A convex lens made of glass `(mu_(g)=3//2)` has focal length f in air. The image of an object placed in front of it is inverted, real and magnified. Now the whole arrangement 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 involving a convex lens made of glass placed in air and then in a liquid (water). We will derive the necessary relationships and determine the distance between the object and the lens.

### Step 1: Understanding the Focal Length of the Lens in Air
The focal length \( f \) of a convex lens in air is given by the lens maker's formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where:
- \( \mu \) is the refractive index of the lens material (glass in this case, \( \mu_g = \frac{3}{2} \)).
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 2: Calculate the Focal Length in Water
When the lens is placed in a liquid (water), the effective focal length \( f' \) can be calculated using the modified lens maker's formula:
\[ \frac{1}{f'} = \left(\frac{\mu}{\mu_{water}} - 1\right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where \( \mu_{water} \approx \frac{4}{3} \). Substituting the values:
\[ \frac{1}{f'} = \left(\frac{\frac{3}{2}}{\frac{4}{3}} - 1\right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 3: Simplifying the Expression
Calculating the term:
\[ \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8} \]
Thus,
\[ \frac{9}{8} - 1 = \frac{9}{8} - \frac{8}{8} = \frac{1}{8} \]
Now we can write:
\[ \frac{1}{f'} = \frac{1}{8} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 4: Relating Focal Lengths
From the previous calculations, we can relate the focal lengths:
\[ f' = 8f \]
This means that the focal length of the lens in water is 8 times the focal length in air.

### Step 5: Object Position and Image Characteristics
In air, the image is inverted, real, and magnified, indicating that the object is placed between \( f \) and \( 2f \). When the lens is placed in water, the focal length becomes \( 8f \).

### Step 6: New Image Characteristics
With the new focal length \( f' = 8f \), the object will now be positioned between the pole and the new focus (which is now much further away). This means that the new image will be virtual and magnified.

### Conclusion
Thus, the distance between the object and the lens can be concluded based on the new focal length and the characteristics of the image formed.