A YDSE is performed in a medium of refractive index `4 // 3`, A light of 600 nm wavelength is falling on the slits having 0.45 nm separation . The lower slit `S_(2)` is covered b a thin glass plate of thickness 10.4 mm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in figure. (All the wavelengths in this problem are for the given medium of refractive index `4 // 3`, ignore absorption.)

The location of the central maximum (bright fringe with zero path difference) on the y-axis will be

Options

A. `4.33mm`
B. `2.56mm`
C. `3.26mm`
D. `5.16mm`

Answer

A

Explanation

Given `1 = 600nm = 6xx10^(-7)m`
`d = 0.45 mm = 0.45xx10^(-3) m , D = 1.5m`
Thickness of glass sheet, `t = 10.4 mm = 10.4xx10^(-6)m`
Refractive index of medium, `mu_(m) = ($)/(3)`
And refractive index of glass sheet, `mu_(g) = 1.5`
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)`
Net path difference will be zero when `Delta x_(1) = Delta x_(2)`
or `(yd)/(D) = ((mu_(g))/(mu_(m))-1)t :. y = ((mu_(g))/(mu_(m)) - 1)t (D)/(d)`
Substituting the value, we have
`y = ((1.5)/(4//3) -1) (10.4xx10^(-6)(1.5))/(0.45xx10^(-3)) , y = 4.33xx10^(-3)m`
or we can say `y = 4.33 mm`