Optics Question 90
Question: A ray of light is incident at the glass?water interface at an angle i, it emerges finally parallel to the surface of water, then the value of $ {\mu _{g}} $ would be
[IIT-JEE (Screening) 2003]
Options:
A) (4/3) sin i
B) 1/sin i
C) 4/3
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
For glass-water interface $ _{g}{\mu _{w}}=\frac{\sin i}{\sin r} $ -(i) For water-air interface $ _{w}{\mu _{a}}=\frac{\sin r}{\sin 90^{o}} $ -(ii)
$ \Rightarrow $ $ _{g}{\mu _{w}}\times _{w}{\mu _{a}}=\frac{\sin i}{\sin r}\times \frac{\sin r}{\sin 90^{o}}=\sin i $
$ \Rightarrow $ $ \frac{{\mu _{w}}}{{\mu _{g}}}\times \frac{{\mu _{a}}}{{\mu _{w}}}=\sin i\Rightarrow {\mu _{g}}=\frac{1}{\sin i} $
BETA
