Power of biconvex lens is 10 diopters and radius of curvature of each surface is 10cm.Then refractive index of material of lens is

Options

A. 43924
B. 44082
C. 43954
D. 43892

Answer

To find the refractive index of the material of a biconvex lens given its power and radius of curvature, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Power of the lens (P) = 10 diopters
- Radius of curvature of each surface (R1 and R2) = 10 cm

2. **Convert Power to Focal Length**:
- The relationship between power (P) and focal length (f) is given by:
\[ P = \frac{1}{f} \]
- Therefore, the focal length (f) can be calculated as:
\[ f = \frac{1}{P} = \frac{1}{10} \text{ m} = 0.1 \text{ m} = 10 \text{ cm} \]

3. **Apply the Lens Maker's Formula**:
- The Lens Maker's Formula for a biconvex lens is:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- For a biconvex lens, R1 is positive and R2 is negative. Thus:
\[ R_1 = +10 \text{ cm}, \quad R_2 = -10 \text{ cm} \]

4. **Substituting Values into the Formula**:
- Substitute the values of f, R1, and R2 into the Lens Maker's Formula:
\[ \frac{1}{10} = (\mu - 1) \left( \frac{1}{10} - \frac{1}{-10} \right) \]
- Simplifying the right side:
\[ \frac{1}{10} = (\mu - 1) \left( \frac{1}{10} + \frac{1}{10} \right) = (\mu - 1) \left( \frac{2}{10} \right) \]
- This simplifies to:
\[ \frac{1}{10} = \frac{(\mu - 1)}{5} \]

5. **Solving for the Refractive Index (μ)**:
- Cross-multiplying gives:
\[ 1 = (\mu - 1) \cdot \frac{1}{2} \]
- Therefore:
\[ \mu - 1 = 2 \implies \mu = 3 \]

6. **Final Result**:
- The refractive index of the material of the lens is:
\[ \mu = 1.5 \]