The first diffraction minima due to a single slit diffraction is at \( \theta = 30^\circ \) for a light of wavelength \( 5000\,\mathring{\mathrm{A}} \). The width of the slit is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1\times10^{-4}\,\mathrm{cm} \)
In single-slit diffraction the minima are at angles given by the condition for a minimum, \( a\sin\theta = n\lambda \), with \( n = 1 \) for the first minimum.
\( \displaystyle a = \frac{\lambda}{\sin\theta} \)
\( \lambda = 5000\,\mathring{\mathrm{A}} = 5000\times10^{-10}\,\mathrm{m} = 5\times10^{-7}\,\mathrm{m} \)
\( \sin 30^\circ = 0.5 \)
\( \displaystyle a = \frac{5\times10^{-7}}{0.5} = 1\times10^{-6}\,\mathrm{m} \)
Converting to centimetres (\( 1\,\mathrm{m} = 100\,\mathrm{cm} \)):
\( \displaystyle a = 1\times10^{-6} \times 100\,\mathrm{cm} = 1\times10^{-4}\,\mathrm{cm} \)