Neet Solved Paper 2015 Question 39
Question: The refracting angle of a prism is A, and refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is
Options:
A) $ {180^{{}^\circ }}-3A $
B) $ {180^{{}^\circ }}-2A $
C) $ {90^{{}^\circ }}-A $
D) $ {180^{{}^\circ }}+2A $
Show Answer
Answer:
Correct Answer: B
Solution:
As, we know that $ \mu =\frac{\sin ( \frac{A+D _{m}}{2} )}{\sin \frac{A}{2}} $
$ \Rightarrow \cot \frac{A}{2}=\frac{\sin ( \frac{A+D _{m}}{2} )}{\sin \frac{A}{2}} $
$ \Rightarrow \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin ( \frac{A+D _{m}}{2} )}{\sin \frac{A}{2}} $ $ \sin ( \frac{\pi }{2}-\frac{A}{2} )=\sin ( \frac{A+D _{m}}{2} ) $
$ \Rightarrow ,\frac{\pi }{2}-\frac{A}{2}=\frac{A}{2}+\frac{D _{m}}{2} $
$ \Rightarrow D _{m}=\pi -2A $ $ D _{m}={180^{{}^\circ }}-2A $
BETA
