Question 366: Ray Optics

Question

For a prism of prism angle $	heta$ = 60$^\circ$, the refractive indices of the left half and the right half are, respectively, n
1
and n
2
(n
2
$\ge$ n
1
) as shown in the figure. The angle of incidence i is chosen such that the incident light rays will have minimum deviation if n
1
= n
2
= n = 1.5. For the case of unequal refractive indices, n
1
= n and n
2
= n + $\Delta$n (where $\Delta$n << n), the angle of emergence e = i + $\Delta$e. Which of the following statement(s) is(are) correct?
![](https://imagex.cdn.examgoal.net/1kuff7s9u/15518681-b483-4327-a533-8711d87a2b15/94de0c20-2697-11ec-abd8-3b6ac251d2e9/file-1kuff7s9v.png)

Accepted Answer

${n_1} = {n_2} = n = {3 \over 2}$ for minimum deviation.
${r_m} = {A \over 2} = {{60} \over 2} = 30^\circ $ and $i = e = 	heta $
$n = {{\sin \left( {{{{\delta _{\min }} + A} \over 2}} ight)} \over {\sin \left( {{A \over 2}} ight)}} $
$\Rightarrow {3 \over 2} = {{\sin 	heta } \over {\sin \left( {{{60} \over 2}} ight)}}$
$ \Rightarrow \sin 	heta  = {3 \over 4} \Rightarrow \cos 	heta  = {{\sqrt 7 } \over 4}$
If ${n_1} = n = {3 \over 2}$ and ${n_2} = n + \Delta n$
$e = 	heta  + \Delta 	heta $
$(1)\sin 	heta  = n\sin 30^\circ $
$(n + \Delta n)\sin 30^\circ  = (1)\sin (	heta  + \Delta 	heta )$
Solving ${1 \over 2}(\Delta n) = \sin (	heta  + \Delta 	heta ) - \sin 	heta $
${{\Delta n} \over 2} = {{\sin (	heta  + \Delta 	heta ) - sin	heta } \over {\Delta 	heta }} 	imes \Delta 	heta $
${{d(\sin 	heta )} \over {d	heta }} = \cos 	heta $
${{\Delta n} \over 2} = (\cos 	heta )(\Delta 	heta ) \Rightarrow {{\Delta n} \over 2} = {{\sqrt 7 } \over 4}(\Delta 	heta )$
$ \Rightarrow {{\Delta n} \over {\Delta 	heta }} = {{\sqrt 7 } \over 2} = {{2.64} \over 2}$
${{\Delta n} \over {\Delta 	heta }} = 1.34 > 1$
$\Rightarrow$ $\Delta$n > $\Delta$$	heta$, so option (a) is incorrect.
$\Delta n = (1.34)\Delta 	heta $
$ \Rightarrow \Delta 	heta  \propto \Delta n \Rightarrow $ option (b) is correct.
${{\Delta n} \over {\Delta 	heta }} = {{\sqrt 7 } \over 2} \Rightarrow {{2.8 	imes {{10}^{ - 3}}} \over {\Delta 	heta }} = {{\sqrt 7 } \over 2}$
$\Delta 	heta  = {{5.6 	imes {{10}^{ - 3}}} \over {\sqrt 7 }} = \sqrt 7  	imes 0.8 	imes {10^{ - 3}}$
$\Delta 	heta  = (2.64 	imes 0.8) 	imes {10^{ - 3}} = 2.11 	imes {10^{ - 3}}$ rad
So, option (c) is correct.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1kx9rkx18/9f4fa9d2-27f8-48cf-aa79-3bedfee29a68/e1da97c0-5edf-11ec-9712-45aafe915683/file-1kx9rkx19.png?format=png)

Answer

The value of $\Delta$e (in radians) is greater than that of $\Delta$n

Answer

$\Delta$e is proportional to $\Delta$n

Answer

$\Delta$e lies between 2.0 and 3.0 milliradians, if $\Delta$n = 2.8 $	imes$ 10
$-$3

Answer

$\Delta$e lies between1.0 and 1.6 milliradians, if $\Delta$n = 2.8 $	imes$ 10
$-$3

Ray Optics

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