Question 365: Ray Optics

Question

A wide slab consisting of two media of refractive indices n
1
and n
2
is placed in air as shown in the figure. A ray of light is incident from medium n
1
to n
2
at an angle $	heta$, where sin$	heta$ is slightly larger than 1/n
1
. Take refractive index of air as 1. Which of the following statement(s) is(are) correct?
![](https://imagex.cdn.examgoal.net/1kucmmt4t/ed5172fb-87de-4fc9-8b6b-7800257a1472/3c9ffad0-250e-11ec-bb3b-1f67b3a78587/file-1kucmmt4u.png)

Accepted Answer

$\sin 	heta  > {1 \over {{n_1}}}$ (Given)
i.e., $\sin {	heta _1} > {1 \over {{n_1}}}$
${n_1}\sin {	heta _1} = {n_2}\sin {	heta _2}$
$\sin {	heta _2} = {{{n_1}\sin {	heta _1}} \over {{n_2}}}$
if ${n_1} = {n_2}$ then ${	heta _2} = {	heta _1}$
${n_2}\sin {	heta _2} = (1)\sin {	heta _3}$
$\sin {	heta _3} = {n_2}\sin {	heta _2}$
$\sin {	heta _3} = {n_1}\sin {	heta _1}$
$\sin {	heta _1} = {{\sin {	heta _3}} \over {{n_1}}} > {1 \over {{n_1}}}$
$\sin {	heta _3} > 1$
${	heta _3} > 90^\circ $
This means ray cannot enter air.
For ${n_1} > {n_2};\sin {	heta _1} = {{{n_2}} \over {{n_1}}}\sin {	heta _2} > {1 \over {{n_1}}}$
$\sin {	heta _2} > {1 \over {{n_2}}}$
For surface 2 - air interface
${n_2}\sin {	heta _2} = \sin {	heta _3}$
$\sin {	heta _2} = {{\sin {	heta _3}} \over {{n_2}}} > {1 \over {{n_2}}}$
${	heta _2} > 90^\circ $
It means ray is reflected back in medium-2
For surface 1 and surface 2 interface
${n_2}\sin {	heta _2} = {n_1}\sin {	heta _1}$
$\sin {	heta _{2C}} = {{{n_1}} \over {{n_2}}}$
${	heta _{2C}}$ : critical angle
For ray to enter medium - 1
${	heta _2}  {n_1}$
${{{n_2}} \over {{n_1}}}\sin {	heta _2} > {{{n_2}} \over {{n_1}}}$
$\sin {	heta _2} > {1 \over {{n_2}}}$
For surface 2 - air interface
${n_2}\sin {	heta _2} = \sin {	heta _3}$
$\sin {	heta _2} = {{\sin {	heta _3}} \over {{n_2}}} > {1 \over {{n_2}}}$
${	heta _2} > 90$
It means ray reflected back in medium - 2
${n_2}\sin {	heta _2} = {n_1}\sin {	heta _1}$
$\sin {	heta _1} = {{{n_2}} \over {{n_1}}}\sin {	heta _2}$
$\sin {	heta _{2C}} = {{{n_1}} \over {{n_2}}};{	heta _{2C}}$ $	o$ critical angle
For ray to enter medium - 1
${	heta _2}  {1 \over {{n_1}}}$
$\sin {	heta _2} > 1 \Rightarrow {	heta _2} = 90^\circ $
$	herefore$ Ray is reflected back in medium.
![](['https://app-content.cdn.examgoal.net/fly/@width/image/1l4lbcerv/24be61fd-0ea5-4cef-99a8-7d9c99f48c3d/4ea779b0-efcf-11ec-bce3-811af8b3515c/file-1l4lbcerw.png?format=png', 'https://app-content.cdn.examgoal.net/fly/@width/image/1l4lbgecu/1b22d095-d86f-4fa9-9781-2caec7acc698/bd917ce0-efcf-11ec-b73b-b1ffe3ba4e96/file-1l4lbgecv.png?format=png', 'https://app-content.cdn.examgoal.net/fly/@width/image/1l4lbrteh/168bac50-bb0f-4b29-83af-0351b9c8029a/fb177fa0-efd0-11ec-b73b-b1ffe3ba4e96/file-1l4lbrtei.png?format=png', 'https://app-content.cdn.examgoal.net/fly/@width/image/1l4lbzffw/809dfd3b-3075-465b-a6a7-ea356cfc3cd9/cec7b2c0-efd1-11ec-bccc-dfad6763b020/file-1l4lbzffx.png?format=png'])

Answer

The light ray enters air if n
2
= n
1

Answer

The light ray is finally reflected back into the medium of refractive index n
1
if n
2
< n
1

Answer

The light ray is finally reflected back into the medium of refractive index n
1
if n
2
> n
1

Answer

The light ray is reflected back into the medium of refractive index n
1
if n
2
= 1

Ray Optics

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