Two rays of light from a point below the surface of water are equally inclined to each other at 60^o in water. What is the angle between the rays when they emerge into air? (Take the refractive index of water to be $\frac{4}{3}$ )

41.8^o

44.1^o

60.0^o

83.6^o

120.0^o

Correct answer is D

$\frac{4}{3}$ = $\frac{sin r_1}{sin 30}$ and $\frac{4}{3}$ = $\frac{sin r_2}{sin 30}$

r₁ = 41.8^o and r₂ = 41.8^o

angle between them = r₁ + r₂

= 83.6^o

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