First diffraction minima is obtained at an angle of \( 30^\circ \) for red light of wavelength \( 6500\,\mathring{\mathrm{A}} \). The width of the slit is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.3\times10^{-6}\,\mathrm{m} \)
In single-slit diffraction the first minimum occurs where \( a\sin\theta = \lambda \), a being the width of the slit.
\( \lambda = 6500\,\mathring{\mathrm{A}} = 6500\times10^{-10}\,\mathrm{m} = 6.5\times10^{-7}\,\mathrm{m} \)
\( \displaystyle a = \frac{\lambda}{\sin\theta} = \frac{6.5\times10^{-7}}{\sin 30^\circ} = \frac{6.5\times10^{-7}}{0.5} = 1.3\times10^{-6}\,\mathrm{m} \)