The half angular width of the central bright maximum in Fraunhofer diffraction pattern, for a slit of width \( 1.2 \times 10^{-3}\,\mathrm{mm} \) illuminated by monochromatic light of wavelength 600 nm is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\pi}{6} \)
In single-slit diffraction the central maximum extends from the first minimum on one side to the first minimum on the other. So its half angular width is the angle of the first minimum:
\( \displaystyle a\sin\theta = \lambda \)
\( a = 1.2\times10^{-3}\,\mathrm{mm} = 1.2\times10^{-6}\,\mathrm{m} \)
\( \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m} \)
\( \displaystyle \sin\theta = \frac{6\times10^{-7}}{1.2\times10^{-6}} = 0.5 \)
\( \theta = 30^\circ = \pi/6 \)
Here \( \lambda/a = 0.5 \) is not small, so the small-angle form \( \theta \approx \lambda/a \) cannot be used; the sine must be kept.