Q.

In Young's double slit experiment the distance between the slits and the screen is doubled. The separation between the slits is reduced to half. As a result the fringe width

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a

is halved

b

becomes four times

c

remains unchanged

d

is doubled

answer is B.

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Detailed Solution

Fringe width, β=λDdwhere D is the distance between slits and screen and d is the distance between the slits. When D is doubled and d is reduced to half, then fringe width becomesβ'=λ(2D)(d/2)=4λDd=4β
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