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In Young's double slit experiment, the f...

In Young's double slit experiment, the fringes are displaced index 1.5 is introduced in the path of one of the beams. When this plate in replaced by another plate of the same thickness, the shift of fringes is `(3//2)x`. The refractive index of the second plate is

A

`1.75`

B

`1.50`

C

`1.25`

D

`1.00`

Text Solution

Verified by Experts

The path difference introduced by a plate of thickness `t` and refractive index `mu` is given by
`Delta=(mu-1)t`
A path difference of `lambda` introduced a phase shift of `beta` where `phi=lambda D//(2d).2d=` separation between slits
So, a path difference of `(mu-1)t` introduces a shift `x` on the screen given by
`x=((mu-1)tbeta)/(lambda)`
for first plate, `x=((mu_(1)-1)tbeta)/(lambda)`
for second plate `(3)/(2)x=((mu_(2)-1)tbeta)/(lambda)`
`:. ((mu_(2)-1))/((mu_(1)-1))=(3)/(2)rArr((mu_(2)-1))/((1.5-1))=(3)/(2)`
`rArrmu_(2)=1.75`
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