Question

In YDSE, light of wavelength `lamda = 5000 Å` is used, which emerges in phase from two a slits distance `d = 3 xx 10^(-7) m` apart. A transparent sheet of thickness `t = 1.5 xx 106(-7) m`, refractive indes `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?)

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

Answer 1

Correct Answer - d
The path difference intorduced due to introduction of transparent sheet is given by `Delta x = (m - 1)t`.
If the central maxima occupies position of nth fringe, then
`(mu - 1)t = n lambda = d sin theta`
`implies sin theta ((mu-1) t)/(d) = ((1.17 - 1) xx 1.5 xx 10^(-7))/(3 xx 10^(-7))`
`=0.085`
Hence, the angular position of central maxima is
`theta = sin^(-1) (0.085)= 4.88^(@)`
For small angles,
`sin theta = theta = tan theta`
`implies tan theta = (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.