A plano-convex lens is made of glass of refractive index 1.6. the radius of curvature of its curved surface is 30 cm. What is the focal length of the lens ?

Options

A. 50 cm
B. 75 cm
C. 100 cm
D. 150 cm

Explanation

To find the focal length of a plano-convex lens, we can use the lens maker's formula, which is given by:

\[ \frac{1}{F} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Where:
- \( F \) is the focal length of the lens,
- \( n \) 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 \( n = 1.6 \)
- Radius of curvature of the curved surface \( R_1 = 30 \, \text{cm} \)
- For a plano-convex lens, the second surface is flat, so \( R_2 = \infty \).

2. **Substitute the values into the lens maker's formula:**
\[ \frac{1}{F} = (1.6 - 1) \left( \frac{1}{30} - \frac{1}{\infty} \right) \]

3. **Calculate \( \frac{1}{\infty} \):**
- Since \( \frac{1}{\infty} = 0 \), the equation simplifies to:
\[ \frac{1}{F} = (0.6) \left( \frac{1}{30} - 0 \right) \]

4. **Simplify the equation:**
\[ \frac{1}{F} = 0.6 \cdot \frac{1}{30} \]

5. **Calculate \( \frac{1}{30} \):**
\[ \frac{1}{30} = 0.0333 \, \text{(approximately)} \]

6. **Multiply:**
\[ \frac{1}{F} = 0.6 \cdot 0.0333 = 0.02 \]

7. **Find \( F \):**
\[ F = \frac{1}{0.02} = 50 \, \text{cm} \]

### Final Answer:
The focal length of the plano-convex lens is \( 50 \, \text{cm} \).