Radii of curvature of a converging lens are in the ratio 1:2. Its focal length is 6cm and refractive index is 1:5. Then, its radii of curvature are…

Options

A. 9cm and 18cm
B. 6cm and 12cm
C. 3cm and 6cm
D. 4.5cm and 9cm

Explanation

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 the Given Values:**
- Focal length (f) = 6 cm
- Refractive index of the lens (μ2) = 1.5
- Refractive index of the surrounding medium (μ1) = 1 (air)
- Ratio of the radii of curvature (R1:R2) = 1:2

2. **Express the Radii of Curvature:**
- Let the first radius of curvature be R1 = r.
- Then, the second radius of curvature will be R2 = 2r.

3. **Apply the Lens Maker's Formula:**
The lens maker's formula is given by:
\[ \frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Substituting the known values:
\[ \frac{1}{6} = \left(\frac{1.5}{1} - 1\right) \left(\frac{1}{r} - \frac{1}{2r}\right) \]

4. **Simplify the Equation:**
- Calculate \(\frac{1.5}{1} - 1 = 0.5\).
- The term \(\frac{1}{r} - \frac{1}{2r} = \frac{2 - 1}{2r} = \frac{1}{2r}\).
- Thus, the equation becomes:
\[ \frac{1}{6} = 0.5 \cdot \frac{1}{2r} \]

5. **Solve for r:**
- Simplifying further:
\[ \frac{1}{6} = \frac{0.5}{2r} = \frac{0.25}{r} \]
- Cross-multiplying gives:
\[ 1 = \frac{0.25 \cdot 6}{r} \implies r = 0.25 \cdot 6 = 1.5 \text{ cm} \]

6. **Calculate the Second Radius of Curvature:**
- Since R2 = 2r:
\[ R2 = 2 \cdot 1.5 = 3 \text{ cm} \]

7. **Final Result:**
- The radii of curvature are:
- R1 = 1.5 cm
- R2 = 3 cm

### Answer:
The radii of curvature of the lens are 1.5 cm and 3 cm.