A slit of width a is illuminated by white light. The first minimum for red light (\( \lambda=6500 \) Å) will fall at \( \theta=30^\circ \) when a is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.3\,\mu\mathrm{m} \)
In single-slit diffraction the minima lie at \( \theta\approx n\lambda/a \), \( n=\pm1,\pm2,\ldots \). This small-angle form comes from the condition that the path difference \( a\sin\theta \) between the waves from the two edges of the slit is \( n\lambda \). At \( 30^\circ \) the angle is not small, so the first minimum (\( n=1 \)) is found from \( a\sin\theta=\lambda \).
With 1 Å = \( 10^{-10}\,\mathrm{m} \), the data are:
\( \lambda=6.5\times10^{-7}\,\mathrm{m} \)
\( \sin30^\circ=0.5 \)
\( \begin{aligned} a &= \frac{\lambda}{\sin\theta} \\ &= \frac{6.5\times10^{-7}\,\mathrm{m}}{0.5} \\ &= 1.3\times10^{-6}\,\mathrm{m} \\ &= 1.3\,\mu\mathrm{m} \end{aligned} \)