The radius of curvature of the convex face of a plano-convex lens is $55 \mathrm{cm}$ and the refractive index of its material is $1.55$. Find the focal length of the lens.

The radii of curvature of the two faces of a convex lens are $0.10 \mathrm{m}$ and $0.15 \mathrm{m}$ respectively. If the focal length of the lens is $0.12 \mathrm{m}$, find the refractive index of the material of the lens.

The radii of curvature of the two surfaces of a convexo-concave lens are $15 \mathrm{cm}$ and $30 \mathrm{cm}$. Find the focal length of the lens. Take refractive index of the lens material to be $1.5$.

A bi-convex spherical lens A has a focal length of $20 \mathrm{cm}$. If it is split along a plane $AB$ into two halves to form two plano convex lenses,then what will be the focal length of each plano-convex lens?

What if the lens be cut along $CD$?

The radii of curvature of a convexo-convex lens of glass $(n = 1.50)$ are in the ratio $1:2$. This lens renders the rays coming from an illuminated filament at a distance of $6.0 \mathrm{cm}$ parallel. Calculate the radii of curvature of its surfaces.

A converging lens has a focal length of $20 \mathrm{cm}$ in air. The refractive index of the material of the lens is $1.6$. If it is immersed in a liquid of refractive index $1.3$, what will be its new focal length?

The refractive index of glass with respect to air is $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ and that of water is $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$. The radii of curvature of a glass convex lens are $20 \mathrm{cm}$ and $30 \mathrm{cm}$ respectively. What will be the focal length of the lens in air?