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