A convex lens of glass `(n=(3)/(2))` has a focal length 20 cm. The lens is immersed in water of refractive index `(4)/(3)`. What is the change in the power of convex lens ?

Options

A. 1.75 D
B. 3.75 D
C. 4.5 D
D. 5.75 D

Explanation

To solve the problem of finding the change in the power of a convex lens when it is immersed in water, we can follow these steps:

### Step 1: Calculate the initial power of the lens in air.
The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{f} \]
where \( f \) is the focal length in meters.

Given that the focal length \( f = 20 \) cm, we convert it to meters:
\[ f = 20 \, \text{cm} = 0.20 \, \text{m} \]
Now, we can calculate the initial power of the lens:
\[ P_{\text{air}} = \frac{1}{0.20} = 5 \, \text{diopters} \]

### Step 2: Calculate the focal length of the lens when immersed in water.
When the lens is immersed in a medium with a different refractive index, the effective focal length \( f' \) can be calculated using the lens maker's formula:
\[ \frac{1}{f'} = (n_l - n_m) \cdot \frac{1}{R} \]
where:
- \( n_l \) is the refractive index of the lens (glass),
- \( n_m \) is the refractive index of the medium (water),
- \( R \) is the radius of curvature of the lens.

However, we can use the simplified formula for power in different media:
\[ P' = P \cdot \frac{n_l}{n_m} \]
Given:
- \( n_l = \frac{3}{2} \) (for glass),
- \( n_m = \frac{4}{3} \) (for water).

Substituting the values:
\[ P' = 5 \cdot \frac{\frac{3}{2}}{\frac{4}{3}} = 5 \cdot \frac{3}{2} \cdot \frac{3}{4} = 5 \cdot \frac{9}{8} = \frac{45}{8} = 5.625 \, \text{diopters} \]

### Step 3: Calculate the change in power.
The change in power \( \Delta P \) is given by:
\[ \Delta P = P' - P_{\text{air}} \]
Substituting the values we found:
\[ \Delta P = 5.625 - 5 = 0.625 \, \text{diopters} \]

### Step 4: Conclusion.
The change in the power of the convex lens when immersed in water is:
\[ \Delta P = 0.625 \, \text{diopters} \]

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