A monochromatic beam of light of 6000 `Å` is used in YDSE set-up. The two slits are covered with two thin films of equal thickness t but of different refractive indices as shown in figure. Considering the intensity of the incident beam on the slits to be `I_(0)`, find the point on the screen at which intensity is `I_(0)` and is just above the central maxima. (Assume that there is no change in intensity of the light after passing through the films.) Consider `t = 6 mu m, d = 1 mm,` and `D = 1 m,` where d and D have their usual meaning. Give your answer in mm.

Answer

`y = 30 mm`

Explanation

`I_(0) = 4 I_(0) cos^(2).(phi)/(2)`
`implies cos. (phi)/(2) = (1)/(2)`
`implies phi = (2 pi)/(3)`
Therefore, path difference of the two beams producing intensity `I_(0)` on the screen is `lambda//3`.
Optical path difference between two beams meeting on the screen is
`Delta x = {(S_(2) P - t + (4)/(3) t) - (S_(1) P - t + (3)/(2) t)}`
` = (S_(2) P - S_(1) P) + t ((4)/(3) - (3)/(2))`
`Delta x = d sin theta - (t)/(6)`
`:. (dy)/(D) - (t)/(6) = (lambda)/(3)`
`implies (dy)/(D) = (lambda)/(3) + (t)/(6)`
`implies y = (D)/(d) [(lambda)/(3) + (t)/(6)]`
` = (1)/(10^(-3)) ((6000 xx 10^(-10))/(3) + (0.6 xx 10^(-6))/(6))`
`= (1)/(10^(-3)) xx 3 xx 10^(-7) = 3 xx 10^(-4) m`
`:. y = 30 mm`.