A thin lens, made of glass of refractive index `3//2,` produces a real and magnified image of an object in air. If the whole system, maintaing the same distance between the object and the lens, is immersed in water `(RI=4//3),` then the image formed will be

Options

A. real, magnified
B. real,diminished
C. virtual, magnified
D. virtual,diminished

Explanation

To solve the problem step by step, we will analyze the situation of the lens in both air and water, and how the change in medium affects the focal length and the characteristics of the image formed.

### Step-by-Step Solution:

1. **Identify the given data:**
- Refractive index of the lens (glass), \( \mu_{lens} = \frac{3}{2} \)
- Refractive index of air, \( \mu_{air} = 1 \)
- Refractive index of water, \( \mu_{water} = \frac{4}{3} \)
- The lens produces a real and magnified image in air.

2. **Calculate the focal length of the lens in air:**
Using the lens maker's formula:
\[ \frac{1}{F} = \left( \mu_{lens} - \mu_{medium} \right) \cdot \frac{1}{R} \]
In air, the focal length \( F \) can be expressed as:
\[ \frac{1}{F} = \left( \frac{3}{2} - 1 \right) \cdot \frac{1}{R} = \frac{1}{2R} \]
Thus, \( F = 2R \).

3. **Calculate the focal length of the lens in water:**
When the system is immersed in water, the new focal length \( F' \) is given by:
\[ \frac{1}{F'} = \left( \mu_{lens} - \mu_{water} \right) \cdot \frac{1}{R} \]
Substituting the values:
\[ \frac{1}{F'} = \left( \frac{3}{2} - \frac{4}{3} \right) \cdot \frac{1}{R} \]
To simplify:
\[ \frac{3}{2} = \frac{9}{6}, \quad \frac{4}{3} = \frac{8}{6} \]
Thus,
\[ \frac{1}{F'} = \left( \frac{9}{6} - \frac{8}{6} \right) \cdot \frac{1}{R} = \frac{1}{6R} \]
Therefore, \( F' = 6R \).

4. **Determine the relationship between the new and old focal lengths:**
From the calculations:
\[ F' = 3F \quad \text{(since \( F = 2R \) and \( F' = 6R \))} \]

5. **Analyze the image formation:**
Initially, the object is placed between the center of curvature (2F) and the focal point (F) to produce a real and magnified image. With the new focal length \( F' = 3F \), the object will now be between the optical center and the new focal point.

6. **Determine the nature of the new image:**
When the object is placed between the optical center (0) and the new focal point (F'), a virtual and magnified image is formed.

### Conclusion:
The image formed when the system is immersed in water will be a **virtual and magnified image**.