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.

Answer

A, B

Explanation

Given that `t=0.02mm`
`=0.02xx10^-3m`
`mu_1=1.45`
`lamda=600nm=600xx10^-9m`
a. `Let I_1=` intensity of source without paper =I
Then `I_2` = intensity of source with paper
`=(1/9)I`
`rarr I_1/I_2=9/4`
`rarr r_1/r_2=3/2 [:. Ipropr^2]`
`where r_1 and r_2` are corresponding amplitudes
`So, I_(max)/I_(min)=((r_1+r_2)^2)/((r_1+r_2)^2)=((3+2)^2)/((3-2)^2)`
`=25/1=25:`
b. No of fringe that will cross the origin is given by
`n=(mu-1)t)/lamda`
`=((1.45-1)xx0.02xx10^-3)`
`=(0.45xx0.02x10^-3)/(6xx10^-7)=15`