Ratio of Magnifications for Mirror Lens System in Different Media

To find the magnitude of the ratio of the magnifications $\left| \frac{M_2}{M_1} \right|$, we analyze the image formation by the mirror-lens combination in both cases step-by-step.

1. Image Formation by the Concave Mirror

The focal length of a spherical mirror depends only on its radius of curvature, so it remains unchanged when immersed in a medium: $f_m = -10 \text{ cm}$

The object is placed in front of the concave mirror at a distance $u_m = -15 \text{ cm}$.

Using the mirror formula: $\frac{1}{v_m} + \frac{1}{u_m} = \frac{1}{f_m}$

Substitute the values: $\frac{1}{v_m} + \frac{1}{-15} = \frac{1}{-10}$ $\frac{1}{v_m} = -\frac{1}{10} + \frac{1}{15} = -\frac{1}{30} \implies v_m = -30 \text{ cm}$

The linear magnification produced by the mirror is: $m_m = -\frac{v_m}{u_m} = -\frac{-30}{-15} = -2$

Since the distance between the mirror and the lens is $d = 50 \text{ cm}$, the image formed by the mirror acts as a real object for the convex lens at a distance: $u_l = -(50 - 30) = -20 \text{ cm}$

---

2. Magnification in Air ($M_1$)

In air ($\mu_{\text{air}} = 1$), the focal length of the convex lens is $f_{l1} = +10 \text{ cm}$.

Using the lens formula: $\frac{1}{v_{l1}} - \frac{1}{u_l} = \frac{1}{f_{l1}}$

$\frac{1}{v_{l1}} - \frac{1}{-20} = \frac{1}{10}$ $\frac{1}{v_{l1}} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20} \implies v_{l1} = +20 \text{ cm}$

The magnification of the lens in air is: $m_{l1} = \frac{v_{l1}}{u_l} = \frac{20}{-20} = -1$

Thus, the total magnification $M_1$ of the system in air is: $M_1 = m_m \times m_{l1} = (-2) \times (-1) = 2$

---

3. Magnification in the Medium ($M_2$)

When the entire setup is placed in a medium of refractive index $\mu_m = \frac{7}{6}$, the focal length of the lens changes.

From Lens Maker's Formula: $\frac{1}{f_{l1}} = \left(\frac{\mu_l}{\mu_{\text{air}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $\frac{1}{10} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \implies \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{5} \text{ cm}^{-1}$

For the medium of refractive index $\mu_m = \frac{7}{6}$: $\frac{1}{f_{l2}} = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $\frac{1}{f_{l2}} = \left(\frac{1.5}{7/6} - 1\right) \left(\frac{1}{5}\right) = \left(\frac{9}{7} - 1\right) \frac{1}{5} = \frac{2}{35} \text{ cm}^{-1}$ $f_{l2} = \frac{35}{2} \text{ cm}$

Now, using the lens formula for the lens in the medium: $\frac{1}{v_{l2}} - \frac{1}{-20} = \frac{2}{35}$ $\frac{1}{v_{l2}} = \frac{2}{35} - \frac{1}{20} = \frac{8 - 7}{140} = \frac{1}{140} \implies v_{l2} = +140 \text{ cm}$

The magnification of the lens in the medium is: $m_{l2} = \frac{v_{l2}}{u_l} = \frac{140}{-20} = -7$

Thus, the total magnification $M_2$ of the system in the medium is: $M_2 = m_m \times m_{l2} = (-2) \times (-7) = 14$

---

4. Ratio of Magnifications

The ratio of the magnitude of magnifications is: $\left| \frac{M_2}{M_1} \right| = \left| \frac{14}{2} \right| = 7$