A glass prism is diped into a liquid of refractive index greater than 1. What should be the change of minimum deviation?

A ray of light is incident at an angle of incidents 40^(@) on a prism having refractive index 1.6 . What should be the value of the angle of the prism for minimum deviation? Given sin 40^(@) = 0.6428 , sin 23^(@)42'=0.4018.

A glass prism is immersed in a liquid of refractive index 1.4. If the refractive index of glass is 1.6 , find the limiting angle of the prism for emergence of ray through it.

The refracting angle of a prism is A, and refractive index of the material of the prism is cot (A // 2) . The angle of minimum deviation is

A convex lens whose redius of curvatures of both surfaces are same, refractive index 1.5 an focal length is 15cm . If it is immerged into a liquid of refractive index 1.7 what will be the focal length of he lesns?

A glass prism of refracting angle 60^(@) and of refractive index 1.6, is immersed in water (refractive index is 1.33). What is its angle of minimum deviation ? [sin 36.87^(@) = 0.6]

The refractive index of the material of a prism is sqrt((3)/(2) and the refractive angle is 90^(@) . What are the values of the angle of minimum deviation of the prism and the angle of the incidence at the minimum deviation position?

The refracting angle of a prism is 60^(@) and its refractive index is sqrt((7)/(3) . What should be the minimum angle of incidence on the first refracting surface so that the ray can emerge somehow from the second refracting surface?

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. By rotating the prism,the angle of minimum deviation is measured to be 40^(@) . What is the refractive index of the material of the prism? If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. Refracting angle of the prism is 60^(@) .