The focal length of a thin biconvex lens is $20 \,cm$. When an object is moved from a distance of $25 \,cm$ in front of it to $50 \,cm$, the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is

If an object is placed at $A$ $(OA > f)$,where $f$ is the focal length of the lens,the image is found to be formed at $B$. $A$ perpendicular is erected at $O$ and $C$ is chosen on it such that the angle $\angle BCA$ is a right angle. Then the value of $f$ will be

$A$ convex lens is made from glass material having a refractive index of $1.4$ with the same radius of curvature on both sides. The ratio of its focal length to its radius of curvature is . . . . . . .

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 $cm$)?

An object approaches a convergent lens from the left of the lens with a uniform speed $5 \ m/s$ and stops at the focus. The image :-

$A$ lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I.$ If an aperture of diameter $\frac{d}{2}$ in the central region of the lens is covered by a black paper,the focal length of the lens and the intensity of the image will be,respectively: