Question

Q. R.I. of water is 1.333 and that of glass is 1.5. If a ray of light is incident at 30° on the water-glass interface, at what angle would it be refracted?

pls answer correctly
thenks​

Answer 1

Given data : Refractive index of water is 1.333 and that of glass is 1.5. If a ray of light is incident at 30° on the water-glass interface.

To find : Refracted angle ?

Solution :

⟹ Refractive index of water, n1 = 1.333

⟹ Refractive index of glass, n2 = 1.5

⟹ Incident angle, ( i ) = 30°

Here we use, Snell's law of refraction

⟹ (sin i)/(sin r) = n2/n1

⟹ (sin 30°)/(sin r) = 1.5/1.333

⟹ (1/2)/(sin r) = 1.5/1.333

⟹ 1/sin r = (1.5/1.333) * 2

⟹ 1/sin r = 3/1.333

⟹ sin r = 1.333/3

⟹ sin r = 0.4443

⟹ r = sin^( - 1 ) 0.4443

⟹ r = 26.3785°

Answer : Hence angle of refraction is sin^( - 1 ) 0.4443 or 26.3785°.