The young's double slit experiment is done in a medium of refractive index `4//3`.A light of `600nm`wavelength is falling on the slits having `0.45` mm separation .The lower slit `S_(2)` is covered by a thin glass sheet of thickness `10.4mu m` and refractive index`1.5`.the intereference pattern is observed ona screen placed `1.5m` from the slits are shown

(a) Find the location of the central maximum (bright fringe with zero path difference)on the y-axis.
(b) Find the light intensity at point `O`relative to the maximum fringe intensity.
(c )Now,if `600nm`light is replaced by white light of range 400 to700 nm find the wavelength of the light that from maxima exactly point `O`.[All wavelengths in this problem are for the given medium of refractive index `4//3` .Ignore dispersion]

Answer

A, B, C, D

Explanation

(a) Let the central maxima is obtained at a distance x below O.[This is because a glass sheet is present is front of `S_(2) which increases its path length to the screen. Therefore the path length of ray from `S_(1)` to the screen should also increase].
Here,
`implies(xs)/(D)=((mu_g)/(mu_m)-1)t`
`impliesx=((mu_g)/(mu__m)-1)txx(D)/(d)=((1.5)/((4)/(3))-1)xx((10.4xx10^(-6))(1.5))/(0.45xx10xx^(-3))`
`=4.33xx10^(-3)`m
(b) For O, path difference `=((mu_g)/mu_m-1)t`
`:'` Phase difference `phi=(2pi)/(lamda)((mu_g)/(mu_m)-1)t=(2xx3.14)/(6xx10^(-7))((1.5)/((4)/(3))-1)(10.4xx10^(-6))` `=6.8rad`
We know that `I=I_(0)cos^(2)(phi)/(2)` `:'``(I)/(I_(0))=cos^(2)(6.8)=0.75`
(c) For maxima, path difference `+=nlamda`
`:'`nlamda=((mu_g)/mu_m-1)t`
`implieslamda=((mu_g)/(mu_m)-1)(t)/(n)=((1.5)/((4)/(3))-1)(10.4xx10^(-6))/(n)`
`=(1.3xx10^(-6))/(n)`
Putting different values of n for find the wave length in the range of `0.4xx10^(-6)`m to `0.7xx10^(-6)`m we get `lamda=0.65xx10^(-6)`m and `0.433x10^(-6)`m