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In a Young's double-slit experiment with light of wavelength lambda, the separation of slits is d and distance of screen is D such that D > > d > > lambda. If the Fringe width is beta, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is: |
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Answer» `BETA/2` `I_("max") = 4I_0` Now `(I_("max"))/(2) = 4I_0 cos^2 (phi)/2` `1/2 = cos^2 (phi)/2 IMPLIES cos (phi)/2 = 1/(sqrt(2))` `(phi)/2 = (PI)/4` `phi = (pi)/2` `Deltax = lambda/(2pi) xx phi` `implies Deltax = lambda/(2 pi) xx pi/2 = lambda/4` `Delta X = lambda/4` `Delta x = (dy)/D = lambda/4` `implies y = (D/d) xx lambda/4` `y = ((lambda D)/d) xx 1/4` `y = beta/4`. |
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