$A$ lens with a focal length of $50 \ cm$ forms an image of a distant object that subtends an angle of $1 \text{ milliradian}$ at the lens. What is the size of the image in $mm$?

When an object is at a distance $u_1$ and $u_2$ from a lens,a real image and a virtual image are formed,respectively,having the same magnification $m$. The focal length of the lens is:

Explain how the image is formed by a thin convex lens and derive the lens maker's formula: $\frac{1}{f} = (n_{21} - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.

An equiconvex lens made of glass of refractive index $\frac{3}{2}$ has focal length $f$ in air. It is completely immersed in water of refractive index $\frac{4}{3}$. The percentage change in the focal length is

$A$ convex lens is placed between an object and a screen which are at a fixed distance $D$ apart. For one position of the lens,the magnification of the image formed on the screen is $m_1$. When the lens is shifted by a distance $d$,the magnification of the image formed on the same screen is $m_2$. The focal length of the lens is:

If the radii of curvature of both convex surfaces of a lens are $20 \, cm$,find the focal length of the lens in the given arrangement. (in $, cm$)