InterviewSolution
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InterviewSolution
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(a) Explain two features to distinguish between the interference patternn in Young's double-slit experiment with the diffraction pattern obtained due to a single-slit. (b) A monochromaticlight of wavelength 500 nm is incident normally on a single-slit of width 0.2mm to produce a diffraction pattern. find the angular width of the central maximum obtained on the screen. Estimate the number of fringes obtained in Young's double-slit experiment with fringe width 0.5mm, which can be accommodated within the region of total angular apread of the central maximum due to single-slit. |
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Answer» Solution :(b) Here wavelength of MONOCHROMATIC light `lamda=500nm=500xx10^(-9)m=5xx10^(-7)` m and slit width `a=0.2mm=0.2xx10^(-3)m=2.0xx10^(-4)m` `THEREFORE` angular width of central diffraction maximum `alpha=(2lamda)/(a)=(2xx5xx10^(-7))/(2.0xx10^(-6))=5xx10^(-3)` radian Iit means that central diffraction maximum extends from `theta=0` to `theta=pm(lamda)/(a)=2.5xx10^(-3)rad`. If .D. be the DISTANCE of screen from the single-slit then it means that central diffraction maximum extends from `y=0` to `y=pmDtheta=2.5xx10^(-3)D`. As fringe width of Young.s double-slit experiment `beta=0.5=0.5xx10^(-4)m=5.0xx10^(-3)m`, hence number of interference fringes on either side of central maxima and lying within the single-slit diffraction maximum are `n=(2.5xx10^(-3)D)/(5.0xx10^(-3))=(D)/(2)`. `therefore` Total number of interference fringes which can be accommodated withhin the region of total angular SPEED of the central maximum due to single-slit `=(2n+1)=((2D)/(2)+1)=(D+1)`. |
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