Interference in Thin Films?

  • Thread starter premed
  • Start date
  • #1
10
0

Homework Statement


Light of wavelength [tex]\lambda[/tex] strikes a pane of glass of thickness T and refractice index n, as shown in figure(I attached a diagram I made). Part of the beam is reflected off the upper surface of the glass, and part is transmitted and then reflected off the lower surface of the glass. Destructive interference between these two beams will occur if :
a) T = [tex]\lambda[/tex] / 2
b) 2T = [tex]\lambda[/tex] / 2
c) T = [tex]\lambda[/tex] / 2n
d) 2T = [tex]\lambda[/tex] / 2n

Homework Equations



2T= m[tex]\lambda[/tex]

The Attempt at a Solution


According to this equation, destructive interference occurs when 2T = [tex]\lambda[/tex] /2. Right? Since [tex]\lambda[/tex] = [tex]\lambda[/tex] [tex]_{0}[/tex] / n, then 2T= [tex]\lambda[/tex] / 2n right? The correct answer is C though. I got answer D. What am i doing wrong. Thanks
 

Attachments

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi premed,

In section 3 you have remarked that destructive interference occurs when [itex]2 T = \lambda/2[/itex]. How did you find this? In your relevant equations section, you had already stated that the conditions for destructive interference was given by [itex]2 T = m \lambda[/itex].

In that equation, m is an integer (with appropriate restrictions) and [itex]\lambda[/itex] is the wavelength in the film; I don't think you can get [itex]2 T = \lambda/2[/itex] from that unless m=1/2 which is not allowed.
 
  • #3
10
0
Sorry. I'll clarify. The book gives the equation for constructice interfence as 2T=m[tex]\lambda[/tex] where m is an integer. For destructice interference 2T = (m+1/2)[tex]\lambda[/tex] . thats why i put in 2T = [tex]\lambda[/tex] /2. The m in this case doesn't matter as long as 2T is not an integral multiple of [tex]\lambda[/tex] because it is destructive interference. I set m = 0 so thats how I got 2T = [tex]\lambda[/tex] /2. Does that make sense?
 
  • #4
alphysicist
Homework Helper
2,238
1
Hi premed,

With thin film interference you have two things you have to keep track of: the effects of the path length difference of the two light rays, and the effects of reflection on the phase of the light rays.

When a light ray reflects off a material with a higher index of refraction, it experiences a phase reversal. Whether [itex]2T = m\lambda_n[/itex] (where [itex]\lambda_n[/itex] is the wavelength in the film) is the constructive or destructive condition depends on how many phase reversals there are in the problem (0, 1 or 2).

So how many phase reversals occur? What does that by itself do to the relative phase of the light rays? Once you have that information, then the condition

[tex]
2 T = m \lambda_n
[/tex]

will leave their relative phase (from the reflection effects) unchanged, and

[tex]
2 T = (m+\frac{1}{2}) \lambda_n
[/tex]

moves them out of phase an additional 180 degrees.
 

Related Threads on Interference in Thin Films?

  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
2K
Top