The focal length of a convex lens of `R.I. 1.5` is `f` when it is placed in air. When it is immersed in a liquid is behaves as a converging lens its focal length becomes `xf(x gt 1)`. The refractive index of the liquid

Options

A. `gt 3//2`
B. `lt (3//2)` and `lt 1`
C. `lt 3//2`
D. All of these

Answer

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.

### Step-by-Step Solution:

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

2. **Focal Length in Air**:
When the lens is in air, we have:
- \( \mu_1 = 1.5 \) (refractive index of the lens),
- \( \mu_2 = 1 \) (refractive index of air).
Thus, the equation becomes:
\[ \frac{1}{f} = (1.5 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

3. **Focal Length in Liquid**:
When the lens is immersed in a liquid, we denote the refractive index of the liquid as \( n \). The lens behaves as a converging lens, and its focal length becomes \( xf \) (where \( x > 1 \)). The formula now becomes:
\[ \frac{1}{xf} = (1.5 - n) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

4. **Setting Up the Equations**:
From the two scenarios, we can set up the following equations:
- From air:
\[ \frac{1}{f} = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \text{(1)} \]
- From liquid:
\[ \frac{1}{xf} = (1.5 - n) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \text{(2)} \]

5. **Dividing the Equations**:
Dividing equation (2) by equation (1):
\[ \frac{1/xf}{1/f} = \frac{(1.5 - n)}{0.5} \]
This simplifies to:
\[ \frac{1}{x} = \frac{(1.5 - n)}{0.5} \]
Rearranging gives:
\[ 1.5 - n = 0.5 \cdot \frac{1}{x} \]
Thus:
\[ n = 1.5 - \frac{0.5}{x} \]

6. **Finding the Refractive Index**:
Since \( x > 1 \), the term \( \frac{0.5}{x} < 0.5 \). Therefore:
\[ n < 1.5 + 0.5 = 2 \]
Also, since \( n \) must be greater than 1 (as it is a liquid), we have:
\[ 1 < n < 2 \]

7. **Conclusion**:
The refractive index of the liquid is:
\[ n < \frac{3}{2} \quad \text{and} \quad n > 1 \]

### Final Answer:
The refractive index of the liquid is \( n \) such that \( 1 < n < \frac{3}{2} \).