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 will use the lens maker's formula and the relationship between the focal lengths of the lens in air and in the liquid.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Refractive index of the lens (glass), \( \mu_{lens} = 1.5 \)
- Focal length of the lens in air, \( f_1 \) (not given directly, but we will assume it to be positive for a convex lens).
- Focal length of the lens in the liquid, \( f_2 = -0.5 \, m \) (since it acts as a concave lens).

2. **Use the Lens Maker's Formula**:
The lens maker's formula states:
\[ \frac{1}{f} = (\mu_{lens} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Here, \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces. For an equi-convex lens, \( R_1 = R \) and \( R_2 = -R \).
Thus,
\[ \frac{1}{f} = (\mu_{lens} - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = 2(\mu_{lens} - 1) \frac{1}{R} \]

3. **Calculate the Focal Length in Air**:
Since we don't have the radius of curvature, we can express \( f_1 \) in terms of \( R \):
\[ f_1 = \frac{R}{2(\mu_{lens} - 1)} = \frac{R}{2(1.5 - 1)} = \frac{R}{2 \times 0.5} = \frac{R}{1} = R \]

4. **Set Up the Equation for Focal Length in Liquid**:
When the lens is dipped in a liquid of refractive index \( \mu_{medium} \), the new focal length \( f_2 \) is given by:
\[ \frac{1}{f_2} = \left( \mu_{lens} - \mu_{medium} \right) \left( \frac{1}{R} + \frac{1}{R} \right) = 2 \left( \mu_{lens} - \mu_{medium} \right) \frac{1}{R} \]
Thus,
\[ f_2 = \frac{R}{2(\mu_{lens} - \mu_{medium})} \]

5. **Relate the Two Focal Lengths**:
We know:
\[ f_1 = R \quad \text{and} \quad f_2 = -0.5 \]
Setting the two equations equal gives:
\[ R = -0.5 \cdot 2(\mu_{lens} - \mu_{medium}) \]
Substituting \( \mu_{lens} = 1.5 \):
\[ R = -0.5 \cdot 2(1.5 - \mu_{medium}) \]
Simplifying:
\[ R = -1(1.5 - \mu_{medium}) \implies R = -1.5 + \mu_{medium} \]

6. **Equate the Two Expressions for R**:
From the earlier steps, we have:
\[ R = -1.5 + \mu_{medium} \]
Since \( R \) can be expressed as \( R = 1.5 \) (from the first focal length calculation):
\[ 1.5 = -1.5 + \mu_{medium} \]
Solving for \( \mu_{medium} \):
\[ \mu_{medium} = 1.5 + 1.5 = 3.0 \]

### Final Answer:
The refractive index of the liquid is \( \mu_{medium} = 3.0 \).