A thin equi-convex lens is made of glass of refractive index `1.5` and its length is `0.2 m.` If it acts as a concave lens of `0.5 m` focal length when dipped in a liquid, the refractive index of the liquid is

Options

A. `17/8`
B. `15/8`
C. `13/8`
D. `9/8`

Explanation

To solve the problem, we need to find the refractive index of the liquid in which the lens is dipped. We will use the lens maker's formula and the information provided in the question.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Refractive index of the lens (glass), \( \mu = 1.5 \)
- Focal length of the lens in air, \( f_{air} = 0.2 \, m \)
- Focal length of the lens in the liquid, \( f_{liquid} = -0.5 \, m \) (since it acts as a concave lens)

2. **Use the Lens Maker's Formula:**
The lens maker's formula for a thin lens is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

3. **Calculate for the Lens in Air:**
For the lens in air:
\[ \frac{1}{f_{air}} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ \frac{1}{0.2} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Rearranging gives:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{0.2 \times 0.5} = \frac{1}{0.1} = 10 \]

4. **Calculate for the Lens in Liquid:**
For the lens in the liquid, we denote the refractive index of the liquid as \( \mu_1 \):
\[ \frac{1}{f_{liquid}} = \left(\frac{\mu_1}{\mu} - 1\right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the known values:
\[ \frac{1}{-0.5} = \left(\frac{\mu_1}{1.5} - 1\right) \left(10\right) \]
This simplifies to:
\[ -2 = \left(\frac{\mu_1}{1.5} - 1\right) \times 10 \]
Dividing both sides by 10:
\[ -0.2 = \frac{\mu_1}{1.5} - 1 \]
Rearranging gives:
\[ \frac{\mu_1}{1.5} = 0.8 \]
Therefore:
\[ \mu_1 = 0.8 \times 1.5 = 1.2 \]

5. **Conclusion:**
The refractive index of the liquid is \( \mu_1 = 1.2 \).