A circular beam of light of diameter `d = 2 cm` falls on a plane refractive of glass. The angle of incidence is `60^(@)` and refractive index of glass is `mu = 3//2`. The diameter of the refracted beam is

A circular beam of light (diameter d) falls on a plane surface of a liquid. The angle of incidence is 45^(@) and refractive index of the liquid is mu . The diameter of the refracted beam is

A circular beam of light (diameter d) falls on a plane surface of a liquid. The angle of incidence is 45^(@) and refractive index of the liquid is mu . The diameter of the refracted beam is

A circular beam of light of diameter (width) falls on a plane surface of glass. The angle of incidence is .I., angle of refraction is .r. and refractive index of glass is mu . Then the diameter of the refracted beam d. is…….

A ray of light is incident on a glass. The reflected ray gets totally polarised. The magnitude of incidence angle is (refractive index of glass = mu )

A beam of light strikes a surface at angle of incidence of 60^(@) and reflected beam becomes completely polarised . The refractive index of glass surface is -

A beam of light strikes a surface at angle of incidence of 60^(@) and reflected beam becomes completely polarised . The refractive index of glass surface is -

A ray of light is incident on a medium of refractive index sqrt(2) at an angle of incidence of 45^(@) . The ratio of the width of the incident beam in air to that of the refracted beam in the medium is

A plane wavefront is made incident at an angle of 30^@ on the surface of the glass . Calculate the angle of refraction if refractive index of glass is 1.5 .