Question

Q-22.11 Unanswered A beam of light traveling in air strikes the surface of mineral oil at an angle of 23.8° with the normal to the surface. If the light travels at 2.17 X 108 m/s in the oil, what are the following?(a) index of refraction of mineral oil and (b) angle of refraction. (a) 1.38 (b) 17.0 deg A (a) 1.88 (b) 19.0 deg. B C (a) 1.48 (b) 10.0 deg. (a) 2.38 (b) 12.0 deg D

Answer 1

a)

c = speed of light in air = 3 x 10⁸ m/s

v = speed of light in oil = 2.17 x 10⁸ m/s

n_oil = index of refraction of oil

Index of refraction of oil is given as

n_oil = c/v

n_oil = (3 x 10⁸)/(2.17 x 10⁸)

n_oil = 1.38

b)

i = angle of incidence = 23.8

r = angle of refraction = ?

n_air = index of refraction of air = 1

n_oil = index of refraction of oil = 1.38

Using snell's law

n_air Sini = n_oil Sinr

1 Sin23.8 = (1.38) Sinr

Sinr = 0.292

r = Sin^-1(0.292)

r = 17.0 deg

So the correct choice is

A) (a) 1.38 (b) 17.0 deg