The focal length of a double convex lens is equal to radius of curvature of either surface. The refractive index of its material is

Options

A. `3//2`
B. `1`
C. `4//3`
D. none of these

Answer

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

### Step-by-step Solution:

1. **Understand the Given Information**:
- We have a double convex lens.
- The focal length (f) is equal to the radius of curvature (R) of either surface.

2. **Recall the Lensmaker's Formula**:
The lensmaker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \( f \) = focal length of the lens
- \( \mu \) = refractive index of the lens material
- \( R_1 \) and \( R_2 \) = radii of curvature of the two surfaces of the lens.

3. **Assign Values to Radii of Curvature**:
For a double convex lens:
- Let \( R_1 = R \) (positive for the first surface)
- Let \( R_2 = -R \) (negative for the second surface)

4. **Substitute Values into the Lensmaker's Formula**:
Substituting \( R_1 \) and \( R_2 \) into the formula gives:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) \]
This simplifies to:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = (\mu - 1) \left( \frac{2}{R} \right) \]

5. **Set Focal Length Equal to Radius of Curvature**:
Since \( f = R \), we can substitute this into the equation:
\[ \frac{1}{R} = (\mu - 1) \left( \frac{2}{R} \right) \]

6. **Cancel \( R \) from Both Sides**:
Since \( R \) is not zero, we can cancel \( R \):
\[ 1 = 2(\mu - 1) \]

7. **Solve for the Refractive Index \( \mu \)**:
Rearranging the equation gives:
\[ 1 = 2\mu - 2 \]
\[ 2\mu = 3 \]
\[ \mu = \frac{3}{2} = 1.5 \]

### Final Answer:
The refractive index of the material of the lens is \( \mu = 1.5 \).