A convex lens of focal legnth `0.2 m` and made of glass `(mu = 1.50)` is immersed in water `(mu = 1.33)`. Find the change in the focal length of the lens.

Options

A. `5.8 m`
B. `0.58 cm`
C. `0.58m`
D. `5.8 cm`

Explanation

To find the change in the focal length of a convex lens when it is immersed in water, we can use the lens maker's formula. Here are the steps to solve the problem:

### Step 1: Identify the given values
- Focal length of the lens in air, \( f_A = 0.2 \, m \)
- Refractive index of the glass lens, \( \mu_{glass} = 1.50 \)
- Refractive index of water, \( \mu_{water} = 1.33 \)

### Step 2: Use the lens maker's formula in air
The lens maker's formula in air is given by:
\[ \frac{1}{f_A} = \left( \mu_{glass} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 3: Calculate \( \frac{1}{R_1} - \frac{1}{R_2} \)
From the formula, we can rearrange it to find \( \frac{1}{R_1} - \frac{1}{R_2} \):
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{f_A} \cdot \frac{1}{\mu_{glass} - 1} \]
Substituting the known values:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{0.2} \cdot \frac{1}{1.50 - 1} = \frac{1}{0.2} \cdot \frac{1}{0.5} = 10 \]

### Step 4: Use the lens maker's formula in water
Now, we apply the lens maker's formula when the lens is immersed in water:
\[ \frac{1}{f_W} = \left( \frac{\mu_{glass}}{\mu_{water}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f_W} = \left( \frac{1.50}{1.33} - 1 \right) \cdot 10 \]

### Step 5: Calculate \( \frac{1.50}{1.33} \)
Calculating \( \frac{1.50}{1.33} \):
\[ \frac{1.50}{1.33} \approx 1.128 \]
Thus,
\[ \frac{1}{f_W} = (1.128 - 1) \cdot 10 = 0.128 \cdot 10 = 1.28 \]

### Step 6: Calculate the focal length in water
Now, we find \( f_W \):
\[ f_W = \frac{1}{1.28} \approx 0.78125 \, m \]

### Step 7: Find the change in focal length
The change in focal length \( \Delta f \) is given by:
\[ \Delta f = f_W - f_A = 0.78125 - 0.2 = 0.58125 \, m \]
Rounding this to two decimal places gives:
\[ \Delta f \approx 0.58 \, m \]

### Final Answer
The change in the focal length of the lens when immersed in water is approximately **0.58 m**.

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