Monochromatic lights of wavelengths 620 nm and 626 nm, respectively are used to study diffraction at a single slit of aperture 0.7 mm. The distance between the slit and the screen is 1.8 m. The separation between the positions of the first maxima of the diffraction pattern obtained in the two cases is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.023 mm
The central maximum lies at the centre for every wavelength, so “first maxima” here means the first secondary maxima. For a single slit of width \( a \) they lie approximately at \( \theta = 3\lambda/2a \), so on a screen at distance \( D \)
\( \displaystyle y_1 = D\theta = \frac{3\lambda D}{2a} \;\Rightarrow\; \Delta y = \frac{3D\,\Delta\lambda}{2a} \)
\( \Delta\lambda = 626 - 620 = 6\,\mathrm{nm} = 6\times10^{-9}\,\mathrm{m} \)
\( a = 0.7\,\mathrm{mm} = 0.7\times10^{-3}\,\mathrm{m} \)
\( D = 1.8\,\mathrm{m} \)
\( \displaystyle \Delta y = \frac{3 \times 1.8 \times 6\times10^{-9}}{2 \times 0.7\times10^{-3}} = \frac{3.24\times10^{-8}}{1.4\times10^{-3}} = 2.31\times10^{-5}\,\mathrm{m} \)
\( \Delta y = 2.31\times10^{-5}\,\mathrm{m} = 0.023\,\mathrm{mm} \)