$A$ lens which has a focal length of $4 \; cm$ and a refractive index of $1.4$ is immersed in a liquid of refractive index $1.6$. What will be the new focal length in $cm$?

$A$ lens has a power of $-4.0 \text{ Diopter}$. This means . . . . . . .

The two surfaces of a concave lens, made of glass of refractive index $1.5$, have the same radii of curvature $R$. It is now immersed in a medium of refractive index $1.75$. Then the lens:

The distance between an object and its image produced by a converging lens is $0.72 \ m$. The magnification is $2$. What will be the magnification when the object is moved by $0.04 \ m$ towards the lens?

The focal length of a convex lens is $f$. An object is placed at a distance $x$ from its first focal point. The ratio of the size of the real image to that of the object is

$A$ screen is placed $100 \,cm$ from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by $20 \,cm$. The focal length of the lens is (in $\,cm$)