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 $

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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}} $



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