A linear aperture of width $0.04 \mathrm{~cm}$ is placed immediately in front of a converging lens o

Question

A linear aperture of width $0.04 \mathrm{~cm}$ is placed immediately in front of a converging lens of focal length $80 \mathrm{~cm}$. The aperture is illuminated normally by a parallel beam of light of wavelength $600 \mathrm{~nm}$. The distance of the second dark band of the diffraction pattern from the centre of the screen is

Professor PC Thomas Classes · Answer

(a) : The slit width is $a = 0.04 \mathrm{~cm} = 4 	imes 10^{-4} \mathrm{~m}$, wavelength $\lambda = 600 \mathrm{~nm} = 6 	imes 10^{-7} \mathrm{~m}$, and the screen distance is the focal length $D = f = 80 \mathrm{~cm} = 0.8 \mathrm{~m}$. The position of the $n$-th minimum is $y_n = \frac{n\lambda D}{a}$. For the second dark band ($n=2$), $y_2 = \frac{2 	imes (6 	imes 10^{-7}) 	imes 0.8}{4 	imes 10^{-4}} = \frac{9.6 	imes 10^{-7}}{4 	imes 10^{-4}} = 2.4 	imes 10^{-3} \mathrm{~m} = 0.24 \mathrm{~cm}$.