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If fringe width is 0.4 mm, the distance between fifth bright and third dark band on same side isA. 1 mmB. 2 mmC. 3 mmD. 4 mm |
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Answer» Correct Answer - A Position of nth bright fringe from central maxima ` x_(n_(1))=(n_(1)lambdaD)/(d)` here `n_(1)=5` `:. x_(n_(1))=(5 lambda D)/(d)` Position of nth dark fringe from central maxima `x_(n_(1))=(n_(1) lambda D)/( d)` here `n_(1)=5 ` ` :. x_(n_(1))=(5)/(2) (lambdaD)/(d)` `x_(n_(1))-x_(n)=(2.5 lambda D)/( d)=2.5 beta ` Given ` beta =0.4 mm` `implies x_(n_(1))-x_(n)=1 mm` |
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