Optics Question 966
Question: A ray of light is incident at an angle i from denser to rare medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection and the angle of refraction are respectively r and r’ , then the critical angle will be
[IIT-JEE 1983; MP PET 1995; CBSE PMT 1996; MP PMT 1985, 99; Pb. PET 2002]
Options:
A) $ {{\sin }^{-1}}(\sin r) $
B) $ {{\sin }^{-1}}(\tan r’) $
C) $ {{\sin }^{-1}}(\tan i) $
D) $ {{\tan }^{-1}}(\sin i) $
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Answer:
Correct Answer: C
Solution:
$ _{D}{\mu _{R}}=\frac{\sin i}{\sin {r}’}{\Rightarrow _{R}}{\mu _{D}}=\frac{\sin {r}’}{\sin \ i}=\frac{1}{\sin C} $
$ \Rightarrow \sin C=\frac{\sin i}{\sin (90-r)}=\frac{\sin i}{\cos r}=\frac{\sin i}{\cos i} $ (as Ði = Ðr)
$ \Rightarrow \sin C=\tan i\Rightarrow C={{\sin }^{-1}}(\tan i) $
