An object is placed at a distance of $10 \, cm$ from a lens. The image is formed at a distance of $20 \, cm$ behind the lens. The power of the lens is ......

$A$ convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it, then the focal length of the plano-convex lens is ($\mu =$ the refractive index of the material of the lens).

$A$ biconvex lens has a focal length $2/3$ times the radius of curvature of either surface. The refractive index of the material is:

$f_v$ and $f_r$ are the focal lengths of a convex lens for violet and red lights respectively,and $F_v$ and $F_r$ are the magnitudes of the focal lengths of a concave lens for violet and red lights respectively. Then:

$A$ thin converging lens of focal length $f=25 \,cm$ forms the image of an object on a screen placed at a distance of $75 \,cm$ from the lens. The screen is moved closer to the lens by a distance of $25 \,cm$. The distance through which the object has to be shifted, so that its image on the screen is sharp again, is (in $\,cm$)

$A$ glass convex lens has a refractive index of $1.55$ with both faces having the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 \,cm$ (in $\,cm$)?