A convex lens of focal length `20 cm` and made of glass `(mu = 1.5)` is immersed in water of `mu = 1.33` Calculate change in focal length of the lens.

Answer

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

### Step 1: Understand the Lens Maker's Formula
The lens maker's formula relates the focal length of a lens to its radii of curvature and the refractive indices of the lens material and the surrounding medium. The formula is given by:

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

Where:
- \( f \) = focal length of the lens
- \( \mu \) = refractive index of the lens material
- \( R_1 \) and \( R_2 \) = radii of curvature of the lens surfaces

### Step 2: Calculate Focal Length in Air
Given:
- Focal length in air, \( f_a = 20 \, \text{cm} \)
- Refractive index of glass, \( \mu_g = 1.5 \)

Using the lens maker's formula in air, we can express it as:

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

### Step 3: Calculate Focal Length in Water
When the lens is immersed in water, we need to use the refractive index of water, \( \mu_w = 1.33 \). The new focal length \( f_w \) 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: Relate the Two Focal Lengths
From the previous step, we can express \( \frac{1}{R_1} - \frac{1}{R_2} \) in terms of \( f_a \):

\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{f_a \cdot (\mu_g - 1)} \]

Substituting this into the equation for \( f_w \):

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

### Step 5: Substitute Values
Now substituting the known values:

- \( \mu_g = 1.5 \)
- \( \mu_w = 1.33 \)
- \( f_a = 20 \, \text{cm} \)

Calculating \( \frac{\mu_g}{\mu_w} - 1 \):

\[ \frac{1.5}{1.33} - 1 \approx 0.128 \]

Now substituting this back into the equation for \( f_w \):

\[ \frac{1}{f_w} = 0.128 \cdot \frac{1}{20 \cdot (1.5 - 1)} \]

Calculating \( 1.5 - 1 = 0.5 \):

\[ \frac{1}{f_w} = 0.128 \cdot \frac{1}{20 \cdot 0.5} = 0.128 \cdot \frac{1}{10} = 0.0128 \]

### Step 6: Calculate \( f_w \)
Taking the reciprocal to find \( f_w \):

\[ f_w = \frac{1}{0.0128} \approx 78.125 \, \text{cm} \]

### Step 7: Calculate Change in Focal Length
Finally, the change in focal length when the lens is immersed in water is:

\[ \Delta f = f_w - f_a = 78.125 \, \text{cm} - 20 \, \text{cm} = 58.125 \, \text{cm} \]

### Final Answer
The change in focal length of the lens when immersed in water is approximately \( 58.125 \, \text{cm} \).

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