Thin Film Interference Problems

In summary: Where L is the thickness of the film, n and m are integers and \lambda_1 and \lambda_2 are the wavelengths for the two minimum reflections. The problem is that with n=2, we get m=2.36, which is not an integer. So it seems that this combination of refractive indices does not have a solution.In summary, the first problem involves understanding phase reversal and constructive and destructive interference in order to calculate the thickness of the metal foil. The second problem is more complicated, involving a thin film of alcohol on a glass plate and requires reducing the ratio of wavelengths to integers and taking into account the index of refraction of the
  • #1
404
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There's two problems I can't seem to figure out...

A fine metal foil separates one end of two pieces of optically flat glass. When light of wavelength 500nm is incident normally, 28 dark lines are observed (with one at each end). How thick is the foil? (answer = 6.75um)
I sort of got the answer, except I used m as 28 instead of 27. So my question is why are you suppose to use m as 27 to multiply instead of 28? Also What do they mean by incident normally, is it like striking the glass at a right angel?

and the other one is ...
A thin film of alcohol (n=1.36) lies on a flat glass plate (n=1.51). When monochromatic light, whose wavelength can be changed, is incident normally, the reflected light is a minimum for wavelength = 512nm, and wavelength = 640nm. What is the thickness of the film? (answer = 471nm)
 
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  • #2
404 said:
I sort of got the answer, except I used m as 28 instead of 27. So my question is why are you suppose to use m as 27 to multiply instead of 28? Also What do they mean by incident normally, is it like striking the glass at a right angel?

The key is to understand that two of the 28 dark lines are at the ends of the glass. The dark line where the two flats touch is dark because there is phase reversal at one surface, but not at the other. The next dark line occurs when the path length across the space and back is exactly one wavelength, the one after that exactly 2 wavelengths, etc. Since there are 28 lines, and one is at the edge, the space at the first line from the touching edge is 1/27 times the space at the last line, which is the thickness of the spacer. So 1/27 times the spacer thickness is 1/2 wavelength. The "normal" incidence does mean perpendicular to the glass flats. It has to be normal so that the extra path length of the light is two times the space. If the light were at an angle, the path length difference would be more than twice the space. You would still see lines, but not as many and the calculation would be more complicated.
 
  • #3
A thin film of alcohol (n=1.36) lies on a flat glass plate (n=1.51). When monochromatic light, whose wavelength can be changed, is incident normally, the reflected light is a minimum for wavelength = 512nm, and wavelength = 640nm. What is the thickness of the film? (answer = 471nm)

First you need to reduce the ratio of the wavelenths to a ratio of integers. Then you need to understand what conditions cause phase reversal for some reflections and not others, and decide if both surfaces reflect the same phase, or differently. When you do that, you will realize that those integers are connected to the number of wavelengths of light that will "fit" into the thin film layer. There will be more of the sorter wavelengths than longer wavelengths in the film. Once you know how many of each wavelength are in the film, you can calculate the film thickness as a multiple of the wavelength. Since the film has an index of refraction other than 1, the wavelength in the film will be shorter than in the air. You will have to deal with that.

Edit

I think this problem may be flawed. For this combination of refractive indices, the wavelength ratio may be impossible. With the ratio given, even if the materials were reversed I'm not getting the answer given. I seem to be getting twice the thickness. Anyone else have an opinion?
 
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  • #4
404 said:
There's two problems I can't seem to figure out...


I sort of got the answer, except I used m as 28 instead of 27. So my question is why are you suppose to use m as 27 to multiply instead of 28?
The first glass/air surface causes the inner surface of the first glass to reflect without a phase shift. At the second air/glass surface, the second glass reflects with a pi phase shift so it provides constructive interference if the space pi/2 ([itex]\lambda/4[/itex]) thick and destructive interference if the space is 0 or pi ([itex]2\lambda/4[/itex]) thick. Thereafter the dark band (destructive interference) occurs at half wavelength intervals: [itex]4\lambda/4, 6\lambda/4, 8\lambda/4... 54\lambda/4[/itex] and light (constructive) at [itex]3\lambda/4, 5\lambda/4, 7\lambda/4... 53\lambda/4[/itex]. So the maximum thickness is 54*500e-9/4 = 6.75e-6 m.

Normal means right angle.

The second one is a little more complicated. Both the surfaces (air/alcohol, alcohol/glass) reflect with a pi phase shift since the index of refraction of the subsequent medium is higher at both surfaces. So there will be minimal reflection when the alcohol is pi/2 or [itex]\lambda/4[/itex]thick. The wavelength changes with the index of refraction of the alcohol (not of the glass). The wavelength in alcohol of the first light is about 376 nm and 471 nm for the second light. Can you work it out from that?

AM
 
  • #5
Andrew Mason said:
The second one is a little more complicated. Both the surfaces (air/alcohol, alcohol/glass) reflect with a pi phase shift since the index of refraction of the subsequent medium is higher at both surfaces. So there will be minimal reflection when the alcohol is pi/2 or [itex]\lambda/4[/itex]thick. The wavelength changes with the index of refraction of the alcohol (not of the glass). The wavelength in alcohol of the first light is about 376 nm and 471 nm for the second light. Can you work it out from that?

AM

Given that both surfaces have the same phase at reflection, your [itex]\lambda/4[/itex] condition is certainly correct for the minimum thickness. The next thickness that will result in destructive interference is an additional [itex]\lambda/2[/itex], and the one after that still another [itex]\lambda/2[/itex]. The sequence being

[itex]\lambda/4[/itex], [itex]3\lambda/4[/itex], [itex]5\lambda/4[/itex], etc

The thickness of the film is the same for both wavelengths, so we should be looking for a solution to the equation

[tex]L = (2n+1)\lambda_1/4 = (2m+1)\lambda_2/4[/tex]

for integer values of m, n. Solving for the ratio of the wavelengths gives

[tex](2n+1)/(2m+1) = \lambda_2/\lambda_1 = 640/512 = 5/4[/tex]

which has no solution for integer n and m. The only way you could have a ratio of wavelengths equal to consecutive integers is if the path length through the film was a multiple of a wavelength, which would be the case if there was phase reversal at only one surface, which would happen if the film layer had a higher index of refraction than the air or the bottom layer. The possible thicknesses would then be

[itex]\lambda/2[/itex], [itex]2\lambda/2[/itex], [itex]3\lambda/2[/itex], etc., leading to

[tex]L = n\lambda_1/2 = m\lambda_2/2[/tex]

for integer values of m, n. Solving for the ratio of the wavelengths gives

[tex]n/m = \lambda_2/\lambda_1 = 640/512 = 5/4[/tex]

This equation has the solution n = 5, m = 4. The thickness of the film must be 5/2 times the shorter wavelength and 4/2 = 2 times the longer wavelength. Using your correct adjustment for the index of refraction of the film, this gives

[tex]L = 5\lambda_1/2 = 4\lambda_2/2 = 5*376 nm/2 = 4*471 nm/2 = 941 nm[/tex]

So, it appears there is no solution to problem as stated, and if you try to salvage the problem by changing the index of refraction of the bottoom layer to less than 1.36, you get a different answer from the one given.

Have I gone wrong somewhere?
 
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  • #6
OlderDan said:
So, it appears there is no solution to problem as stated, and if you try to salvage the problem by changing the index of refraction of the bottoom layer to less than 1.36, you get a different answer from the one given.

Have I gone wrong somewhere?
I agree with you. On the other hand, if the wavelength was at a maximum at 640 and a minimum at 512, it would work. If the question said: "at a maximum at wavelength = 640nm", the answer would be d=471 nm.

AM
 
  • #7
Andrew Mason said:
I agree with you. On the other hand, if the wavelength was at a maximum at 640 and a minimum at 512, it would work. If the question said: "at a maximum at wavelength = 640nm", the answer would be d=471 nm.

AM

Yes. Maybe that's what it was supposed to say. Nice catch.
 
  • #8
thin coating (t=340) is placed on glass .which visible wavelength
(400 <wavelength<700 nm )
will be absent in the reflected beam if the glass has an index of refraction n =1.35
 

Related to Thin Film Interference Problems

1. What is thin film interference?

Thin film interference is a phenomenon that occurs when light waves reflect off the top and bottom surfaces of a thin film, causing them to interfere with each other. This can result in the appearance of colors or a change in the intensity of the reflected light.

2. How does the thickness of the film affect interference?

The thickness of the film plays a crucial role in thin film interference. When the thickness is equal to a certain fraction of the wavelength of light, constructive interference occurs and the reflected light is more intense. When the thickness is different, destructive interference occurs and the reflected light is less intense.

3. What materials are commonly used in thin film interference experiments?

Transparent materials such as glass, water, and air are commonly used in thin film interference experiments. These materials allow light to pass through and create the interference patterns. Other materials such as soap films and thin metal sheets can also be used.

4. How is thin film interference used in real-world applications?

Thin film interference has many practical applications, such as in anti-reflective coatings for glasses and camera lenses, in the production of optical filters, and in the technology of liquid crystal displays (LCDs). It is also used in the study of thin films and their properties in materials science and engineering.

5. What factors can affect the interference pattern in thin film interference?

The interference pattern in thin film interference can be affected by factors such as the angle of incidence of the light, the refractive index of the film and surrounding medium, and the wavelength of the incident light. Temperature and humidity can also play a role in altering the interference pattern. Additionally, imperfections or irregularities in the film can cause deviations from the expected interference pattern.

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