The radius of curvature for a convex lens is 40 cm , for each surface. Its refractive index is 1.5. The focal length will be

Options

A. 40 cm
B. 20 cm
C. 80 cm
D. 30 cm

Answer

To find the focal length of a convex lens given its radius of curvature and refractive index, we can use the lens maker's formula:

\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Where:
- \( f \) is the focal length of the lens,
- \( \mu \) is the refractive index of the lens material,
- \( R_1 \) is the radius of curvature of the first surface,
- \( R_2 \) is the radius of curvature of the second surface.

### Step-by-step Solution:

1. **Identify the given values:**
- Refractive index, \( \mu = 1.5 \)
- Radius of curvature for both surfaces, \( R_1 = +40 \, \text{cm} \) (convex surface is positive)
- \( R_2 = -40 \, \text{cm} \) (the second surface is concave relative to the incoming light)

2. **Substitute the values into the lens maker's formula:**
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{40} - \frac{1}{-40} \right) \]

3. **Calculate \( \mu - 1 \):**
\[ \mu - 1 = 1.5 - 1 = 0.5 \]

4. **Calculate \( \frac{1}{R_1} - \frac{1}{R_2} \):**
\[ \frac{1}{R_1} = \frac{1}{40} \quad \text{and} \quad \frac{1}{R_2} = \frac{1}{-40} = -\frac{1}{40} \]
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{40} - \left(-\frac{1}{40}\right) = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20} \]

5. **Combine the results:**
\[ \frac{1}{f} = 0.5 \cdot \frac{1}{20} = \frac{0.5}{20} = \frac{1}{40} \]

6. **Find the focal length \( f \):**
\[ f = 40 \, \text{cm} \]

### Final Answer:
The focal length of the convex lens is \( 40 \, \text{cm} \).

---