A) \[{{\cos }^{-1}}(\mu /2)\]
B) \[2{{\cos }^{-1}}(\mu /2)\]
C) \[2{{\sin }^{-1}}(\mu )\]
D) \[2{{\sin }^{-1}}\,(\mu /2)\]
Correct Answer: B
Solution :
[b] By using \[\mu =\frac{\sin i}{\sin r}\]\[\Rightarrow \]\[\mu =\frac{\sin 2r}{\sin r}=\frac{2\sin r\cos r}{\sin r}\] |
\[(\sin 2\theta =2\,\sin \theta \cos \theta )\]\[\Rightarrow \] \[r={{\cos }^{-1}}\left( \frac{\mu }{2} \right)\] |
So, \[i=2r={{\cos }^{-1}}\,\left( \frac{\mu }{2} \right)\] |
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