The focal length of a glass convex lens of refractive index 1.5 is 20 cm. What is the focal power of the lens when immersed in a liquid of refractive index of 1.25 ?

Options

A. 4 D
B. 3 D
C. 2 D
D. 5 D

Explanation

To solve the problem, we need to determine the focal power of a glass convex lens when it is immersed in a liquid with a different refractive index. Here’s a step-by-step solution:

### Step 1: Understanding the Given Data
- The refractive index of the glass lens (μg) = 1.5
- The focal length of the lens in air (f) = 20 cm
- The refractive index of the liquid (μl) = 1.25

### Step 2: Calculate the Focal Length of the Lens in Air
The focal length (f) of a lens is given as:
\[ f = 20 \, \text{cm} \]

### Step 3: Using the Lens Maker's Formula
The lens maker's formula in air is given by:
\[ \frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
However, since we are not given the radii of curvature (R1 and R2), we will focus on the change in the focal length when the lens is immersed in a liquid.

### Step 4: Focal Length of the Lens in Liquid
When the lens is immersed in a liquid, the new focal length (f_cl) can be calculated using:
\[ \frac{1}{f_{cl}} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Since \(\frac{1}{R_1} - \frac{1}{R_2}\) remains the same, we can express the relationship as:
\[ \frac{1}{f_{cl}} = \left( \frac{\mu_g - 1}{\mu_l - 1} \right) \cdot \frac{1}{f} \]

### Step 5: Substitute the Values
Now substituting the values:
- \(\mu_g = 1.5\)
- \(\mu_l = 1.25\)
- \(f = 20 \, \text{cm} = 0.2 \, \text{m}\)

Calculating:
\[ \frac{1}{f_{cl}} = \left( \frac{1.5 - 1}{1.25 - 1} \right) \cdot \frac{1}{0.2} \]
\[ = \left( \frac{0.5}{0.25} \right) \cdot 5 = 2 \cdot 5 = 10 \]
Thus,
\[ f_{cl} = \frac{1}{10} = 0.1 \, \text{m} = 10 \, \text{cm} \]

### Step 6: Calculate the Focal Power
The power (P) of the lens is given by:
\[ P = \frac{1}{f_{cl}} \quad \text{(in meters)} \]
Substituting the focal length:
\[ P = \frac{1}{0.1} = 10 \, \text{diopters} \]

### Conclusion
The focal power of the lens when immersed in the liquid is **10 diopters**.