A convex glass lens `(mu_(g) = 1.5)` has a focal length of 8 cm when placed in air. What is the focal length of the lens when it is immersed in water ?
`(mu_(omega) = (4)/(3))`

Options

A. 32 cm
B. 6 cm
C. 16 cm
D. 30 cm

Answer

To find the focal length of a convex lens when it is immersed in water, we can use the lens maker's formula. The formula for the focal length \( f \) of a lens in a medium is given by:

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

Where:
- \( \mu_l \) is the refractive index of the lens material.
- \( \mu_m \) is the refractive index of the medium.
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 1: Calculate the focal length in air
Given:
- \( \mu_g = 1.5 \) (refractive index of the glass lens)
- \( \mu_a = 1 \) (refractive index of air)
- \( f_1 = 8 \, \text{cm} \)

Using the lens maker's formula for air:
\[ \frac{1}{f_1} = \left( \frac{\mu_g}{\mu_a} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{8} = \left( \frac{1.5}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Calculating:
\[ \frac{1}{8} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ \frac{1}{8} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Thus,
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{4} \]

### Step 2: Calculate the focal length in water
Now, we need to find the focal length when the lens is immersed in water.
Given:
- \( \mu_w = \frac{4}{3} \) (refractive index of water)

Using the lens maker's formula for water:
\[ \frac{1}{f_2} = \left( \frac{\mu_g}{\mu_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f_2} = \left( \frac{1.5}{\frac{4}{3}} - 1 \right) \left( \frac{1}{4} \right) \]
Calculating:
\[ \frac{1}{f_2} = \left( \frac{1.5 \times 3}{4} - 1 \right) \left( \frac{1}{4} \right) \]
\[ \frac{1}{f_2} = \left( \frac{4.5}{4} - 1 \right) \left( \frac{1}{4} \right) \]
\[ \frac{1}{f_2} = \left( 1.125 - 1 \right) \left( \frac{1}{4} \right) \]
\[ \frac{1}{f_2} = 0.125 \left( \frac{1}{4} \right) \]
\[ \frac{1}{f_2} = \frac{0.125}{4} = \frac{0.125}{4} = \frac{0.03125} \]
Thus,
\[ f_2 = \frac{1}{0.03125} = 32 \, \text{cm} \]

### Final Answer
The focal length of the lens when it is immersed in water is \( 32 \, \text{cm} \).