The Young's double-slit experiment is done in a medium of refractive index `4//3`. A light of 600 nm wavelength isfalling on the slits having 0.45 mm separation. The lower shift `S_(2)` is covered by a thin glass sheet of refractive index. 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in Figure
a. Find the location of central maximum (bright fringe with zero path difference) on the y-axis.
b. Find the light intensity of point O relative to the maximum fringe intensity.
c. Now , if 600 nm light is replaced by white light of range 400 - 700 nm, find the wavelengths of the light that from maxima exaclty at point O.
(All wavelength in the problem are for the given medium of refractive index `4//3` Ignoe dispersion.)

Explanation

Given `lambda nm = 6 xx 10^(-7) m, d = 0.45 mm = 0.45 xx 10^(-3) m, D = 1.5`. Thickness of glass sheet,
`t = 10.4 mu m = 10.4 xx 10^(-6)m`.
Refractive index of glass sheet, `mu_(g) = 1.5`.
a. Let central maximum is obtained at a distance y below point O.Then,
`Delta x_(1) = S_(1) P - S_(2) P = (yd)/(D)`
Path difference due to glass sheet,
`Delta x_(2) = (mu_(g)/(mu_(m) - 1 )) t`
New path diffenence will be zero when
`Deltax_(1) = Delta x_(2)`
`implies (yd)/(d) = ((mu_(g))/(mu_(m)) - 1) t`
`implies y = ((mu_(g))/(mu_(m)) - 1) t (D)/(d)`
Subsituting the value, we have
`y = ((mu_(g))/(mu_(m))-1)t (D)/(d)`
or we can say `y = 4.33 mm.`
a. At O, `Delta x_(1) = 0` and `Delta x (2) = ((mu_(g))/(mu_(m)) - 1)t`
`:.` Net path difference `Delta x = Delta x_(2)`
Corresponding phase difference, `Detla phi` or simply
` phi = (2 pi)/(lambda) Delta x`.
Substituting the values, we have
`phi = (2 pi)/(6 xx 10^(-7)) ((1.5)/(4//3) -1) (10.4 xx 10^(-6)) = ((13)/(3)) pi`
Now,
`I (phi) = I_(max) cos^(2) ((phi)/(2))`
`I = I_(max) cos^(2) ((13 pi)/(6))`
`= (3)/(4) I_(max)`
At O, path difference is `Delta x = Delta x_(2) = ((mu_(g)) /(mu_(m) - 1)) t`
For maximum intensity at O,
` Delta x = n lambda` (here `n = 1, 2, 3,...)`
`lambda = (Delta x)/(1), (Delta x)/(2), (Delta x)/(3),....` and so on
`Delta x = ((1.5)/(4//3) - 1) (10.4 xx 10^(-6) m)`
`= ((1.5)/(4//3) - 1) (10.4 xx 10^(-3) nm`
`= 1300 nm`
Maximum intensity will be corresponding to
`lambda 1300 nm, (1300)/(2) nm, (1300)/(3) nm, (1300)/(4) nm,...`
`= 1300 nm, 650 nm, 433.33 nm, 325 nm,...`
The wavelength in the range 400 to 700 nm are 650 nm and 433.33 nm.