A slit of width 'd' is illuminated by red light of wavelength 650 nm. What should be the value of slit width 'd' so that first minimum fall at an angle of diffraction of \( \frac{\pi}{6} \)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.3\times10^{-6} \) m
In single-slit diffraction the minima occur at angles given by \( d\sin\theta = n\lambda \) (\( n = 1, 2, \ldots \)). For the first minimum \( n = 1 \) and \( \theta = \frac{\pi}{6} = 30^\circ \).
\( \displaystyle d = \frac{\lambda}{\sin\theta} = \frac{650\times10^{-9}\,\mathrm{m}}{\sin30^\circ} = \frac{650\times10^{-9}}{0.5} = 1.3\times10^{-6}\,\mathrm{m} \)