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)))`

What is the focal length of a convex lens ( mu = 1.5) with radii of curvature R.

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

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

Focal lens of plane convex lens (mu=3/2) is f and radius of curvature of curved surface is R. So write the relation between f and R.

What is the focal length of a convex lens (mu=1*5) with radii of curvature R?

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.