Question

Consider a concave mirror and a convex lens (refractive index = 1. 5) of focal length 10 cm each , separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 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 7/6, the magnification becomes $M_2$. The magnitude $\left|\frac{M_2}{M_1}\right|$ is

Answer 1

Answer: 7

In vacuum
For mirror $u=-15,f=-10$
then $\frac{1}{v}+\frac{1}{-15}=\frac{1}{-10}$ (According to mirror formula)
$v=-30$
$m_1=-\left(\frac{-30}{-15}\right)=-2$
ie $1^{st}$ image is formed at 30 cm before mirror. Now it will behave as object for lens.
hence for lens u $=-20,f=10$
$\frac{1}{v}-\frac{1}{-20}=\frac{1}{10}$
$v=20$
$m_2=\left(\frac{20}{-20}\right)=-1$
overall magnification $M_1=m_1{m}_2=2$
In vacuum
$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{10}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{5}$
In medium for lens- $\left(\mu =\frac{7}{6}\right)$
$\frac{1}{f^{\prime }}=\left(\frac{\frac{3}{2}}{\frac{7}{6}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{f^{\prime }}=\frac{2}{7}\times \frac{1}{5}$
$f^{\prime }=\frac{35}{2}$
there is no change for mirror. hence for mirror
$u=-15,f=-10$
then $v=-30$
$m_1=-2$
For Lens
$u=-20,f=\frac{35}{2}$
$\frac{1}{v}-\frac{1}{-20}=\frac{2}{35}$
$v=140$
$m_2=\left(\frac{140}{-20}\right)=-7$
Overall magnification $M_2=m_1{m}_2=14$
Now $\left(\frac{M_2}{M_1}\right)=\frac{14}{2}=7$