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The width of an interference fringe is 1...

The width of an interference fringe is 1.5 mm . What would be the width of the fringe if the separation between the slits is made twice the original value?

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In the first case `y =(D)/(2d) lambda` =1.5 mm
where 2d = distance between the slits
In the second case the distance between the slits = `2 xx (2d)`
`:. Y= (D)/(2xx2d)lambda=(1)/(2)((D)/(2d)lambda)=(1)/(2)xx1.5=0.75 ` mm
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