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.
B
quadrupled
C
halved.
D
onethird

Solution

## The correct option is B quadrupledWe 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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