In a Young's double slit experiment ,a thin sheet of refractive index 1.6 is used to cover one slit while a thin the sheet of refractive index 1.3 is used to cover the second slit.
The thickness of both the sheets are same and the wavelength of light used is 600 nm.
If the central point on the screen in now occupied by what had been the 10th bring fringe (m=10 ) , then the thickness of covering sheets is

Options

A. `50 mum `
B. `8 mu m`
C. `20 mum`
D. `40 mum`

Answer

C

Explanation

Given refractive index of first sheet `mu_1 =1.6` refractive index of second sheet `mu_2 =1.3` and wavelength of light `lambda = 600 nm = 600 xx 10^(-9)` m
A young's double slit experiment is shown below in which two sheets are covered the slits.
So , the path difference introduced in slit l .
`Delta X_1 =(mu_1 -1) t=(1.6-1) t = 0.6 t`

Similarly path difference introduced in slit 2,
`Deltax_2 =(mu_2 -1) t=(1.3)t=0.3 t`
so, the net path difference introduced in central maxima ,
`Delta x_("central maxima") = Delta x_1 - Deltax_2 = 0.6 t - 0.3 t = 0.3 t`
For central maxima, which occupied the `10 ^(th)` bright frige,
`Deltax_(cen) = 10 lambda`
`rArr t = (10xx 600xx 10^(-9))/(0.3) = 20 mum`