A thin lens is made with as material having refractive index `mu=1.5.` both the sides are convex. It is dipped in water `(mu=1.33)`. It will behave like

Options

A. a convergent lens
B. b divergent lens
C. c rectangular slab
D. d prism

Answer

To determine how a thin lens behaves when submerged in water, we can analyze the situation step by step.

### Step-by-Step Solution:

1. **Identify the Refractive Indices**:
- The refractive index of the lens material, \( \mu_L = 1.5 \).
- The refractive index of water, \( \mu_W = 1.33 \).

2. **Determine the Lens Type**:
- The lens is a thin lens with both sides convex. In air, a convex lens converges light.

3. **Calculate the Focal Length of the Lens in Air**:
- The formula for the focal length \( f_L \) of a lens in air is given by:
\[ \frac{1}{f_L} = (\mu_L - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- For a thin lens with both surfaces convex, we can assume \( R_1 = +R \) and \( R_2 = -R \):
\[ \frac{1}{f_L} = (1.5 - 1) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) \]
\[ \frac{1}{f_L} = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R} \]
- Thus, \( f_L = R \).

4. **Calculate the Focal Length of the Lens in Water**:
- When the lens is submerged in water, we need to recalculate the focal length using the refractive index of water:
\[ \frac{1}{f_W} = \left(\frac{\mu_L}{\mu_W} - 1\right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- Here, \( \frac{\mu_L}{\mu_W} = \frac{1.5}{1.33} \approx 1.13 \):
\[ \frac{1}{f_W} = (1.13 - 1) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) \]
\[ \frac{1}{f_W} = 0.13 \left( \frac{2}{R} \right) = \frac{0.26}{R} \]
- Thus, \( f_W = \frac{R}{0.26} \).

5. **Determine the Nature of the Lens**:
- Since \( f_W \) is positive, the lens remains a converging lens even when submerged in water.

6. **Conclusion**:
- The lens behaves like a converging lens when submerged in water.

### Final Answer:
The thin lens behaves like a **converging lens** when submerged in water.