Question

A very thin sheet of plastic (n = 1.60) covers one slit of a double-slit apparatus illumi- nated by 640 nm light. The center point on the screen, instead of being a maximum, is dark. What is the minimum thickness of the plastic?

Answer 1

Given

   plastic refractive index n = 1.6

  

wavelength of the light is Lambda = 640 nm

the optical path difference is Delta = (n-1)t

due to plastic sheet , minimum firnge formed instead of maximum on screen so t

the condition for minimum is mimumum delta = Lambda/2

   so (n-1)t = lambda /2

       t = lambda/2(n-1)
       t = (640*10^-9) /2(1.6-1) m

       t = 5.33333*10^-7 m
  
       t = 533.33 nm