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

The refractive index of the material of a double convex lens is $1.5$ and its focal length is $5 \,cm$. If the radii of curvature are equal, the value of the radius of curvature (in $cm$) is

If the central portion of a convex lens is wrapped in black paper as shown in the figure,

$A$ convex lens of glass $(\mu_g = 1.45)$ has a focal length $f_a$ in air. The lens is immersed in a liquid of refractive index $(\mu_l) = 1.3$. The ratio of the $f_l / f_a$ is

$A$ 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 $\,cm$)

$A$ card sheet divided into squares each of size $1 \, mm^2$ is being viewed at a distance of $9 \, cm$ through a magnifying glass (a converging lens of focal length $10 \, cm$) held close to the eye. What is the magnification produced by the lens?