A convex lens made of glass (`mu_(g)`= 3/2) has focal length f in air. The image of an object placed . infront of it is inverted real and magnified. Now the whole arrangement is immersed in water (`mu_(w)` = 4/3) without changing the distance between object and lens. Then

Options

A. the new focal length will become 4f
B. the new focal length will become f/4
C. new image will be virtual and magnified
D. new image will be real inverted and smaller in size.

Answer

To solve the problem step by step, we will analyze the situation of a convex lens first in air and then in water.

### Step 1: Understanding the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{\mu_2}{\mu_1} - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where:
- \( f \) = focal length of the lens
- \( \mu_2 \) = refractive index of the lens material (glass)
- \( \mu_1 \) = refractive index of the medium outside the lens (air or water)
- \( R_1 \) and \( R_2 \) = radii of curvature of the lens surfaces

### Step 2: Focal Length in Air
In air, the refractive index \( \mu_1 \) is approximately 1. The refractive index of glass is given as \( \mu_g = \frac{3}{2} \). Thus, we can write:
\[ \frac{1}{f} = \frac{\frac{3}{2}}{1} - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Let this be Equation (1).

### Step 3: Focal Length in Water
Now, when the lens is immersed in water, the refractive index of water \( \mu_w = \frac{4}{3} \). The lens formula now becomes:
\[ \frac{1}{f'} = \frac{\mu_g}{\mu_w} - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f'} = \frac{\frac{3}{2}}{\frac{4}{3}} - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Calculating \( \frac{\frac{3}{2}}{\frac{4}{3}} \):
\[ \frac{3}{2} \cdot \frac{3}{4} = \frac{9}{8} \]
Thus:
\[ \frac{1}{f'} = \left( \frac{9}{8} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{1}{f'} = \frac{1}{8} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 4: Relating \( f' \) to \( f \)
To relate \( f' \) to \( f \), we can divide the two equations:
\[ \frac{1/f}{1/f'} = \frac{\frac{1}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)}{\frac{1}{8} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \]
This simplifies to:
\[ \frac{f'}{f} = \frac{8}{2} = 4 \]
Thus:
\[ f' = 4f \]

### Step 5: Nature of the New Image
Initially, the image was inverted, real, and magnified, which indicates that the object was placed between \( f \) and \( 2f \). Now, since \( f' = 4f \), the new focal length is greater than \( 2f \). Therefore, the object distance \( u \) is less than \( f' \).

When the object is placed between the focal point and the lens, the nature of the image changes to:
- Virtual
- Erect
- Magnified

### Final Answer
1. The new focal length \( f' \) is \( 4f \).
2. The nature of the new image is virtual, erect, and magnified.