A) \[56.4{}^\circ \]
B) \[28.2{}^\circ \]
C) \[50.4{}^\circ \]
D) \[39.4{}^\circ \]
Correct Answer: A
Solution :
The relation between critical angle (0 and refractive index (u) is \[\sin C=\frac{1}{\mu }\] ... (i) Also, \[\mu =\frac{{{v}_{r}}}{{{v}_{d}}}\] .... (ii) From Eqs. (i) and (ii), we get \[\sin \,C=\frac{{{v}_{d}}}{{{v}_{r}}}\] Given, \[{{v}_{d}}=2\times {{10}^{8}}m/s\], \[{{v}_{d}}=2.4\times {{10}^{8}}m/s\] \[\therefore \] \[\sin C=\frac{2\times {{10}^{8}}}{2.4\times {{10}^{8}}}=\frac{5}{6}\] \[\Rightarrow \] \[\Rightarrow C={{\sin }^{-1}}\left( \frac{5}{6} \right)={{56.4}^{o}}\]
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