Question 63: Ray Optics

Question

A light beam is travelling from Region I to Region IV 
 (Refer figure). The refractive index in Regions I, II, III and IV are $${n_0},{{{n_0}} \over 2},{{{n_0}} \over 6}$$ and $${{{n_0}} \over 8}$$, respectively. The angle of incidence $$	heta$$ for which the beam just misses entering Region IV is

Accepted Answer

From Snell's laws
$${\mu _1}\sin 	heta  = {\mu _2}\sin {	heta _2}$$
Case I : Region I
$${n_0}\sin 	heta  = {{{n_0}} \over 2}\sin \alpha $$ ...... (i)
Case II : Region II
$${{{n_0}} \over 2}\sin \alpha  = {{{n_0}} \over 6}\sin {	heta _c}$$ ..... (ii) ($$	herefore$$ $$\beta  = {	heta _c}$$)
Case II : Region III
$${{{n_0}} \over 6}\sin {	heta _c} = {{{n_0}} \over 8}\sin 90^\circ $$
$$ \Rightarrow {{{n_0}} \over 6}\sin {	heta _c} = {{{n_0}} \over 8}$$ ..... (iii)
From equation (i), (ii) and (iii), we have,
$${n_0}\sin 	heta  = {{{n_0}} \over 8}$$
$$ \Rightarrow \sin 	heta  = {1 \over 8}$$
$$ \Rightarrow 	heta  = {\sin ^{ - 1}}\left( {{1 \over 8}} ight)$$

Answer

$${\sin ^{ - 1}}\left( {{3 \over 4}} \right)$$

Answer

$${\sin ^{ - 1}}\left( {{1 \over 8}} \right)$$

Answer

$${\sin ^{ - 1}}\left( {{1 \over 4}} \right)$$

Answer

$${\sin ^{ - 1}}\left( {{1 \over 3}} \right)$$

Ray Optics

A light beam is travelling from Region I to Region IV (Refer figure). The refractive index in Regions I, II, III and IV are ${n_0},{{{n_0}} \over 2},{{{n_0}} \over 6} and {{{n_0}} \over 8}, respectively. The angle of incidence \theta$ for which the beam just misses entering Region IV is

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