A) \[{{\sin }^{-1}}\left( \frac{1}{\mu } \right)\]
B) \[{{\tan }^{-1}}\left( \frac{1}{\mu } \right)\]
C) \[{{\sin }^{-1}}\left( \frac{\mu -1}{\mu } \right)\]
D) \[{{\cos }^{-1}}\left( \frac{1}{\mu } \right)\]
Correct Answer: D
Solution :
If \[\alpha \]= maximum value of base angle for which light is totally reflected form hypotenuse. \[(90{}^\circ -\alpha )\]= C = minimum value of angle of incidence at hypotenuse for total internal reflection \[\sin (90{}^\circ -\alpha )=\sin C=\frac{1}{\mu }\]\[\Rightarrow \cos \alpha =\frac{1}{\mu }\]\[\Rightarrow \alpha ={{\cos }^{-1}}\left( \frac{1}{\mu } \right)\]You need to login to perform this action.
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