A biconvex thin lens is prepared from glass of refractive index `mu_(2)=(3)/(2)`. The two conducting surfaces have equal radii of 20 cm each. One of the the surface is silvered from outside to make it reflecting. It is placed in a medium of refractive index `mu_(1)=(5)/(3)`. It acts as a

Options

A. converging mirror
B. diverging mirror
C. concave mirror of focal length 12.5 cm
D. convex mirror of focal length 12.5 cm

Answer

To solve the problem step by step, we need to determine whether the given biconvex lens with one silvered surface acts as a converging or diverging mirror and calculate its focal length.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Refractive index of the lens (glass), \( \mu_2 = \frac{3}{2} \)
- Refractive index of the medium, \( \mu_1 = \frac{5}{3} \)
- Radius of curvature of both surfaces, \( R_1 = R_2 = 20 \, \text{cm} \)

2. **Determine the Type of Mirror**:
- Since one surface of the biconvex lens is silvered, it behaves like a concave mirror. A concave mirror converges light rays, hence it acts as a converging mirror.

3. **Calculate the Focal Length of the Lens**:
- We will use the Lens Maker's Formula:
\[ \frac{1}{F_L} = \left( \mu_2 - \mu_1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- Substituting the values:
\[ \frac{1}{F_L} = \left( \frac{3}{2} - \frac{5}{3} \right) \left( \frac{1}{20} - \left(-\frac{1}{20}\right) \right) \]

4. **Calculate the Difference in Refractive Indices**:
- Convert \( \frac{3}{2} \) and \( \frac{5}{3} \) to a common denominator:
\[ \frac{3}{2} = \frac{9}{6}, \quad \frac{5}{3} = \frac{10}{6} \]
- Thus,
\[ \frac{3}{2} - \frac{5}{3} = \frac{9}{6} - \frac{10}{6} = -\frac{1}{6} \]

5. **Calculate the Curvature Terms**:
- The curvature terms:
\[ \frac{1}{20} - \left(-\frac{1}{20}\right) = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \]

6. **Substitute Back into the Lens Formula**:
- Now substituting back:
\[ \frac{1}{F_L} = -\frac{1}{6} \cdot \frac{1}{10} = -\frac{1}{60} \]
- Therefore, the focal length of the lens is:
\[ F_L = -60 \, \text{cm} \]

7. **Calculate the Focal Length of the Mirror**:
- The focal length of the concave mirror formed by the silvered surface is half the radius of curvature:
\[ F_M = \frac{R}{2} = \frac{20}{2} = 10 \, \text{cm} \]

8. **Calculate the Equivalent Focal Length**:
- The equivalent focal length of the system (acting as a mirror) can be calculated as:
\[ \frac{1}{F_{equiv}} = \frac{1}{F_L} + \frac{1}{F_M} \]
- Substituting the values:
\[ \frac{1}{F_{equiv}} = -\frac{1}{60} + \frac{1}{10} = -\frac{1}{60} + \frac{6}{60} = \frac{5}{60} = \frac{1}{12} \]
- Therefore, the equivalent focal length is:
\[ F_{equiv} = 12 \, \text{cm} \]

### Final Answer:
The biconvex lens with one silvered surface acts as a **converging mirror** and has an equivalent focal length of **12 cm**.