The focal length of lens of refractive index `1.5` in air is `30 cm` When it is immersed in water of refractive index `(4)/(3)` ,then its focal length will be

Options

A. 0.15m
B. 0.30m
C. 0.45m
D. 1.20m

Explanation

To find the new focal length of a lens when it is immersed in water, we can use the lens maker's formula. Here’s a step-by-step solution:

### Step 1: Understand the given data
- The refractive index of the lens (μ2) = 1.5
- The refractive index of air (μ1) = 1.0
- The focal length of the lens in air (f) = 30 cm
- The refractive index of water (μ1') = 4/3

### Step 2: Use the lens maker's formula
The lens maker's formula is given by:
\[ \frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

### Step 3: Calculate the radii of curvature using the initial conditions
For the lens in air:
\[ \frac{1}{30} = \left(\frac{1.5}{1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
\[ \frac{1}{30} = (0.5) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Let \( \frac{1}{R_1} - \frac{1}{R_2} = x \), then:
\[ \frac{1}{30} = 0.5x \implies x = \frac{1}{15} \]
Thus, we have:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{15} \quad \text{(Equation 1)} \]

### Step 4: Calculate the new focal length in water
Now, when the lens is immersed in water, we use the new refractive index:
\[ \frac{1}{f'} = \left(\frac{1.5}{\frac{4}{3}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Calculating \( \frac{1.5}{\frac{4}{3}} \):
\[ \frac{1.5 \times 3}{4} = \frac{4.5}{4} = 1.125 \]
So,
\[ \frac{1}{f'} = (1.125 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
\[ \frac{1}{f'} = 0.125 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Substituting \( \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{15} \) from Equation 1:
\[ \frac{1}{f'} = 0.125 \times \frac{1}{15} \]
\[ \frac{1}{f'} = \frac{0.125}{15} = \frac{1}{120} \]
Thus, the new focal length \( f' \) is:
\[ f' = 120 \text{ cm} \]

### Conclusion
The focal length of the lens when immersed in water is **120 cm**.

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