A slit is illuminated by light of wavelength \( 6500\,\mathring{\mathrm{A}} \). The first minimum is observed at \( \theta = 30^\circ \). The width of the slit is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.3\,\mu\mathrm{m} \)
In single-slit diffraction the minima occur where \( a\sin\theta = n\lambda \); the first minimum has \( n = 1 \).
\( \displaystyle a = \frac{\lambda}{\sin\theta} \)
\( \lambda = 6500\,\mathring{\mathrm{A}} = 6500\times10^{-10}\,\mathrm{m} = 6.5\times10^{-7}\,\mathrm{m} \)
\( \sin30^\circ = 0.5 \)
\( \displaystyle a = \frac{6.5\times10^{-7}\,\mathrm{m}}{0.5} = 1.3\times10^{-6}\,\mathrm{m} = 1.3\,\mu\mathrm{m} \)