In the single slit experiment of diffraction, the aperture of the slit is 2 mm. If a monochromatic light of wavelength 600 nm is incident normally on the slit, the distance between 2nd order minima and the 4th order maxima on the screen placed 2 m away from the slit will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.5 mm
In single-slit diffraction the minima lie at \( a\sin\theta = n\lambda \), and the secondary maxima lie approximately midway between them, at \( a\sin\theta = (n + \frac{1}{2})\lambda \). For small angles the distance from the centre of the pattern on the screen is \( y = D\sin\theta \).
\( a = 2\,\mathrm{mm} = 2\times10^{-3}\,\mathrm{m} \)
\( \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m} \)
\( D = 2\,\mathrm{m} \)
\( \displaystyle \frac{\lambda D}{a} = \frac{6\times10^{-7} \times 2}{2\times10^{-3}} = 6\times10^{-4}\,\mathrm{m} = 0.6\,\mathrm{mm} \)
\( y_{\text{2nd min}} = 2 \times 0.6 = 1.2\,\mathrm{mm} \)
\( y_{\text{4th max}} = 4.5 \times 0.6 = 2.7\,\mathrm{mm} \)
Taking both on the same side of the central maximum:
\( \displaystyle \Delta y = 2.7 - 1.2 = 1.5\,\mathrm{mm} \)