A lens behaves as a converging lens in air and a diverging lens in water. The refractive index of the material is (refractive index of water `=1.33`)

Options

A. equal to unity
B. equal to `1.33`
C. between unity and `1.33`
D. greater than `1.33`

Explanation

To solve the problem, we will use the lens maker's formula, which relates the focal length of a lens to its refractive index and the radii of curvature of its surfaces. The formula is given by:

\[ \frac{1}{f} = \left(\frac{\mu_L}{\mu_M} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Where:
- \( f \) = focal length of the lens
- \( \mu_L \) = refractive index of the lens
- \( \mu_M \) = refractive index of the medium
- \( R_1 \) and \( R_2 \) = radii of curvature of the lens surfaces

### Step-by-Step Solution:

1. **Condition in Air (Converging Lens)**:
- In air, the refractive index \( \mu_M = 1 \).
- The lens behaves as a converging lens, so the focal length \( f \) is positive.
- Using the lens maker's formula:
\[ \frac{1}{f} = \left(\mu_L - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \text{(Equation 1)} \]

2. **Condition in Water (Diverging Lens)**:
- In water, the refractive index \( \mu_M = 1.33 \).
- The lens behaves as a diverging lens, so the focal length \( f \) is negative.
- Using the lens maker's formula:
\[ \frac{1}{f} = \left(\frac{\mu_L}{1.33} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \text{(Equation 2)} \]

3. **Setting Up the Equations**:
- From Equation 1:
\[ \frac{1}{f} = \left(\mu_L - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
- From Equation 2:
\[ -\frac{1}{f} = \left(\frac{\mu_L}{1.33} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

4. **Equating the Two Expressions**:
- Set the two expressions for \(\frac{1}{f}\) equal to each other:
\[ \left(\mu_L - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = -\left(\frac{\mu_L}{1.33} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
- Let \( x = \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \):
\[ (\mu_L - 1)x = -\left(\frac{\mu_L}{1.33} - 1\right)x \]
- Assuming \( x \neq 0 \) (since the lens has curvature), we can divide both sides by \( x \):
\[ \mu_L - 1 = -\left(\frac{\mu_L}{1.33} - 1\right) \]

5. **Solving for \( \mu_L \)**:
- Rearranging gives:
\[ \mu_L - 1 = -\frac{\mu_L}{1.33} + 1 \]
- Combine like terms:
\[ \mu_L + \frac{\mu_L}{1.33} = 2 \]
- Factor out \( \mu_L \):
\[ \mu_L \left(1 + \frac{1}{1.33}\right) = 2 \]
- Calculate \( 1 + \frac{1}{1.33} \):
\[ 1 + \frac{1}{1.33} = \frac{1.33 + 1}{1.33} = \frac{2.33}{1.33} \]
- Thus:
\[ \mu_L \cdot \frac{2.33}{1.33} = 2 \]
- Solving for \( \mu_L \):
\[ \mu_L = \frac{2 \cdot 1.33}{2.33} \approx 1.14 \]

### Final Answer:
The refractive index of the lens material is approximately \( \mu_L \approx 1.14 \).