$A$ 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 $cm$?

$A$ bulb is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance $d$ ahead of the second wall. The required focal length will be:

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 $20\;cm$?

An object placed at a distance of $16 \ cm$ from a convex lens produces an image of magnification $m$ $(m > 1)$. If the object is moved towards the lens by $8 \ cm$,then again an image of magnification $m$ is obtained. The numerical value of the focal length of the lens is (in $cm$)

The radius of curvature of the convex surface of a thin plano-convex lens is $15 \, cm$ and the refractive index of its material is $1.6$. The power of the lens will be:

$A$ turnip sits before a thin converging lens,outside the focal point of the lens. The lens is filled with a transparent gel so that it is flexible; by squeezing its ends toward its center,you can change the curvature of its front and rear sides. Suppose that the image must be formed on a card which is at a certain distance behind the lens,while you move the turnip away from the lens,then you should