In a Young's double slit experiment `lamda= 500nm, d=1.0 mm andD=1.0m`. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.
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From `I_(R ) = 4 I_(0) "cos"^(2)(phi)/(2) = (1)/(2)(4 I_(0))` `"cos"^(2)(phi)/(2) = (1)/(2), "cos"(phi)/(2) = (1)/(sqrt(2)) , (phi)/(2) = 45^(@), phi = 90^(@)` Minimum distance from central maximum corresponding to `phi = pi//2` would be : `(phi)/(2pi) xx beta` `x = (pi//2)/(2pi) xx ((lambda D)/(d)) = (1)/(4) xx (500 xx 10^(-9) xx 1)/(1 xx 10^(-3)) x` `= 125 xx 10^(-6)m = 1.25 xx 10^(-4)m`
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