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Red light of wavelength 6500 Å from a distant source falls on a slit 0.50 mm wide. What is the distance between the two dark bands on each side of the central bright band of the diffraction pattern observed on a screen placed 1.8 m from the slit. |
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Answer» `2.2 mm` `a = 0.50 mm = 0.5 xx 10^(-3)` m D = 1.8 m The distance between the dark bands on each SIDE of CENTRAL bright band is the WIDTH of central maximum. `thereforebeta_("central") = (2lambdaD)/(a) = (2 xx 6500xx 10^(-10)xx 18)/(0.5 xx 10^(-3))` `= 4.68 xx 10^(-3)`m `= 4.68 mm` For circular aperature, we get circular fringes and the width of central bright band in this CASE is `beta_("central") = 2(1.22 lambda)/(a)D = (2 xx 1.22 xx 6500 xx 10^(-10) xx 1.8)/(0.5 xx 10^(-3))` ` = 5.71 xx 10^(-3) m = 5.71 mm`. |
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