Question

A pin is placed 10 cm in front of a convex lens of focal length 20 cm and made of a material of refractive index 1.5. The convex surface of the lens farther away from Jhe pin is silvered and has a radius of curvature of 22 cm. Determine the position of the final image. Is the image real or virtual ?

Answer 1

The given system behaves like a mirror of power given by $P=2P_L+P_M$ $or-\frac{1}{F}=2\left(\frac{1}{0.2}\right)+\left(\frac{-2}{0.22}\right)$ As $P_L=\frac{1}{f(m)}$ and $P_M=\frac{-1}{f(m)}=frac-2R(m)$ Solving this equation, we get F = -1.1 m = - 110cm i.e. the system behaves as a concave mirror of focal length 18.33 cm Using the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ we have $\frac{1}{v}-\frac{1}{10}=-\frac{1}{110}$ or $c=11cm$ i.e. virtual image will be formed at a distance of 11 cm.