Question

Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length $10\text{cm}$ each, separated by a distance of $50\text{cm}$ in air as shown in the figure. An object is placed at a distance of $15\text{cm}$ from the mirror. Its erect image formed by this combination has magnification $M_1$.When the set up is kept in a medium of refractive index $\frac{7}{6}$, the magnification becomes $M_2$. The magnitude of $\left|\frac{M_2}{M_1}\right|$ is

Answer 1

From the mirror formula,
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u},u=-15\text{cm},f=-10\text{cm}$
$\frac{1}{-10}=\frac{1}{v}+\frac{1}{-15}$
So, $v=-30\text{cm}$ (right).
Magnification due to mirror is $M_{1m}=\frac{-v}{u}=-2$
Now $u$ for the lens $=20\text{cm}=2f_{lens}$.
From the lens formula , $v_{lens}=20\text{cm}$ (right)
$\Rightarrow$ magnification by the lens $|M_{1L}|=1\text{(in air)}$.
Net Magnification $M_1=-2\left(1\right)=-2$

When kept in liquid $({\mu }^{\prime }=\frac{7}{6})$,
Change of medium does not affect $f$ for mirror and hence magnification as well.
$M_{2m}=-2$
The focal length of the lens is changed from $f=10\text{cm}$ to $f^{\prime }$
$\frac{\frac{1}{f}}{\frac{1}{f^{\prime }}}=\frac{({\mu }_g-1)(\frac{1}{R_1}-\frac{1}{R_2})}{(\frac{{\mu }_g}{{\mu }^{\prime }}-1)(\frac{1}{R_1}-\frac{1}{R_2})}=\frac{(\frac{3}{2}-1)}{(\frac{3}{2}\frac{6}{7}-1)}\Rightarrow f^{\prime }=\frac{35}{2}\text{cm}$.

So, for lens, $u=20\text{cm}$ and $\frac{1}{v^{\prime }}-\frac{1}{-30}=\frac{2}{35}\Rightarrow v^{\prime }=140\text{cm}$.
$\therefore M_{2L}=\frac{v^{\prime }}{u^{\prime }}=\frac{140}{20}=7$.
Net magnification $M_2=-2\left(7\right)=-14$
Now, for the system, $\frac{M_2}{M_1}=7$.