Two symmetric double convex lenses A and B have same focal length but the radii of curvature differ so that `R_(A) = 0.9R_(B)`. If refractive index of A is 1.63 find the refractive index of B.

A double convex lens has focal length 25 cm. The radius of curvature of one of the surfaces is double of the other. Find the radii if the refractive index of the material of the lens is 1.5.

A double convex lens has focal length 25 cm. The radius of curvature of one of the surfaces is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.

A double convex lens has focal length 25 cm. The radius of curvature of one of the surfaces is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.

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

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

(ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15cm. Its focal length is 12 cm. What is the refractive index of glass ? (Refractive index of air-water = 1.33. Refractive index for ar - glass = 1.5.)

A biconvex lens of focal length f and radii of curvature of both the surface R is made of a material of refractive index n_(1) . This lens is placed in a liquid of refractive index n_(2) . Now this lens will behave like

The focal length of a double convex lens is equal to radius of curvature of either surface. The refractive index of its material is

A double convex thin lens made of the refractive index 1.6 has radii of curvature 15 cm each. What is the focal length (in cm) of this les when immersed in a fluid of refractive index 1.63?