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 (refractive index $=1$) 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 $\left|\frac{M_2}{M_1}\right|$ is

Answer 1

The distance of the first image formed by reflection from the spherical concave mirror can be found using mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}....\left(1\right)$

For concave mirror: $u=-15\text{cm},f=-10\text{cm}$

Substituting in equation $\left(1\right)$,

$\frac{1}{v}+\frac{1}{-15}=\frac{1}{-10}$

$\Rightarrow \frac{1}{v}=\frac{10-15}{150}$

$\therefore v=-30\text{cm}$

The magnification is given as,

$m_1=-\frac{v}{u}=-\frac{-30}{-15}=-2$

The first image will act as the object for the convex lens and distance of image formed by the lens can be found using lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}....\left(2\right)$

For convex lens:
$u=-20\text{cm},f=10\text{cm}$

Substituting in equation $\left(2\right)$,

$\frac{1}{v}-\frac{1}{-20}=\frac{1}{10}$

$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{2-1}{20}$

$\therefore v=20\text{cm}$

The magnification is given as,

$m_2=\frac{v}{u}=\frac{20}{-20}=-1$

Therefore, total magnification,

$M_1=m_1\times m_2=-2\times -1=2$

Now, the whole setup is immersed in liquid of $\mu =\frac{7}{6}$. The reflection from concave mirror will remain same as there is no change in object position and focal length.

The change in focal length of convex lens can be found using lens maker formula.

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

Now, when the arrangement is placed in air,

$\frac{1}{f_{air}}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2}).....\left(3\right)$

& when the arrangement is placed in the 2nd medium,
$\frac{1}{f_{\text{medium}}}=(\frac{1.5}{\frac{7}{6}}-1)(\frac{1}{R_1}-\frac{1}{R_2}).....\left(4\right)$

Dividing (3) by (4),

$\Rightarrow \frac{f_{\text{medium}}}{f_{air}}=\frac{\frac{1}{2}}{\frac{3}{2}\times \frac{6}{7}-1}$

$\Rightarrow \frac{f_{\text{medium}}}{f_{air}}=\frac{7}{4}$

$\therefore f_{\text{medium}}=f_{air}\times \frac{7}{4}=10\times \frac{7}{4}=\frac{70}{4}\text{cm}$

Now, position of the final image formed can be found using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{medium}}.....\left(4\right)$

For convex lens: $u=-20\text{cm},f=\frac{70}{4}\text{cm}$

Substittuting in equation (4)

$\frac{1}{v}-\frac{1}{-20}=\frac{4}{70}$

$\Rightarrow \frac{1}{v}=\frac{4}{70}-\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{80-70}{1400}$

$\therefore v=140\text{cm}$

The magnification is given as $m_3=\frac{v}{u}=\frac{140}{-20}=-7$

Total magnification in this case $M_2=m_1\times m_3=-2\times -7=14$

$\therefore \frac{M_2}{M_1}=\frac{14}{2}=7$