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In Young's double-slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of wavefront 450 nm. The screen is 1.0 m away from the slits. (a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum. (b) How will the fringe pattern change if the screen is moved away form the slits? |
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Answer» Solution :It is given that distance between the two SLITS d=0.15mm=`1.5xx10^(-4)m`, wavelength of light used `lamda=450nm=450xx10^(-9)m` and distance of screen from double-slit `D=1.0m` (a) (i) Distance of the second bright (n=2) FRINGE from the central maximum `x=(nDlamda)/(d)=(2Dlamda)/(d)=(2xx1.0xx450xx10^(-9))/(1.5xx10^(-4))m=6xx10^(-3)m or 6mm`. (ii) Distance of the second dark (n=2) fringe from the central maximum `x=((2n-1)Dlamda)/(2d)=(3Dlamda)/(2d)=(3xx1.0xx450xx10^(-9))/(2xx1.5xx10^(-4))=4.5xx10^(-3)m or 4.5`mm (b) If the screen is moved away from the slits, the fringe WIDTH increases PROPORTIONATELY (since fringe width `beta=(lamdaD)/(d)`) and we obtain broader fringes. however, INTENSITY of fringes is reduced. |
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