A double convex lens has focal length 25 cm. The radius of curvature of one of the surfaces is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Focal length (f) = 25 cm
- Refractive index of the lens (μ) = 1.5
- The radius of curvature of one surface (R1) is double that of the other surface (R2), i.e., R1 = 2R2.

2. **Write the Lens Maker's Formula:**
The lens maker's formula is given by:
\[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

3. **Substitute Known Values:**
Substitute the known values into the formula:
\[ \frac{1}{25} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Simplifying the left side:
\[ \frac{1}{25} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

4. **Express R1 in terms of R2:**
Since R1 = 2R2, we can substitute R1 into the equation:
\[ \frac{1}{25} = 0.5 \left( \frac{1}{2R_2} - \frac{1}{R_2} \right) \]

5. **Simplify the Equation:**
The term inside the parentheses can be simplified:
\[ \frac{1}{2R_2} - \frac{1}{R_2} = \frac{1 - 2}{2R_2} = \frac{-1}{2R_2} \]
Substituting this back into the equation:
\[ \frac{1}{25} = 0.5 \left( \frac{-1}{2R_2} \right) \]
This simplifies to:
\[ \frac{1}{25} = \frac{-0.5}{2R_2} = \frac{-0.25}{R_2} \]

6. **Solve for R2:**
Rearranging gives:
\[ R_2 = -0.25 \times 25 = -6.25 \text{ cm} \]
Since R2 is negative (as per sign convention), we take the absolute value for the radius:
\[ R_2 = 6.25 \text{ cm} \]

7. **Find R1:**
Now, using R1 = 2R2:
\[ R_1 = 2 \times 6.25 = 12.5 \text{ cm} \]

8. **Final Answer:**
The radii of curvature are:
- R1 = 12.5 cm
- R2 = 6.25 cm