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 that causes it to act as a converging lens. We will use the lens maker's formula and the properties of refractive indices.

### Step-by-Step Solution:

1. **Understanding the Lens Maker's Formula**:
The lens maker's formula is given by:
\[ \frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
where:
- \( f \) is the focal length of the lens,
- \( \mu_2 \) is the refractive index of the lens material (glass),
- \( \mu_1 \) is the refractive index of the surrounding medium (liquid),
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

2. **Identifying the Signs of Radii**:
For a concave lens:
- The radius of curvature \( R_1 \) (for the first surface) is negative,
- The radius of curvature \( R_2 \) (for the second surface) is positive.
Thus, we can denote:
\[ R_1 = -r \quad \text{and} \quad R_2 = +r \]

3. **Substituting into the Lens Maker's Formula**:
Substituting the values of \( R_1 \) and \( R_2 \) into the lens maker's formula gives:
\[ \frac{1}{f} = \left(\frac{\mu_g}{\mu_l} - 1\right) \left(-\frac{1}{r} - \frac{1}{r}\right) = \left(\frac{\mu_g}{\mu_l} - 1\right) \left(-\frac{2}{r}\right) \]

4. **Condition for Convergence**:
For the lens to behave as a converging lens, the focal length \( f \) must be positive. This implies:
\[ \left(\frac{\mu_g}{\mu_l} - 1\right) \left(-\frac{2}{r}\right) > 0 \]
Since \(-\frac{2}{r}\) is negative (as \( r > 0 \)), we need:
\[ \frac{\mu_g}{\mu_l} - 1 < 0 \quad \Rightarrow \quad \frac{\mu_g}{\mu_l} < 1 \]
This means:
\[ \mu_g < \mu_l \]

5. **Conclusion**:
Since \( \mu_g \) is the refractive index of the glass and \( \mu_l \) is the refractive index of the liquid, we conclude:
\[ \mu_{ag} < \mu_{al} \]
Therefore, the correct answer is that the refractive index of glass with respect to air is less than that of the liquid with respect to air.

### Final Answer:
\[ \mu_{ag} < \mu_{al} \]