Two sources give interterence pattern which is observed on a screen, placed at a distance D from the sources. The fringe width is given as 2W. If the distance of sources from the screen is doubled, then the fringe width will
A
Become W/2
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B
Remains the same
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C
Become W
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D
Become 4W
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Solution
The correct option is D Become 4W Given, W1=2W, Fringe width is given by W=λDd ⇒2W=λDd When D′=2D, we get W2=λD′d ⇒W2=λ2Dd=2W1 W2=2×2W=4W.