Statement-1 : Focal length of an equiconvex lens of `mu = 3//2` is equal to radius of curvature of each surface.
Statement-2 : it follows from
`(1)/(f) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2)))`

The focal length of an equiconvex lens placed in air is equal to radius of curvature of either surface. Is it true ?

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

The focal length of an equiconvex lens is equal to radius of curvature of either surface. What is the refractive index of the material of the prism ?

A convex lens, mu = 1.4 both sides R_1 , R_2 to radius of curvature of both sides. Focal length = ?

STATEMENT- 1 Power of a lens depends on nature of material of lens, medium in which it is placed and radii of curvature of its surface. STATEMENT 2 it follows the relation p=(1)/(f)=((mu_(2))/(mu_(1))-1)[(1)/(R_(1))-(1)/(R_(2))]

A biconvex lens (mu=1.5) has radius of curvature 20cm (both). find its focal length.

A plano-convex lens mu = 1.5 has focal length of 18 cm in air. Calculate the radius of curvature of the spherical surface.

What is the focal length of double convex lens for which radius of curvature of either of the surfaces is 30 cm? [ ""_(a) mu_(g)=1.5]

Statement-1 : A glass convex lens (refractive index, mu_(1)=1.5 ), when immersed in a liquid of refractive index mu_(2)=2 , functions as a diverging lens. Statement-2 : The focal length of a lens is given by (1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2))) , where mu is the relative refractive index of the material of the lens with respect to the surrounding medium.