question_answer

A light beam is travelling from Region I to Region IV (Refer Figure). The refractive index in Regions I, II, III and IV are \[{{n}_{0}},\,\frac{{{n}_{0}}}{2},\,\frac{{{n}_{0}}}{6}\] and \[\frac{{{n}_{0}}}{8},\] respectively. The angle of incidence 6 for which the beam just misses entering Region IV is

Region I		Region II		Region III		Region IV
		\[\frac{{{n}_{0}}}{2}\]		\[\frac{{{n}_{0}}}{6}\]		\[\frac{{{n}_{0}}}{8}\]
	0		0.2 m		0.6 m

A) \[{{\sin }^{-1}}\left( \frac{3}{4} \right)\]

B) \[{{\sin }^{-1}}\left( \frac{1}{8} \right)\]

C) \[{{\sin }^{-1}}\left( \frac{1}{4} \right)\]            

D) \[{{\sin }^{-1}}\left( \frac{1}{3} \right)\]

Correct Answer: B

Solution :

[b] As the beam just suffers TIR at interface of region III and IV

\[{{n}_{0}}\sin \theta =\frac{{{n}_{0}}}{2}\,\sin {{\theta }_{1}}=\frac{{{n}_{0}}}{6}\sin \,{{\theta }_{2}}=\frac{{{n}_{0}}}{8}\,\sin {{90}^{o}},\] \[\sin \theta =\frac{1}{8}\Rightarrow \theta ={{\sin }^{-1}}\left( \frac{1}{8} \right)\]