An equiconvex lens of refractive index 1.6 has power 4D in air. Its power in water is:

Options

A. 1.5 D
B. 2.0 D
C. 1.3 D
D. 3.2 D

Answer

To solve the problem, we will use the lensmaker's formula and the concept of power of a lens in different media.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- The refractive index of the lens (μ_l) = 1.6
- The power of the lens in air (P_air) = 4 D (diopters)
- The refractive index of air (μ_air) = 1.0
- The refractive index of water (μ_water) = 1.33

2. **Using the Lensmaker's Formula:**
The power of a lens is given by the formula:
\[ P = \left( \mu_l - \mu_m \right) \cdot \frac{1}{f} \]
where:
- \( P \) is the power of the lens,
- \( \mu_l \) is the refractive index of the lens,
- \( \mu_m \) is the refractive index of the medium,
- \( f \) is the focal length of the lens.

3. **Finding the Focal Length in Air:**
From the power in air:
\[ P_{air} = \left( \mu_l - \mu_{air} \right) \cdot \frac{1}{f} \]
Substituting the values:
\[ 4 = (1.6 - 1) \cdot \frac{1}{f} \]
\[ 4 = 0.6 \cdot \frac{1}{f} \]
Rearranging gives:
\[ f = \frac{0.6}{4} = 0.15 \text{ m} = 15 \text{ cm} \]

4. **Finding Power in Water:**
Now we need to find the power of the lens when it is placed in water:
\[ P_{water} = \left( \mu_l - \mu_{water} \right) \cdot \frac{1}{f} \]
Substituting the values:
\[ P_{water} = (1.6 - 1.33) \cdot \frac{1}{0.15} \]
\[ P_{water} = 0.27 \cdot \frac{1}{0.15} \]
\[ P_{water} = 0.27 \cdot 6.67 \approx 1.8 \text{ D} \]

5. **Final Calculation:**
Thus, the power of the lens in water is approximately:
\[ P_{water} \approx 1.8 \text{ D} \]

### Conclusion:
The power of the lens in water is approximately **1.8 D**.

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