrow \sin \theta = n \times \sin \left( {90 - C} \right);$ from $(1)$

$ \Rightarrow \sin \theta = n\,\cos C$

$\therefore$ $\theta = {\sin ^{ - 1}}\left[ {{2 \over {\sqrt 3 }} \times \cos {{60}^0}} \right]$

or, $\theta = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$

This graph obeys the lens equation

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

where $f$ is a positive constant for a given convex lens.

Power of combination is given by

$P = {P_1} + {P_2} = \left( { - 15 + 5} \right)D$ $\,\,\,\,\,\,\,\,\, = - 10D.$

Now, $P = {1 \over f} \Rightarrow f = {1 \over P} = {1 \over { - 10}}$ metre

$\therefore$ $f = - \left( {{1 \over {10}} \times 100} \right)cm = - 10\,cm.$

For a thin prism, $D = \left( {\mu - 1} \right)A$

Since ${\lambda _b} < {\lambda _r} \Rightarrow {\mu _r} < {\mu _b} \Rightarrow {D_1} < {D_2}$

$\sin {\theta _c} = {1 \over \mu } = {3 \over 4}$

or $\tan {\theta _c} = {3 \over {\sqrt {16 - 9} }} = {3 \over {\sqrt 7 }} = {R \over {12}}$



$ \Rightarrow R = {{36} \over {\sqrt 7 }}\,cm$

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

${1 \over {{f_m}}} = \left( {{{{\mu _g}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

${1 \over {{f_m}}} = \left( {{{1.5} \over {1.6}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Dividing $(i)$ by $(ii)$, ${{{f_m}} \over {{f_a}}} = \left( {{{1.5 - 1} \over {{{1.5} \over {1.6}} - 1}}} \right) = - 8$

${P_a} = - 5 = {1 \over {{f_a}}} \Rightarrow {f_a} = - {1 \over 5}$

$ \Rightarrow {f_m} = - 8 \times {f_a} = - 8 \times - {1 \over 5} = {8 \over 5}$

${P_m} = {\mu \over {{f_m}}} = {{1.6} \over 8} \times 5 = 1D$

The incident angle is ${45^ \circ }$

Incident angle $ > $ critical angle, $i > {i_c}$

$\therefore$ $\sin i > \sin {i_c}$ or $\sin \,45\, > \sin \,{i_c},$ $\sin {i_c} = {1 \over n}$

$\therefore$ $\sin \,{45^ \circ } > {1 \over n}$ or ${1 \over {\sqrt 2 }} > {1 \over n} \Rightarrow n > \sqrt 2 $

KEY CONCEPT : The focal length $\left( F \right)$ of the final mirror

is ${1 \over F} = {2 \over {f\ell }} + {1 \over {{f_m}}}$

Here ${1 \over {{f_\ell }}} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ = \left( {1.5 - 1} \right)\left[ {{1 \over \alpha } - {1 \over { - 30}}} \right] = {1 \over {60}}$

$\therefore$ ${1 \over F} = 2 \times {1 \over {60}} + {1 \over {30/2}} = {1 \over {10}}$

$\therefore$ $F=10cm$

The combination acts as a converging mirror. For the object to be of the same size of mirror,

$u = 2F = 20cm$

A real, inverted and enlarged image of the object is formed by the objective lens of a compound microscope.

When $\theta = {90^ \circ }$ then ${{360} \over \theta } = {{360} \over {90}} = 4$

is an even number. The number of images formed is given by

$n = {{360} \over \theta } - 1 = {{360} \over {90}} - 1 = 4 - 1 = 3$

KEY CONCEPT : When two plane mirrors are inclined a each other at an angle $\theta $ then the number of the images of a point object placed between the plane mirrors is

${{{{360}^ \circ }} \over \theta } - 1,\,\,if{{{{360}^ \circ }} \over \theta }$ is even

$\therefore$ Number of images formed

$ = {{{{360}^ \circ }} \over {{{60}^ \circ }}} - 1 = 5$

The resolving power (RP) of an optical instrument is inversely proportional to the wavelength (λ) of light used. So, if we denote the resolving powers corresponding to λ₁ and λ₂ as RP₁ and RP₂ respectively, we can express this relationship as :

RP $ \propto $ ${1 \over \lambda }$

Therefore, the ratio of the resolving powers corresponding to λ₁ and λ₂ is given by the inverse of the ratio of the wavelengths. In LaTeX notation, this relationship can be written as :

$\frac{RP_1}{RP_2} = \frac{\lambda_2}{\lambda_1}$

Substituting the given wavelengths into this equation, we get :

$\frac{RP_1}{RP_2} = \frac{5000 \, \text{Å}}{4000 \, \text{Å}} = \frac{5}{4}$

So, the correct answer is :

Option D : 5 : 4.

KEY CONCEPT : The resolving power of a telescope

$R.P = {D \over {122\lambda }}$ where $D=$ diameter of the objective lens

$\lambda = $ wavelength of light.

Clearly, larger the aperture, larger is the value of $D,$

more is the resolving power or resolution.

Optical fibers work on the principle of total internal reflection. When light is transmitted through the fiber, it is reflected off the inner walls of the fiber in such a way that it remains within the fiber, allowing it to carry the light signal over great distances with minimal loss.

So, the correct answer is :

Option A : total internal reflection.