A) \[{{\cos }^{-1}}\left( \frac{\mu }{2} \right)\]
B) \[2{{\cos }^{-1}}\left( \frac{\mu }{2} \right)\]
C) \[2{{\sin }^{-1}}(\mu )\]
D) \[2{{\sin }^{-1}}\left( \frac{\mu }{2} \right)\]
E) none of these
Correct Answer: B
Solution :
Given, \[i=2r,\mu =\frac{\sin i}{\sin r}\frac{\sin 2r}{\sin r}=\frac{2\sin r\cos r}{\sin r}\] \[\cos r=\frac{\mu }{2}\]or \[r={{\cos }^{-1}}\left( \frac{\mu }{2} \right)\] \[i=2r=2{{\cos }^{-1}}\left( \frac{\mu }{2} \right)\]
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