A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a YDSE. The paper transimits `4//9` of the light energy falling on it.
a. Find the ratio of maximum intensity to the minimum intensity in interference pattern.
b. How many fringes will cross through the center if an indentical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.

Explanation

a. `(I_(max))/(I_(min)) = (sqrt I_(1) + sqrt I_(2))^(2) /(sqrt I_(1) - sqrt I_(2))^(2) = (1 + 2//3)^(2) /(1 - 2//3)^(2) = (5^(2)) /(1) = 25`
Fringe width, `beta = (D lambda)/(d)`
shifting of fringe pattern due to paper sheet
`s = (t(mu - 1) D)/(d)`
If we paste paper sheet on the order slit, the fringe pattern will shift by same amount on the side.
Hence, number of fringes crossing the center is
`n = (s)/(beta) = (t(mu - 1))/(lambda) =(0.02(1.45 - 1))/(600 xx 10^(-9)) xx 10^(-3)`
`= (0.02 xx 0.45 xx 10^(6))/(600) = (90)/(6) = 15`.