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

Concept: For single slit diffraction, the angular width of the central maximum is \[ \theta = \frac{2\lambda}{a} \] where \(\lambda\) = wavelength of light, \(a\) = slit width. 
Step 1:Convert units \[ \lambda = 628 \text{ nm} = 628 \times 10^{-9} \text{ m} \] \[ a = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m} \] 
Step 2:Calculate angular width in radians \[ \theta = \frac{2\lambda}{a} \] \[ = \frac{2 \times 628 \times 10^{-9}}{2 \times 10^{-4}} \] \[ = 6.28 \times 10^{-3} \text{ radians} \] 
Step 3:Convert to degrees} \[ 1 \text{ rad} = \frac{180}{\pi} \text{ degrees} \] \[ \theta = 6.28 \times 10^{-3} \times \frac{180}{\pi} \] \[ \theta \approx 0.36^\circ \] 
Step 4:Express in required form\[ 0.36^\circ = 36 \times 10^{-2} \text{ degrees} \] Thus \[ \boxed{\alpha = 36} \]