If a convex lens of refractive index `mu _l` is placed in a medium of refractive index `mu_m` Such that `mu_l > mu_m > mu_(air)` then

Options

A. The nature of the lens is diverging
B. The nature of the lens is converging
C. The nalure of the lens may be converging or diverging
D. The nature of the lens can't be determined

Answer

To solve the problem step by step, we will use the lens maker's formula and analyze the conditions given in the question.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have a convex lens with a refractive index \( \mu_l \).
- The lens is placed in a medium with a refractive index \( \mu_m \).
- The conditions given are \( \mu_l > \mu_m > \mu_{air} \).

2. **Lens Maker's Formula**:
- The lens maker's formula is given by:
\[ \frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- Here, \( f \) is the focal length of the lens, \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

3. **Analyzing the Condition**:
- Since \( \mu_l > \mu_m \), we can say:
\[ \frac{\mu_l}{\mu_m} > 1 \]
- This implies that:
\[ \frac{\mu_l}{\mu_m} - 1 > 0 \]
- Therefore, the term \( \left( \frac{\mu_l}{\mu_m} - 1 \right) \) is positive.

4. **Nature of the Lens**:
- For a convex lens, the term \( \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \) is also positive because \( R_1 \) (the radius of curvature of the first surface) is positive and \( R_2 \) (the radius of curvature of the second surface) is negative.
- Thus, both terms in the lens maker's formula are positive, leading to:
\[ \frac{1}{f} > 0 \]
- This means that the focal length \( f \) is positive.

5. **Conclusion**:
- A positive focal length indicates that the lens is converging. Therefore, the nature of the lens is converging.

### Final Answer:
The nature of the lens is converging.