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A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular widthof central maximum obtainedon the screen, 1 m away. Estimate the number of fringes obtained in YDSE with fringe width 0.5 mm, which can be accommodated within the region of total angular spread of the central maximum due to a single slit. |
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Answer» Solution :Calculation of angular width of central maxima Estimation of number of fringes. Angular width of central maximum `OMEGA=(2lamda)/a` `=(2xx5xx10^(-9))/(0.2xx10^(-3))` radin `=5xx10^(-3)` radin `beta=(lamdaD)/d` Lines width of central maxima in the diffraction pattern 2XD `omega'=(2lamdaD)/a` Let n be the number of interference fringes which can be accommodated in the central maxima. `:. nxxbeta= ` [Award the LAST 5 mark if the student WRITES the answer as 2 (taking `d=a`), or just attempts to do these calculation.] |
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