Question

Find the angle of diffraction (in degrees) for first secondary maximum of the pattern due to diffraction at a single slit. The width of the slit and wavelength of light used are 0.55 mm and 550 nm, respectively.

Hint:

For secondary maxima in single slit diffraction, use the approximate formula \(\sin \theta = \frac{(m + \frac{1}{2}) \lambda}{a}\), where \(m = 1, 2, 3, \dots\). Always convert units of wavelength and slit width to meters before calculation.

Answer 1

The angular positions of secondary maxima in single-slit diffraction are approximated by the formula: \[\sin \theta = \frac{(m + \frac{1}{2})\lambda}{a}\] where \( m = 1 \) for the first secondary maximum, \( \lambda = 550 \ \text{nm} = 550 \times 10^{-9} \ \text{m} \), and \( a = 0.55 \ \text{mm} = 0.55 \times 10^{-3} \ \text{m} \). Substituting these values: \[\sin \theta = \frac{(1 + \frac{1}{2}) \cdot 550 \times 10^{-9}}{0.55 \times 10^{-3}} = \frac{1.5 \cdot 550 \times 10^{-9}}{0.55 \times 10^{-3}}\] \[= \frac{825 \times 10^{-9}}{0.55 \times 10^{-3}} = 1.5 \times 10^{-3}\] Therefore: \[\theta = \sin^{-1}(1.5 \times 10^{-3}) \approx 0.086^\circ\] Final Answer: Approximately \( \boxed{0.086^\circ} \)