The radii of curvature of the two surfaces of a lens are 20 cm and 30 cm and the refractive index of the material of the lens is 1.5. If the lens is concave -convex, then the focal length of the lens is

Options

A. 24 cm
B. 10 cm
C. 15 cm
D. 120 cm

Explanation

To find the focal length of a concave-convex lens given the radii of curvature of its two surfaces and the 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:**
- \( R_1 = 20 \, \text{cm} \) (for the first surface, which is convex)
- \( R_2 = -30 \, \text{cm} \) (for the second surface, which is concave; note that the radius for a concave surface is taken as negative)
- \( \mu = 1.5 \)

2. **Substitute the values into the lens maker's formula:**
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-30} \right) \]

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

4. **Calculate \( \frac{1}{R_1} - \frac{1}{R_2} \):**
\[ \frac{1}{R_1} = \frac{1}{20} = 0.05 \]
\[ \frac{1}{R_2} = \frac{1}{-30} = -\frac{1}{30} \approx -0.0333 \]
\[ \frac{1}{R_1} - \frac{1}{R_2} = 0.05 - (-0.0333) = 0.05 + 0.0333 = 0.0833 \]

5. **Substitute back into the formula:**
\[ \frac{1}{f} = 0.5 \times 0.0833 = 0.04165 \]

6. **Calculate the focal length \( f \):**
\[ f = \frac{1}{0.04165} \approx 24 \, \text{cm} \]

### Final Answer:
The focal length of the lens is approximately \( 24 \, \text{cm} \).