A plano-convex lens when silvered ono the plane side behaves like a concave mirror of focal length 60 cm. However, when silvered on the convex side, it behaves like a concave mirror of focal length 20cm. Then, the refractive index of the lens is

Options

A. `3.0`
B. `1.5`
C. `1.0`
D. `2.0`

Explanation

To solve the problem, we need to analyze the behavior of a plano-convex lens when it is silvered on different sides. We will use the lens and mirror formulas to find the refractive index of the lens.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a plano-convex lens.
- When silvered on the plane side, it behaves like a concave mirror with a focal length \( f_e = 60 \) cm.
- When silvered on the convex side, it behaves like a concave mirror with a focal length \( f_e' = 20 \) cm.
- We need to find the refractive index \( \mu \) of the lens.

2. **Using the Mirror Formula for the First Case:**
- For a mirror, the focal length \( f_m \) is related to the lens focal length \( f_l \) and the focal length of the plane mirror (which is infinite).
- The formula is given by:
\[ \frac{1}{f_e} = \frac{2}{f_l} + \frac{1}{f_m} \]
- Here, \( f_m = \infty \), so \( \frac{1}{f_m} = 0 \).
- Plugging in the values:
\[ \frac{1}{60} = \frac{2}{f_l} + 0 \]
- Rearranging gives:
\[ \frac{2}{f_l} = \frac{1}{60} \]
- Therefore:
\[ f_l = 120 \text{ cm} \]

3. **Using the Mirror Formula for the Second Case:**
- Now, we consider the case when the lens is silvered on the convex side.
- The formula remains the same:
\[ \frac{1}{f_e'} = \frac{2}{f_l} + \frac{1}{f_m'} \]
- Here, \( f_m' \) is the focal length of the concave mirror formed by silvering the convex side.
- Plugging in the values:
\[ \frac{1}{20} = \frac{2}{120} + \frac{1}{f_m'} \]
- Simplifying:
\[ \frac{1}{20} = \frac{1}{60} + \frac{1}{f_m'} \]
- Rearranging gives:
\[ \frac{1}{f_m'} = \frac{1}{20} - \frac{1}{60} \]
- Finding a common denominator (60):
\[ \frac{1}{f_m'} = \frac{3 - 1}{60} = \frac{2}{60} = \frac{1}{30} \]
- Therefore:
\[ f_m' = 30 \text{ cm} \]

4. **Relating Focal Length to Radius of Curvature:**
- The focal length of a concave mirror is related to its radius of curvature \( R \) by:
\[ f_m' = \frac{R}{2} \]
- Thus, for \( f_m' = 30 \) cm:
\[ R = 2 \times 30 = 60 \text{ cm} \]

5. **Using the Lens Maker's Formula:**
- The lens maker's formula relates the focal length of the lens to its refractive index \( \mu \):
\[ \frac{1}{f_l} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- For a plano-convex lens:
- \( R_1 = R = 60 \) cm (convex side)
- \( R_2 = \infty \) (plane side, so \( \frac{1}{R_2} = 0 \))
- Plugging in the values:
\[ \frac{1}{120} = (\mu - 1) \left( \frac{1}{60} - 0 \right) \]
- This simplifies to:
\[ \frac{1}{120} = (\mu - 1) \frac{1}{60} \]
- Rearranging gives:
\[ \mu - 1 = \frac{1}{120} \times 60 = \frac{1}{2} \]
- Therefore:
\[ \mu = 1 + \frac{1}{2} = 1.5 \]

### Final Answer:
The refractive index \( \mu \) of the lens is **1.5**.