Optics Question 767
Question: A thin oil layer floats on water. A ray of light making an angle of incidence of $ 40{}^\circ $ shines on oil layer. The angle of refraction of light ray in water is ( $ {\mu _{oil}}=1.45,{\mu _{water}}=1.33 $ )
[MP PMT 1993]
Options:
A) $ 36.1{}^\circ $
B) $ 44.5{}^\circ $
C) $ 26.8{}^\circ $
D) $ 28.9{}^\circ $
Show Answer
Answer:
Correct Answer: D
Solution:
Refraction at air-oil point $ {\mu _{oil}}=\frac{\sin i}{\sin r _{1}} $
$ \therefore $ $ \sin r _{1}=\frac{\sin 40}{1.45}=0.443 $
Refraction at oil-water point $ _{oil}{\mu _{water}}=\frac{\sin r _{1}}{\sin r} $
$ \therefore $ $ \frac{1.33}{1.45}=\frac{0.443}{\sin r} $ or $ \sin r= $
$ \frac{0.443\times 1.45}{1.33} $
Therefore $ r={{28.9}^{o}} $
BETA
