A thin plano-convex lens acts like a concave mirror of radius of curvature `20 cm` when its plane surface is silvered. The radius of curvature of the curved surface if index of refraction of its matarial is `1.5` will be

Options

A. 40 cm
B. 30 cm
C. 10 cm
D. 20 cm

Explanation

To solve the problem, we need to find the radius of curvature of the curved surface of a plano-convex lens that acts like a concave mirror when its plane surface is silvered. Given that the radius of curvature of the concave mirror is -20 cm (since it is a concave mirror), and the refractive index of the lens material is 1.5, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of curvature of the concave mirror, \( R_m = -20 \, \text{cm} \)
- Focal length of the concave mirror, \( F_m = \frac{R_m}{2} = \frac{-20}{2} = -10 \, \text{cm} \)
- Refractive index of the lens material, \( \mu = 1.5 \)

2. **Use the Lens Maker's Formula:**
The formula for the focal length \( F \) of a thin lens is given by:
\[ \frac{1}{F} = \left( \frac{\mu - 1}{R_1} - \frac{\mu - 1}{R_2} \right) \]
For a plano-convex lens:
- The flat surface (plane surface) has \( R_2 = \infty \) (which means it does not contribute to the curvature).
- The curved surface has \( R_1 = R \) (which we need to find).

3. **Substituting Values into the Formula:**
Since \( R_2 = \infty \), the term involving \( R_2 \) becomes zero:
\[ \frac{1}{F} = \frac{\mu - 1}{R} \]
Substituting \( F = -10 \, \text{cm} \) and \( \mu = 1.5 \):
\[ \frac{1}{-10} = \frac{1.5 - 1}{R} \]
\[ \frac{1}{-10} = \frac{0.5}{R} \]

4. **Cross-Multiplying to Solve for \( R \):**
\[ R = -10 \times 0.5 = -5 \, \text{cm} \]

5. **Finding the Radius of Curvature:**
Since we are looking for the radius of curvature of the curved surface, we take the absolute value:
\[ R = 20 \, \text{cm} \]

### Final Answer:
The radius of curvature of the curved surface of the plano-convex lens is \( 20 \, \text{cm} \).