A single slit of width 0.01 mm gives a diffraction pattern using a light of wavelength 400 nm. The first secondary maximum occurs at an angle
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6 × 10−2 radians
In single-slit diffraction the minima lie at \( a\sin\theta = n\lambda \), and the secondary maxima lie roughly midway between them, at \( a\sin\theta \approx (n + 1/2)\lambda \). For the first secondary maximum (n = 1), with the small-angle approximation \( \sin\theta \approx \theta \):
\( \displaystyle \theta \approx \frac{3\lambda}{2a} \)
\( \lambda = 400\,\mathrm{nm} = 4\times10^{-7}\,\mathrm{m} \)
\( a = 0.01\,\mathrm{mm} = 1\times10^{-5}\,\mathrm{m} \)
\( \displaystyle \theta \approx \frac{3 \times (4\times10^{-7}\,\mathrm{m})}{2 \times (1\times10^{-5}\,\mathrm{m})} = 6\times10^{-2}\,\mathrm{rad} \)