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?)

Options

A. (a) `(D(mu-1)t)/(2d)`
B. (b) `(2D(mu-1)t)/(d)`
C. (c) `(D(mu+1)t)/(d)`
D. (d) `(D(mu-1)t)/(d)`

Answer

D

Explanation

The path difference introduced due to introduction of transparent sheet is given by `Deltax=(m-1)t`.
If the central maxima occupies position of nth fringe, then
`(mu-1)t=nlambda=dsintheta`
`sin theta=((mu-1)t)/(d)=((1.17-1)xx1.5xx10^-7)/(3xx10^-7)=0.085`
Hence the angular position of central maxima is
`theta=sin^-1(0.085)=4.88^@`
For small angles `sin theta~=theta~=tantheta`
`tantheta=y/D`
`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.