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

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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 \(\frac{4}{3}\))