Interference of Light - Thin Film

In summary: If they are in-phase with each other then it's constructive interference, otherwise it's destructive interference.
  • #1
GreenPrint
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Homework Statement



Plan plates of glass are in contact along one side and held apart by a wire 0.05 mm in diameter, parallel to the edge in contact and 20 cm distant. Using filtered green mercury light ([itex]λ = 546 nm[/itex]), directed normally on the air film between plates, interference fringes are seen. Calculate the separation of the dark fringes. How many dark fringes appear between the edge and the wire?

Homework Equations





The Attempt at a Solution



I'm trying to follow along with the solutions and I seem to get confused early on

http://imageshack.us/a/img844/8884/ix2m.png [Broken]

I understand that

[itex]Δ = Δ_{p} + Δ_{r} = 2nt = (m + \frac{1}{2})λ[/itex]

Here [itex]n = 1[/itex] because of air and you use [itex]m + \frac{1}{2}[/itex] instead of just [itex]m[/itex] because you want a dark fringe. I believe the solution is using [itex]Δ_{r} = \frac{λ}{2}[/itex]. I'm however not really sure where this is coming from or why it's included. Apparently the geometrical path difference is [itex]\frac{λ}{2}[/itex] but I'm not really sure why.
 
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  • #2
GreenPrint said:
http://imageshack.us/a/img844/8884/ix2m.png [Broken]

Looks like they should have said "destructive" instead of "constructive" here.

I understand that

[itex]Δ = Δ_{p} + Δ_{r} = 2nt = (m + \frac{1}{2})λ[/itex]

Here [itex]n = 1[/itex] because of air and you use [itex]m + \frac{1}{2}[/itex] instead of just [itex]m[/itex] because you want a dark fringe. I believe the solution is using [itex]Δ_{r} = \frac{λ}{2}[/itex]. I'm however not really sure where this is coming from or why it's included. Apparently the geometrical path difference is [itex]\frac{λ}{2}[/itex] but I'm not really sure why.

You should have covered the idea that sometimes a wave "flips over" when it reflects. I think ##\Delta_r## is taking this into account.
 
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  • #3
Yep - the wave inverts on reflection from one of the interfaces.
Which one?

The incoming light reflects of the top surface, but some gets through to reflect off the bottom surface, a distance t away, returns to pass through the top surface again. We are interested in how these two light waves interfere at the top surface.

They interfere constructively if they are in-phase with respect to each other.
This depends on the path difference and the wavelength. One wave travels 2t farther than the other, since it traverses the thickness t twice.

For constructive interference, the path difference is ##m\lambda = 2t## if the wave did not invert.
Since it does, after a path difference of ##n\lambda##, the returning wave will be pi radians out of phase with the wave reflected from the top surface - pi radians is a half wavelength: so the path difference needs to be an extra half-wavelength long to get them back "in phase" - to get constructive interference.

Hence ##(m+\frac{1}{2})\lambda=2t## for constructive interference.
 
  • #4
My book doesn't mention this until Michelson interferometer. It doesn't explain why your supposed to use lambda/2. It appeared in my lecture in thin film interference. I can't seem to find though any explanation as to where it comes from. I also have searched the internet for some sort of explanation on it but can't find anything. Do you know of any text on the internet that explains it? I mean I don't necessarily need to know but I need to know when to take it into consideration and when not to at the very least.

*reading post above*
 
  • #5
Simon Bridge said:
Yep - the wave inverts on reflection from one of the interfaces.
Which one?

The incoming light reflects of the top surface, but some gets through to reflect off the bottom surface, a distance t away, returns to pass through the top surface again. We are interested in how these two light waves interfere at the top surface.

They interfere constructively if they are in-phase with respect to each other.
This depends on the path difference and the wavelength. One wave travels 2t farther than the other, since it traverses the thickness t twice.

For constructive interference, the path difference is ##m\lambda = 2t## if the wave did not invert.
Since it does, after a path difference of ##n\lambda##, the returning wave will be pi radians out of phase with the wave reflected from the top surface - pi radians is a half wavelength: so the path difference needs to be an extra half-wavelength long to get them back "in phase" - to get constructive interference.

Hence ##(m+\frac{1}{2})\lambda=2t## for constructive interference.

This is interesting no one has ever explained this to me. How do I know when they invert and how do I know when they don't?

[itex]Δ = Δ_{p} + Δ_{r}[/itex]

The inverted wave travels 2t as far since it travels through the thickness twice, so then why do we use 2nt as apposed to just 2t for thin film interference?
 
  • #6
See here for a discussion of thin film interference which states a general rule for how to decide if a wave inverts at reflection.
 
  • #7
This is interesting no one has ever explained this to me. How do I know when they invert and how do I know when they don't?
It's the same as for pulses traveling on a string or a slinky spring - when it has to go from one string to another with different properties. If you've never seen this demonstrated, you should.

Animations showing the detail:

http://www.acs.psu.edu/drussell/Demos/reflect/reflect.html

Real life is messier:
http://www.animations.physics.unsw.edu.au/jw/waves_superposition_reflection.htm

Where light moves from a high-speed medium to a low-speed medium, the reflected wave is inverted.

In your case, the light, 1st, goes from glass-to-air, which is slow-to-fast, so the reflected wave is not inverted.
The transmitted wave goes through air until it meets an air-to-glass (fast-to-slow) interface, where the reflected part of the wave is inverted.
So the light wave that traverses the gap gets inverted half-way through it's journey.

It does not have to be air in the gap, and the plates don't have to have the same refractive index (they don't even have to be glass.) So you can get different relations depending on what happens at the different interfaces.
 
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  • #8
Thanks for all of the help guys. It make since to me now. I clicked on all of the links, but was shocked by the link in #6 that went to a AP Physics B class website. I took that class four years ago and learned this material then, apparently I have forgotten all about it in such a small amount of time. Makes me wonder if that'll be the case with the things I am currently learning in my undergraduate a couple of years after I graduate.
 
  • #9
Forgetting that sort of lesson so fast is a sign that you didn't understand it first time around.
Perhaps you skimmed the material at the time and focussed on memorizing equations for the exam?

But it is not too unusual for knowledge you don't practice to get demoted to the back of your memory ... one of the advantages of answering questions on PF is that you keep practicing ;)
Everyone returns to basics sometime.
 

1. What is the interference of light in thin films?

The interference of light in thin films refers to the phenomenon where light waves reflected from the top and bottom surfaces of a thin film interfere with each other, producing a distinct pattern of light and dark bands.

2. How is the interference pattern in thin films created?

The interference pattern in thin films is created when the light waves reflected from the top and bottom surfaces of the film have a phase difference due to the difference in the distance they travel. When the waves meet, they either amplify or cancel each other out, resulting in a pattern of constructive and destructive interference.

3. What factors affect the interference pattern in thin films?

The interference pattern in thin films is affected by the thickness of the film, the angle of incidence of light, and the refractive index of the film material. These factors determine the phase difference between the reflected waves and therefore, the resulting interference pattern.

4. What is the practical application of interference of light in thin films?

The interference of light in thin films has various practical applications, such as in anti-reflective coatings for lenses and solar panels, in optical filters, and in the production of colorful effects in soap bubbles and oil slicks.

5. How is the concept of interference of light in thin films useful in understanding other phenomena?

The concept of interference of light in thin films can be applied to understand other phenomena, such as the colors observed in iridescent materials, the formation of rainbows, and the colors seen in thin layers of oil on water. It is also essential in the study of wave interference and diffraction in general.

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