In YDSE, light of wavelength `lamda = 5000 Å` is used, which emerges in phase from two slits distance `d = 3 xx 10^(-7) m` apart. A transparent sheet of thickness `t = 1.5 xx 10^(-7) m`, refractive index `n = 1.17`, is placed over one of the slits. Where does the central maxima of the interference now appear from the center of the screen? (Find the value of y?)

Explanation

The path differene introduced due to introduction of transparent sheet is given by `Delta x = (mu - 1)t`.

If the central maxima ocupies position of nth fringe,
Then `(mu-1)t=n lambda = d sin theta`
`sin theta =((mu-1)t)/(d)=((1.17-1)xx1.5xx10^(-7))/(3xx10^(-7))=0.085`
Hence is angular position of central maxima is
`theta = sin^(-1)(0.085)=4.88^(@)`
For small angles `sin theta ~= theta ~= tan theta`
As `tan theta = (y)/(D)`
so `(y)/(D)=((mu-1)t)/(d)`
Shift of central maxima is
`Y=(D(mu-1)t)/(d)`.
This formula can be used if D is given.