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Waves travelling through mediums, destructive interference.

  • Thread starter Ush
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  • #1
Ush
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Homework Statement



light travels through air, hits a film (n=1.3) and underneath that film is glass (n=1.5). What thickness of film would result in destructive interference
The lights wavelength is 500nm.

Homework Equations



n = c/v

V = F*λ

The Attempt at a Solution



I drew some diagrams, and my attempt is also on the diagram,
the diagram is attached

Also, I was wondering what an increase in n signifies?
is it increase in density?
and also, I believe a wave flips when it reflects off something with higher n? (fixed boundary) and moves as is through something with less n? (free boundary)

could someone confirm this please?
 

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Answers and Replies

  • #2
Doc Al
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Also, I was wondering what an increase in n signifies?
is it increase in density?
Just think of n as telling you the speed of the wave in a given medium.
and also, I believe a wave flips when it reflects off something with higher n? (fixed boundary) and moves as is through something with less n? (free boundary)
When light reflects from a medium with higher n, it gets a 180 degree phase shift. Reflection from a medium of lower index of refraction has no phase shift.

The key point here is that the light that reflects from the bottom of the film travels an additional distance of 2t compared to the light that reflects from the top surface of the film. For those two reflections to destructively interfere, how many wavelengths must 2t equal? What's the wavelength of the light in the film?
 
  • #3
Ush
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hmm.

λfilm = λair / nfilm = 384.62nm

for minimum thickness, should t = 1/2 λ ?

if so

2t = λfilm
t = 192.3nm =S
 
  • #4
Doc Al
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λfilm = λair / nfilm = 384.62nm
Good.

for minimum thickness, should t = 1/2 λ ?
No, 2t must equal 1/2 λ.
 
  • #5
Ush
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that would mean T is 384.62nm thick?

why would it not be half of the wavelength?
if you fit in 1/2 a wavelength into the film, and it reflects as an inverted wave, wouldn't it still cancel out?

doesn't 2t = 1/2 λ suggest that an entire wavelength goes in, and then reflects on itself?
 
  • #6
Doc Al
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that would mean T is 384.62nm thick?
No.

why would it not be half of the wavelength?
The total phase difference between the two waves must be half a wavelength. The phase difference comes from one of the traveling the extra distance of 2t.
 
  • #7
Ush
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Ohhh okay.

total phase difference is 2t,
a quarter of the wave gets in and reflects??

T = 96.155nm?

I'm having a bit of trouble visualizing this, I drew another diagram, could you see if it's correct?
 

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  • #8
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No your first diagram is correct. These types of problems treat light as rays, not waves.

Think back to Young's Double Slit Experiment

[tex] n\lambda = dsin(\theta) [/tex]

The right side of the equation solved for the path length difference between rays from the slits. In order for constructive interference to occur, the path length difference must equal an integer multiple of a wavelength (crests fall on crests). Similarly, destructive interference happened when the path length difference equaled an integer plus a half multiple of wavelength (0.5x, 1.5x, 2.5x...) (crests fall on troughs). However, notice in Young's Double Slit Experiment the light was coherent (in phase). In this problem the light is possibly not coherent as rays will undergo a 180 degree phase shift whenever they travel from a lower index of refraction to a higher index of refraction.

Notice in this case that the ray reflected from the top of the film has a 180 degree phase shift and the ray reflected from the bottom of the film also has a 180 degree phase shift. Therefore the two rays are actually for all intents and purposes in phase.

So the path length difference between the two rays (for near normal incidence) is 2t and this should therefore be equal to an integer plus a half times the wavelength of light for destructive interference. In other words,

[tex] 2t = (n + \frac{1}{2})\lambda_{in film} [/tex]
 

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