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Light of wavelength `lambda` is incident on a slit of width d and distance between screen and slit is D. Then width of maxima and width of slit will be equal if D is ––A. `(2lambda^(2))/d`B. `d^(2)/(2lambda)`C. `d/lambda`D. `(2lambda)/d` |
Answer» Correct Answer - B In diffraction fringe width `beta=(lambdaD)/d` when `d/2=beta" then"d/2=(lambdaD)/drArrD=d^(2)/(2lambda)` |
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