Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index 1.25 will be

Options

A. `10 cm`
B. `2.5 cm`
C. `5 cm`
D. `7.5 cm`

Answer

To solve the problem, we need to find the new focal length of a convex lens when it is immersed in a liquid with a different refractive index. Here's the step-by-step solution:

### Step 1: Understand the relationship between focal length and refractive index
The focal length \( f \) of a lens is related to its refractive index \( \mu \) and the radii of curvature \( r_1 \) and \( r_2 \) of the lens surfaces by the lens maker's formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]

### Step 2: Calculate the initial focal length
Given:
- Focal length \( f_1 = 2 \, \text{cm} \)
- Refractive index of the lens \( \mu = 1.5 \)

Using the formula:
\[ \frac{1}{f_1} = (\mu - 1) A \]
Substituting the known values:
\[ \frac{1}{2} = (1.5 - 1) A \]
\[ \frac{1}{2} = 0.5 A \]
From this, we can find \( A \):
\[ A = \frac{1}{2 \times 0.5} = 1 \, \text{cm} \]

### Step 3: Calculate the new focal length when immersed in the liquid
When the lens is immersed in a liquid of refractive index \( \mu' = 1.25 \), we use the modified lens maker's formula:
\[ \frac{1}{f_2} = \left( \frac{\mu}{\mu'} - 1 \right) A \]
Substituting the values:
\[ \frac{1}{f_2} = \left( \frac{1.5}{1.25} - 1 \right) A \]
Calculating \( \frac{1.5}{1.25} \):
\[ \frac{1.5}{1.25} = 1.2 \]
Thus,
\[ \frac{1}{f_2} = (1.2 - 1) A = 0.2 \times 1 \]
\[ \frac{1}{f_2} = 0.2 \]
Now, finding \( f_2 \):
\[ f_2 = \frac{1}{0.2} = 5 \, \text{cm} \]

### Conclusion
The focal length of the lens when immersed in a liquid of refractive index 1.25 is \( 5 \, \text{cm} \).