A glass concave lens is placed in a liquid in which it behaves like a convergent lens. If the refractive indices of glass and liquid with respect to air are `""_(a)mu_(g)and""_(a)mu_(l)` respectively, then

Options

A. `""_(a)mu_(g)=5""_(a)mu_(l)`
B. `""_(a)mu_(g)gt""_(a)mu_(l)`
C. `""_(a)mu_(g)lt""_(a)mu_(l)`
D. `""_(a)mu_(g)=2""_(a)mu_(l)`

Answer

To solve the problem, we need to analyze the behavior of a concave lens when placed in a liquid, and how it can behave like a convergent lens under certain conditions.

### Step-by-Step Solution:

1. **Understanding the Lens Types**:
- A concave lens is a diverging lens, which means it usually has a negative focal length.
- A convergent lens, on the other hand, is a convex lens with a positive focal length.

2. **Lens Maker's Formula**:
- The focal length \( f \) of a lens can be determined using the lens maker's formula:
\[ \frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
- Here, \( \mu_2 \) is the refractive index of the lens material (glass), \( \mu_1 \) is the refractive index of the surrounding medium (liquid), \( R_1 \) is the radius of curvature of the first surface, and \( R_2 \) is the radius of curvature of the second surface.

3. **Assigning Values**:
- For a concave lens, \( R_1 \) is negative (since it is a diverging surface) and \( R_2 \) is positive (since it is a converging surface).
- Thus, we can write:
\[ \frac{1}{f} = \left(\frac{\mu_g}{\mu_l} - 1\right) \left(-\frac{1}{|R|} + \frac{1}{|R|}\right) \]

4. **Condition for Convergence**:
- For the lens to behave as a convergent lens, the focal length \( f \) must be positive. This means:
\[ \frac{\mu_g}{\mu_l} - 1 > 0 \]
- Rearranging this gives:
\[ \frac{\mu_g}{\mu_l} > 1 \implies \mu_g > \mu_l \]

5. **Conclusion**:
- Therefore, for the concave lens to behave like a convergent lens in a liquid, the refractive index of glass \( \mu_g \) must be greater than the refractive index of the liquid \( \mu_l \).

### Final Answer:
\[ \mu_g > \mu_l \]