If the refractive index from air to glass is $\frac{3}{2}$ and that from air to water is $\frac{4}{3}$,then the ratio of focal lengths of a glass lens in water and in air is

The radii of curvature of both the surfaces of a convex lens of focal length $f$ and focal power $P$ are equal. One of the surfaces is made plane by grinding. The new focal length and focal power of the lens are:

The refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $40 \ cm$. The ratio of the power of the lens when immersed in a liquid of refractive index $1.25$ to that when placed in air is:

$A$ biconvex lens $(R_{1}=R_{2}=20 \ cm)$ has a focal length equal to the focal length of a concave mirror. The radius of curvature of the concave mirror is (Refractive Index of glass lens $= 1.5$) (in $cm$)

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 $20\;cm$?

$A$ biconvex lens of focal length $f$ and radii of curvature of both surfaces $R$ is made of a material of refractive index $n_{1}$. This lens is placed in a liquid of refractive index $n_{2}$. How will this lens behave?