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Question

In Young's double slit experiment, separation between the slits is halved and distance between slits and screen is doubled. The fringe width is


A
same.
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B
quadrupled
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C
halved.
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D
onethird
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Solution

The correct option is B quadrupled
We have fringe width, 
$$\beta = \dfrac {D \lambda} {d}$$
where $$D$$ is the distance between the screen and the slits; $$d$$ is the distance between the slits and $$\lambda$$ is the wavelength.
Given that the separation between the slits is halved, i.e., $$d' = \dfrac {d}{2}$$ and distance between slits and screen is doubled, i.e., $$D' = 2D$$.
Hence now the modified fringe width ($$ \beta' $$) is given by
$$\beta' = \dfrac {2 D \lambda} {(\dfrac {d}{2})} = \dfrac {4 D \lambda} {d} = 4 \beta$$

Physics

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