A thin flake of transparent material (n = 1.44) is used to cover one slit of...
A thin flake of transparent material (n = 1.44) is used to cover one slit of a double-slit interference arrangement. The central point on the screen is now occupied by what had been the 3th bright side fringe (m = 3). If λ = 574 nm, what is the thickness of the flake in meters?
Homework Answers
in a double slit normally the path difference = d sin 
in this case 1 slit is covered by snow of thickness t (say), due to this additional path difference = t (n-1)
total path difference = d sin
+ t x (n-1)
for location of original central maximum ,
=0
so at
= 0 , total path difference = 0+ t (n -1)= m x 
= 3 x 574 x 10^(-9) m ( since m=3)
t = 3 x 574 x 10^(-9) m / (1.44 - 1)= 3.91 micro meters
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