Question

In a `YDSE` experiment, the distance between the slits `&` the screen is `100 cm`. For a certain distance between the slits, an interference pattern is observed on the screen with the fringe width `0.25 mm`. When the distance between the slits is increased by `Deltad=1.2 mm,` the fringe width decreased to `n=2//3` of the original value. In the final position, a thin glass plate of refractive index `1.5` is kept in front of one of the slits `&` the shift of central maximum is observed to be `20` fringe width. Find the thickness of the plate `&` wavelength of the incident light.

Answer 1

Correct Answer - `mu = 600 nm, t =24 mum`
Clearly `beta_(initial)=(lambdaD)/(d_(i))`
`rArr 0.25xx10^(-3)=(lambda)/(d)xx1 m`
`rArr (lambda)/(d)=2.5xx10^(-4) …………..(1)`
Afterwards :
`(2)/(3) beta_(i) =(lambdaD)/((d+Deltad))`
`rArr (lambda)/(d+Deltad)=(2)/(3)xx(2.5xx10^(-4))/(1) ..............(2)`
Dividing `(1) and (2)`
`(d+Deltad)/(d)=(3)/(2) rArr d=2 (Deltad) =2.4 mm`.
`& lambda=2.4xx2.5xx10^(-7)m = 600 nm`.
Now `P` becomes central maxima.
`rArr` for point `P : d sintheta =(mu-1)t`
`rArr d. tantheta approx (mu-1)t`
`rArr d(20beta)/(D) =(mu-1)t`
`rArr (d)/(D)xx(20xxlambdaD)/(d) =(1.5-1)t`
`rArr t=(20lambda)/(0.5) =(20xx600xx10^(-9))/(0.5)=24 mum`.