A) \[{{\sin }^{-1}}\left( \frac{5}{6} \right)\]
B) \[{{\sin }^{-1}}\left( \frac{5}{12} \right)\]
C) \[{{\sin }^{-1}}\left( \frac{1}{\sqrt{2}} \right)\]
D) \[{{\sin }^{-1}}\left( \frac{1}{2} \right)\]
Correct Answer: A
Solution :
The critical angle is the minimum angle of incidence at which total internal reflection occurs. The angle of incidence is measured with respect to the normal at the refractive boundary. The critical angle \[({{\theta }_{C}})\] is given by \[{{\theta }_{C}}={{\sin }^{-1}}\,\,\left( \frac{{{n}_{2}}}{{{n}_{1}}} \right)\] where \[{{n}_{2}}\] is refractive index of less dense medium and \[{{n}_{1}}\] is refractive index of the denser medium. Also, \[\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{{{v}_{2}}}{{{v}_{1}}}\] \[\therefore \] \[{{\theta }_{C}}={{\sin }^{-1}}\left( \frac{{{v}_{2}}}{{{v}_{1}}} \right)\] Given, \[{{v}_{2}}=2\times {{10}^{8}}m/s,\] \[{{v}_{1}}=2.4\times {{10}^{8}}m/s,\] \[{{\theta }_{C}}={{\sin }^{-1}}\left( \frac{2\times {{10}^{8}}}{2.4\times {{10}^{8}}} \right)\] \[{{\theta }_{C}}={{\sin }^{-1}}\left( \frac{5}{6} \right)\]
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