The focal length of an equi-concave lens is `3/4` times of radius of curvature of its surfaces. Find the refractive index of the material of lens. Under what condition will this lens behave as a converging lens?

The power of a biconvex lens is 10 dioptre and the radius of curvature of each surface is 10 cm. Then the refractive index of the material of the lens is

Consider an equiconvex lens of radius of curvature R and focal length f. If fgtR , the refractive index mu of the material of the lens

A biconvex lens has focal length (2)/(3) times the radius of curvature of either surface. Calculate refraction index f material of the lens.

A biconvex lens has focal length (2)/(3) times the radius of curvature of either surface. Calculate refraction index f material of the lens.

A bicovex lens has radii of cuvature 20 cm each. If the refractive index of the material of the lens is 1.5, what is its focal length?

For a plano convex lens, the radius of curvature of convex surface is 10 cm and the focal length is 30 cm. The refractive index of the material of the lens is

If in a planoconvex lens, the radius of curvature of the convex surface is 10cm and the focal length is 30 cm , the refractive index of the material of the lens will be

The radius of curvature of the curved surface of a plano-convex lens is 20 cm . If the refractive index of the material of the lens be 1.5 , it will

A converging bundle of rays is intercepted by a biconcave lens. The radii of curvature of both surfaces are 20cm and the refractive index of the material of the lens is 1.5. If the rays originially converged to a point 10cm in front of the lens, where will they now converge after passing through the lens?

If in a plano-convex lens, the radius of curvature of the convex surface is 10 cm and the focal length of the lens is 30 cm , then the refractive index of the material of lens will be