Question

In single slit diffraction pattern, the wavelength of light used is \(628\) nm and slit width is \(0.2\) mm. The angular width of central maximum is \(\alpha \times 10^{-2}\) degrees. The value of \(\alpha\) is ____.

Answer 1

Correct Answer: 36

Step 1: Understanding the Concept:
In single slit diffraction, the central maximum extends between the first minima on either side. The condition for the first minimum is \(d \sin \theta = \lambda\). For small angles, \(\sin \theta \approx \theta\). The angular width of the central maximum is \(2\theta\).
Step 2: Key Formula or Approach:
1. Angular position of first minimum: \(\theta = \frac{\lambda}{d}\) (in radians).
2. Angular width: \(W_{\theta} = 2\theta = \frac{2\lambda}{d}\).
3. Conversion: \(\text{Degrees} = \text{Radians} \times \frac{180}{\pi}\).
Step 3: Detailed Explanation:
Given:
\(\lambda = 628 \text{ nm} = 628 \times 10^{-9} \text{ m}\).
\(d = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}\).
Calculate angular width in radians:
\[ W_{\text{rad}} = \frac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}} = \frac{1256 \times 10^{-9}}{2 \times 10^{-4}} = 628 \times 10^{-5} \text{ rad} \]
Now, convert to degrees (using \(\pi \approx 3.14\)):
\[ W_{\text{deg}} = \left( 628 \times 10^{-5} \right) \times \frac{180}{3.14} \]
\[ W_{\text{deg}} = \frac{6.28 \times 10^{-3} \times 180}{3.14} = 2 \times 10^{-3} \times 180 = 0.36^{\circ} \]
Given the width is \(\alpha \times 10^{-2}\) degrees:
\[ 0.36 = \alpha \times 10^{-2} \implies 36 \times 10^{-2} = \alpha \times 10^{-2} \implies \alpha = 36 \]
Step 4: Final Answer:
The value of \(\alpha\) is 36.