For convex lens, the radii of curvatures of the first and second surface are `R_(1) and R_(2)` respectively. Using new Cartesian sign conventions they are

A biconvex lens made of material with refractive index n_(2) . The radii of curvatures of its left surface and right surface are R_(1) and R_(2) . The media on its left and right have refreactive indices n_(1) and n_(3) respectively. The first and second focal lengths of the lens are respectively f_(1) and f_(2) . The ratio, f_(1)//f_(2) , of the two focal lengths is equal to

One surface of a lens is convex and the other is concave, if the radii of curvature are r_(1) and r_(2) respectively, the lens will be convex if

Focal length of the plano-convex lens is when its radius of curvature of the surface is R and n is the refractive index of the lens.

Calculate the focal length of a convex lens whose radii of curvature of two surfaces is 10cm and 15cm respectively and its refractive index is 1.5.

A convex lens has the same radius of curvature (R ) for both the surfaces. For what value of the refractive index of the material of the convex lens, the numerical values of f and R are equal ?

If R_(1)andR_(2) are the radii of curvature of the spherical surfaces of a thin lens and R_(1)gtR_(2) , then this lens can