Calculate the radius of the curvature an equi-concave lens refrective index 1.5, when it is kept in a medium refrecation index 1.4, to have a power of -5D ?

An equiconvex lens of refractive index 1.6 has power 4D in air. Its power in water is:

A plano convex lens of radius of curvature 30 cm and refractive index 1.5 is kept in air. Find its focal length (in cm).

The radius curvature of each surface of a convex lens of refractive index 1.5 is 40 cm. Calculate its power.

A convex lens made up of glass of refractive index 1.5 is dippedin turn (i) in a medium of refractive index 1.65 (ii) in a medium of refractive index 1.33 Then Will it behave as converging or diverging lens in the two cases ?

A convex lens made up of glass of refractive index 1.5 is dippedin turn (i) in a medium of refractive index 1.65 (ii) in a medium of refractive index 1.33 (a) Will it behave as converging or diverging lens in the two cases ? (b) How will its focal length changes in the two media ?

The focal length of equiconvex lens of retractive index, mu=1.5 and radius of curvature, R is

A lens is made of flint glass (refractive index =1.5 ). When the lens is immersed in a liquid of refractive index 1.25 , the focal length:

A convex lens of refractive index 3/2 has a power of 2.5^(@) . If it is placed in a liqud of refractive index 2,the new power of the lens is

The radii of curvatures of a convex lens are 0.04 m and 0.04m. Its refractive index is 1.5. Its focal length is

A thin symmettrical double convex lens of refractive index mu_(2)-1.5 is placed between a medium of refractiv eindex mu_(1)=1.4 to the left and another medium of refractive index mu_(3)=1.6 to the right.Then the system behaves as