A film of oil of refractive index 1.20, lies on water of refractive index 1.33. A light ray is incident at `30^(@)` in the oil on the oil-water boundary. Calculate the angle of refraction in water.

A prism of refractive index 1.53 is placed in water of refractive index 1.33 . If the angle of prism is 60^@ , calculate the angle of minimum deviation in water.

A cubical block of glass of refractive index n_1 is in contact with the surface of water of refractive index n_2. A beam of light is incident on vertical face of the block. After refraction a total internal reflection at the base and refraction at the opposite face take place. The ray emerges at angle theta as shown. The value of theta is given by

If refractive index of glass is 1.50 and of water is 1.33, then criticle angle is

A plane mirror PQ is help normally to water surface of refractive idex 1.33. A ray of light is incident at an angle of 50^@ with the nirror surface. After reflection the ray is refracted into water. The angle of refraction r is

A glass prism of refractive index 1.5 is placed in water of refractive index 1.33. The minimum value of the angle of the prism so that it will not be possible to have any emergent ray is

A glass prism has a refractive angle of 90^(@) and a refractive index of 1.5. A ray is incident at an angle of 30^(@) . The ray emerges from an adjacent face at an angle of

Refractive indices of water an glass are 4//3 and 3//2 respectively. A ray of light travelling in water is incident on the water glass interface at 30^@ . Calculate the angle of refraction.

A prism of refractive angle 60^(@) and refractive index 1.5 is kept in water of refractive index 1.33 . What is the angle of minimum deviation for a monochromatic ray of light in water ? (Given sin 34^(@) = 0.56 )