Focal length of a double convex lens is 20 cm in air. If this lens is dipped in water of refractive index `4//3` then what will be the new focal length ? (Take `mu_(glass) = 1.5`)

Options

A. 60 cm
B. 80 cm
C. 40 cm
D. 20 cm

Explanation

To find the new focal length of a double convex lens when it is dipped in water, we can use the lensmaker's formula. The formula relates the focal length of a lens to its refractive index and the refractive index of the medium in which it is placed.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Focal length of the lens in air, \( f_1 = 20 \, \text{cm} \)
- Refractive index of the lens (glass), \( \mu_{\text{glass}} = 1.5 \)
- Refractive index of air, \( \mu_{\text{air}} = 1 \)
- Refractive index of water, \( \mu_{\text{water}} = \frac{4}{3} \)

2. **Use Lensmaker's Formula in Air:**
The lensmaker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
In air, we can write:
\[ \frac{1}{f_1} = (\mu_{\text{glass}} - \mu_{\text{air}}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the known values:
\[ \frac{1}{20} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Simplifying this gives:
\[ \frac{1}{20} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Therefore:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{10} \quad \text{(Equation 1)} \]

3. **Use Lensmaker's Formula in Water:**
Now, when the lens is dipped in water, we use the lensmaker's formula again:
\[ \frac{1}{f_2} = (\mu_{\text{glass}} - \mu_{\text{water}}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the known values:
\[ \frac{1}{f_2} = (1.5 - \frac{4}{3}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Simplifying \( 1.5 - \frac{4}{3} \):
\[ 1.5 = \frac{3}{2} \quad \Rightarrow \quad \frac{3}{2} - \frac{4}{3} = \frac{9}{6} - \frac{8}{6} = \frac{1}{6} \]
So we have:
\[ \frac{1}{f_2} = \frac{1}{6} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
From Equation 1, we know \( \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{10} \):
\[ \frac{1}{f_2} = \frac{1}{6} \cdot \frac{1}{10} = \frac{1}{60} \]

4. **Calculate the New Focal Length:**
Now, taking the reciprocal to find \( f_2 \):
\[ f_2 = 60 \, \text{cm} \]

### Final Answer:
The new focal length of the lens when dipped in water is **60 cm**.