The focal length of a biconvex lens of refractive index 1.5 is 0.06m. Radii of curvature are in the ratio 1:2. Then radii of curvature of two lens surfaces are

Options

A. `0.045 m, 0.09m`
B. `0.09m, 0.18m`
C. `0.04m, 0.08m`
D. `0.06m, 0.12m`

Answer

To solve the problem, we will use the lens maker's formula, which relates the focal length of a lens to its radii of curvature and refractive index.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Focal length \( f = 0.06 \, \text{m} \)
- Refractive index \( \mu = 1.5 \)
- Ratio of radii of curvature \( R_1 : R_2 = 1 : 2 \)

2. **Assign Variables**:
- Let \( R_1 = R \)
- Let \( R_2 = 2R \) (Since \( R_2 \) is for the second surface of a biconvex lens, it will be negative in the lens maker's formula.)

3. **Lens Maker's Formula**:
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values we have:
\[ \frac{1}{0.06} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-2R} \right) \]

4. **Simplify the Equation**:
- Calculate \( \mu - 1 \):
\[ \mu - 1 = 0.5 \]
- Substitute this into the equation:
\[ \frac{1}{0.06} = 0.5 \left( \frac{1}{R} + \frac{1}{2R} \right) \]
- Combine the terms inside the parentheses:
\[ \frac{1}{R} + \frac{1}{2R} = \frac{2 + 1}{2R} = \frac{3}{2R} \]
- Now the equation becomes:
\[ \frac{1}{0.06} = 0.5 \cdot \frac{3}{2R} \]

5. **Solve for \( R \)**:
- Rearranging gives:
\[ \frac{1}{0.06} = \frac{1.5}{2R} \]
- Cross-multiplying:
\[ 2R = 0.06 \cdot 1.5 \]
- Calculate \( 0.06 \cdot 1.5 \):
\[ 2R = 0.09 \]
- Therefore:
\[ R = \frac{0.09}{2} = 0.045 \, \text{m} \]

6. **Calculate the Radii of Curvature**:
- \( R_1 = R = 0.045 \, \text{m} \)
- \( R_2 = 2R = 2 \cdot 0.045 = 0.09 \, \text{m} \)

### Final Answer:
- The radii of curvature of the two lens surfaces are:
- \( R_1 = 0.045 \, \text{m} \)
- \( R_2 = 0.09 \, \text{m} \)