A single slit of width b is illuminated by a coherent monochromatic light of wavelength `lambda`. If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum ? (i.e., distance between first minimum on either side of the central maximum)

Options

A. 4.5 cm
B. 1.5cm
C. 6.0cm
D. 3.0cm

Answer

To solve the problem, we need to find the width of the central maximum in a single slit diffraction pattern, given the positions of the second and fourth minima.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a single slit of width \( b \) illuminated by monochromatic light of wavelength \( \lambda \).
- The second minimum is at a distance \( y_2 = 3 \) cm from the central maximum.
- The fourth minimum is at a distance \( y_4 = 6 \) cm from the central maximum.
- The distance from the slit to the screen is \( D = 1 \) m.

2. **Using the Condition for Minima**:
- The condition for minima in a single slit diffraction pattern is given by:
\[ b \sin \theta_n = n \lambda \]
- For small angles, we can approximate \( \sin \theta \approx \tan \theta \approx \frac{y_n}{D} \).

3. **Setting Up Equations**:
- For the second minimum (\( n = 2 \)):
\[ b \frac{y_2}{D} = 2 \lambda \quad \text{(1)} \]
- For the fourth minimum (\( n = 4 \)):
\[ b \frac{y_4}{D} = 4 \lambda \quad \text{(2)} \]

4. **Substituting Known Values**:
- Substitute \( y_2 = 3 \) cm and \( y_4 = 6 \) cm into equations (1) and (2):
\[ b \frac{3 \text{ cm}}{100 \text{ cm}} = 2 \lambda \quad \text{(1)} \]
\[ b \frac{6 \text{ cm}}{100 \text{ cm}} = 4 \lambda \quad \text{(2)} \]

5. **Rearranging the Equations**:
- From equation (1):
\[ b \cdot \frac{3}{100} = 2 \lambda \implies \lambda = \frac{b \cdot 3}{200} \quad \text{(3)} \]
- From equation (2):
\[ b \cdot \frac{6}{100} = 4 \lambda \implies \lambda = \frac{b \cdot 6}{400} \quad \text{(4)} \]

6. **Equating the Two Expressions for \( \lambda \)**:
- Set equations (3) and (4) equal to each other:
\[ \frac{b \cdot 3}{200} = \frac{b \cdot 6}{400} \]
- Cross-multiplying gives:
\[ 400 \cdot 3 = 200 \cdot 6 \]
\[ 1200 = 1200 \quad \text{(True)} \]
- This confirms our equations are consistent.

7. **Finding the Width of the Central Maximum**:
- The width of the central maximum is defined as the distance between the first minima on either side of the central maximum.
- For the first minimum (\( n = 1 \)):
\[ b \frac{y_1}{D} = 1 \lambda \]
- Using \( \lambda = \frac{b \cdot 3}{200} \):
\[ b \frac{y_1}{100} = 1 \cdot \frac{b \cdot 3}{200} \]
- Cancel \( b \) (assuming \( b \neq 0 \)):
\[ \frac{y_1}{100} = \frac{3}{200} \implies y_1 = \frac{3 \cdot 100}{200} = 1.5 \text{ cm} \]
- Therefore, the total width of the central maximum is:
\[ 2y_1 = 2 \cdot 1.5 \text{ cm} = 3 \text{ cm} \]

### Final Answer:
The width of the central maximum is **3 cm**.