Question

A candle is placed in front of a concave spherical mirror on its principal axis at a distance of $\left(\frac{4}{3}\right)$ $f$ from the pole of the mirror (here $f$ is the focal length of the mirror). The candle is arranged at right angles to the axis. The image of the candle in the concave mirror impinges upon a convex mirror of focal length $2f$. The distance between the mirrors is $3f$ and their axis coincide. The image of the candle in the first mirror plays the part of a virtual object with respect to the second mirror and gives a real image arranged between the two mirrors. Then the total linear magnification of the system is (only magnitude)

Answer 1

For $1^{st}$ reflection
$u=\frac{-4f}{3}$ and $f=-f$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}+\frac{3}{-4f}=\frac{1}{-f}$
$\frac{1}{v}=\frac{3}{4f}-\frac{1\times 4}{f\times 4},v=-4f$
$m_1=-\frac{(-4f)}{-4f}\times 3=-3$
For $2^{nd}$ reflection
$u=f$ and $f=2f$
$\frac{1}{v}+\frac{1}{f}=\frac{1}{2f}$
$\frac{1}{v}=\frac{1}{2f}-\frac{1}{f}$
$v=-2f$
$m_2=-\frac{(-2f)}{f}=+2$
Total linear magnification = $m_1{m}_2$
= $m_1{m}_2=-3\times 2=-6$
$|m_1{m}_2|=6$