A convex lens of focal length `20` cm made of glass of refractive index `1.5` is immersed in water having refractive index `1.33`. The change in the focal length of lens is

Options

A. `62.2` cm
B. `5.82` cm
C. `58.2` cm
D. `6.22` cm

Explanation

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

### Step 1: Understand the initial conditions
The convex lens has a focal length (f) of 20 cm when it is in air. The refractive index of the glass (μg) is 1.5, and the refractive index of water (μw) is 1.33.

### Step 2: Use the lens maker's formula
The lens maker's formula for a convex lens in air is given by:

\[ \frac{1}{f} = \left( \mu_g - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Where:
- \( f \) = focal length of the lens in air
- \( \mu_g \) = refractive index of the glass
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 3: Calculate the focal length in water
When the lens is immersed in water, the new focal length (fw) can be calculated using the modified lens maker's formula:

\[ \frac{1}{f_w} = \left( \frac{\mu_g}{\mu_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 4: Set up the equations
From the first equation, we can express \(\frac{1}{R_1} - \frac{1}{R_2}\):

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

Substituting \( f = 20 \) cm:

\[ \frac{1}{20} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

This implies:

\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{10} \]

### Step 5: Substitute into the new focal length equation
Now, substituting this into the equation for \( f_w \):

\[ \frac{1}{f_w} = \left( \frac{1.5}{1.33} - 1 \right) \left( \frac{1}{10} \right) \]

Calculating \( \frac{1.5}{1.33} \):

\[ \frac{1.5}{1.33} \approx 1.126 \]

Thus:

\[ \frac{1}{f_w} = (1.126 - 1) \left( \frac{1}{10} \right) = 0.126 \left( \frac{1}{10} \right) = 0.0126 \]

### Step 6: Calculate \( f_w \)
Now, we can find \( f_w \):

\[ f_w = \frac{1}{0.0126} \approx 79.37 \text{ cm} \]

### Step 7: Calculate the change in focal length
The change in focal length (\( \Delta f \)) is given by:

\[ \Delta f = f_w - f = 79.37 - 20 = 59.37 \text{ cm} \]

### Conclusion
Thus, the change in the focal length of the lens when immersed in water is approximately **59.37 cm**.