Question 15: 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, n1 and n2 (n2 $$\ge$$ n1) as shown in the figure. The angle of incidence i is chosen such that the incident light rays will have minimum deviation if n1 = n2 = n = 1.5. For the case of unequal refractive indices, n1 = n and n2 = n + $$\Delta$$n (where $$\Delta$$n << n), the angle of emergence e = i + $$\Delta$$e. Which of the following statement(s) is(are) correct?

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.

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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