In a diffraction from a single slit of width 0.30 mm, first order maxima is found with light of wavelength 6000 Angstrom. The angular spread of first order maxima on either side of the central maxima is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3 \times 10^{-3}\,\mathrm{rad} \)
In single-slit diffraction the dark fringes lie at \( a\sin\theta = n\lambda \) and the secondary (bright) maxima lie roughly midway between them, at \( a\sin\theta \approx (n + \tfrac{1}{2})\lambda \). The first-order maximum is the first secondary maximum, \( n = 1 \).
\( a = 0.30\,\mathrm{mm} = 3\times10^{-4}\,\mathrm{m} \)
\( \lambda = 6000\,\mathrm{\mathring{A}} = 6\times10^{-7}\,\mathrm{m} \)
\( \displaystyle \theta \approx \frac{3}{2}\,\frac{\lambda}{a} = \frac{3}{2} \times \frac{6\times10^{-7}}{3\times10^{-4}} = 3\times10^{-3}\,\mathrm{rad} \)
Note: The wording “angular spread of first order maxima” can be read in more than one way. The first minimum is at \( \lambda/a = 2\times10^{-3}\,\mathrm{rad} \) (Option 3), the full angular width of the central maximum is \( 2\lambda/a = 4\times10^{-3}\,\mathrm{rad} \) (Option 4), and the span between the first secondary maxima on the two sides is \( 3\lambda/a = 6\times10^{-3}\,\mathrm{rad} \) (Option 1). The answer given here is the angular position of the first secondary maximum on either side of the centre, which is Option (2).