A lens is made of flint glass (refractive index `=1.5`). When the lens is immersed in a liquid of refractive index `1.25` , the focal length:

Options

A. increases by a factor of `1.25`
B. increases by a factor of `2.25`
C. increases by a factor of `1.2`
D. decreases by a factor of `1.2`

Explanation

To solve the problem of finding the new focal length of a lens made of flint glass when immersed in a liquid, we will use the Lensmaker's formula. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Lensmaker's Formula
The Lensmaker's formula is given by:
\[ \frac{1}{f} = (\mu_g - \mu_a) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \( f \) = focal length of the lens
- \( \mu_g \) = refractive index of the glass
- \( \mu_a \) = refractive index of the medium (air or liquid)
- \( R_1 \) and \( R_2 \) = radii of curvature of the lens surfaces

### Step 2: Calculate the Focal Length in Air
Given:
- Refractive index of flint glass, \( \mu_g = 1.5 \)
- Refractive index of air, \( \mu_a = 1.0 \)

Using the formula:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ \frac{1}{f} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 3: Calculate the Focal Length in Liquid
Now, when the lens is immersed in a liquid with a refractive index \( \mu_l = 1.25 \), we need to recalculate using the new medium:
\[ \frac{1}{f'} = (\mu_g - \mu_l) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f'} = (1.5 - 1.25) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ \frac{1}{f'} = 0.25 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 4: Relate the Two Focal Lengths
From Step 2, we have:
\[ \frac{1}{f} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Let \( k = \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \). Then:
\[ \frac{1}{f} = 0.5k \quad \text{and} \quad \frac{1}{f'} = 0.25k \]

### Step 5: Express \( f' \) in terms of \( f \)
From \( \frac{1}{f} = 0.5k \), we can express \( k \) as:
\[ k = \frac{2}{f} \]
Substituting this into the equation for \( f' \):
\[ \frac{1}{f'} = 0.25 \left( \frac{2}{f} \right) = \frac{0.5}{f} \]
Thus:
\[ f' = \frac{2f}{1} = 2f \]

### Conclusion
The focal length of the lens when immersed in the liquid is increased by a factor of 2. Therefore, the correct answer is that the focal length increases by a factor of 2.