$A$ plano-convex lens is made of refractive index of $1.6$. The radius of curvature of the curved surface is $60 \,cm$. The focal length of the lens is (in $\,cm$)

$A$ convex lens of focal length $0.15 \,m$ is made of a material of refractive index $\frac{3}{2}$. When it is placed in a liquid, its focal length is increased by $0.225 \,m$. The refractive index of the liquid is

$A$ concave lens (refractive index $n_l = 1.5$) has both surfaces of the same radius of curvature $R$. If it is immersed in a liquid of refractive index $n_m = 1.75$,it will act as a:

Double-convex lenses are to be manufactured from a glass of refractive index $1.55$,with both faces of the same radius of curvature. What is the radius of curvature (in $cm$) required if the focal length is to be $20\;cm$?

In order to obtain a real image of magnification $2$ using a converging lens of focal length $20 \ cm,$ where should an object be placed in $cm$?

Two point sources $S_1$ and $S_2$ are $24 \ cm$ apart. Where should a convex lens of focal length $9 \ cm$ be placed in between them,so that the images of both sources are formed at the same place (in $cm$)?