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

$A$ concave-convex lens of refractive index $1.5$ has radii of curvature of its surfaces as $30 \ cm$ and $20 \ cm$,respectively. The concave surface is facing upwards and is filled with a liquid of refractive index $1.3$. The focal length of the liquid-glass combination will be

$A$ thin lens made of glass of refractive index $\mu_g = 1.5$ has a focal length of $12 \ cm$ in air. If it is immersed in water of refractive index $\mu_w = 4/3$,what will be its new focal length in $cm$?

The graph between the lateral magnification $(m)$ produced by a lens and the distance of the image $(v)$ is given by

The radii of curvature of a double convex lens are $4 \ cm$ and $8 \ cm$. If the refractive index of the material of the lens is $1.5$,the focal length of the lens is nearly (in $cm$)

$A$ student measures the focal length of a convex lens by placing an object pin at a distance '$u$' from the lens and measuring the distance '$v$' of the image pin. The graph between '$u$' and '$v$' plotted by the student should look like: