SATHEE: Optics Question 767

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 ( $μ_{o i l} = 1.45, μ_{w a t e r} = 1.33$ )

[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 $μ_{o i l} = \frac{\sin i}{\sin r_{1}}$

$∴$ $\sin r_{1} = \frac{\sin 40}{1.45} = 0.443$

Refraction at oil-water point $_{o i l} μ_{w a t e r} = \frac{\sin r_{1}}{\sin r}$

$∴$ $\frac{1.33}{1.45} = \frac{0.443}{\sin r}$ or $\sin r =$

$\frac{0.443 \times 1.45}{1.33}$

Therefore $r = {28.9}^{o}$

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Optics Question 767

Question: A thin oil layer floats on water. A ray of light making an angle of incidence of 40∘ shines on oil layer. The angle of refraction of light ray in water is ( μoil=1.45,μwater=1.33 )

Options:

Answer:

Solution:

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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 ( $μ_{o i l} = 1.45, μ_{w a t e r} = 1.33$ )