Optics Question 452
Question: A prism of refractive index $ \mu $ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of $ \mu $ is
Options:
A) $ {{\sin }^{-1}}( \frac{\mu }{2} ) $
B) $ {{\sin }^{-1}}\sqrt{\frac{\mu -1}{2}} $
C) $ 2{{\cos }^{-1}}( \frac{\mu }{2} ) $
D) $ {{\cos }^{-1}}( \frac{\mu }{2} ) $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given $ {\delta _{m}}=A, $ as $ \mu =\frac{\sin ( \frac{A+{\delta _{m}}}{2} )}{\sin ( \frac{A}{2} )} $
$ \Rightarrow \mu =\frac{\sin ( \frac{A+A}{2} )}{\sin ( \frac{A}{2} )}=2\cos \frac{A}{2} $
$ \Rightarrow A=2{{\cos }^{-1}}( \frac{\mu }{2} ) $
