Please show your work 1.) The dark fringe for m = 0 in a Young\'s double-slit ex
ID: 1400513 • Letter: P
Question
Please show your work
1.) The dark fringe for m = 0 in a Young's double-slit experiment is located at an angle of = 7°. What is the angle that locates the dark fringe for m = 3?
3.) You are standing in air and are looking at a flat piece of glass
(n = 1.52)
on which there is a layer of transparent plastic
(n = 1.61).
Light whose wavelength is 649 nm in vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the thickness of the layer.
smaller value __________nm larger value ____________nmExplanation / Answer
1]
we have the equation d*sin = [m-1/2] * lambda
The dark fringe for m = 0 in a Young's double-slit experiment is located at an angle of = 7°
d* sin [70]= 1/2* lambda
d/ lambda= 0.5/0.123= 4.10
The angle that locates the dark fringe for m = 3
d*sin = [3-1/2] * lambda
d*sin = 2.5* lambda
sin =2.5* lambda/ d
sin =2.5 / 4.10= 0.6097
= sin-1[0.6097]= 37.570
2]
Given:-
X7- X1= 0.023m
D=1.1 m
d=1.40*10-4 m
lambda=?
we have the equation ; Xn= [n-1/2]* lambda *D/d
X7= 6.5* lambda *1.1/1.40*10-4 = 5.11*104* lambda
X1= 0.5* lambda *1.1/1.40*10-4 = 0.393*104* lambda
X7-X1= 5.11*104* lambda- 0.393*104* lambda
X7-X1= 4.72*104* lambda
lambda =X7-X1/ 4.72*104 = 0.023/4.72*104 =4.87*10-7m
The wavelength of the light being used is 4.87*10-7m
3]
For the minimum nonzero thickness, m = 1.
destructive interference (because the plastic looks dark) --> 2nt=m*wavelength
t= m*wavelength/2n
t= 1*(649*10-9)/(2*1.61) = 201.6 *10-9m= 201.6 nm------------------smallest value
t= 2*201.6 nm= 403.2 nm----------------------------------second smallest possible value