What Color Will a 320nm Soap Film Appear in Reflected Light?

[revacious]

Interference with thin film :(

Homework Statement

A soap film in air is 320nm. thick. If it is illuminated with white light at normal incidence, what colour will it appear to be in reflected light (assume that the index of the soap film is 1.33).

Homework Equations

The condition for destructive interference is given by:
2tn_f = (m + ½ )λ

The condition for constructive interference is given by:
2tn_f = mλ

The Attempt at a Solution

Well... the light is going from air (n=1) to film (n_f=1.33) so there will be a phase change of Pi/2? Not really sure how to start to be honest.

Sorry for amateur question!

 

[Doc Al]

Well... the light is going from air (n=1) to film (n_f=1.33) so there will be a phase change of Pi/2?

For the reflection at the outer air/film boundary there will be a phase change of pi. Revise your equations accordingly.

 

[revacious]

Well, I got the answer by formula monkeying it and subbing m=1 into the destructive interference equation. But the real problem is I have no idea what I'm doing even though I can get the answer.

My interpretation of the question: When white light (mixture of wavelengths) strikes the air film barrier, some white light is reflected and some is transmitted. The light that is transmitted undergoes a phase change of Pi/2 (??), as it is going from a rarer to a denser medium. After traversing a thickness t, it then strikes the film air barrier (without undergoing a phase change) and is reflected back to the 1st surface of the film.

The optical path taken for the transmitted --> reflected ray would then be 2*t*n_f. This would make a phase difference of 2tn_f(2Pi/λ).

The extent to which the 2nd reflected ray phase differs from the reflected ray determines whether it is constructive or non constructive. If it is destructive, the phases differ by Pi/2 + 2NPi where N is an integer. If it is constructive, the phases differ by 0 + 2NPi.

I still do not know how to do the question though lol.

 

[Doc Al]

Well, I got the answer by formula monkeying it and subbing m=1 into the destructive interference equation.

What you are calling the 'destructive interference equation' only gives destructive interference if there's no phase change upon reflection to worry about. Since we have a phase change at the first surface, that equation actually states the condition for constructive interference (which is what we need here).

My interpretation of the question: When white light (mixture of wavelengths) strikes the air film barrier, some white light is reflected and some is transmitted.

OK.

The light that is transmitted undergoes a phase change of Pi/2 (??), as it is going from a rarer to a denser medium.

The light that is reflected at the first boundary will have a phase shift of pi (λ/2) not pi/2.

After traversing a thickness t, it then strikes the film air barrier (without undergoing a phase change) and is reflected back to the 1st surface of the film.

You have light reflected at each surface. The light reflected from the first surface has a phase shift; light reflected at the second surface does not. Compare the phase difference of these two reflections to find what gives constructive interference.

 

[revacious]

thanks for clearing those things up!

i will try to understand again tomorrow morning; I am falling asleep at the books here! hopefully it will come to me in a dream. But it probably won't and you will probably find me posting again in a few hours.

brb!

thanks for clearing those things up!

i will try to understand again tomorrow morning; I am falling asleep at the books here! hopefully it will come to me in a dream. But it probably won't and you will probably find me posting again in a few hours.

brb!

OK. I think I made some progress understanding-wise...

The second reflected ray travels an extra optical pathlength of 2tn_f. So the difference in phase (Δa) between the second ray and the first ray is 2tn_f(2Pi/λ) + Pi, accounting for the phase change the first ray undergoes upon reflection.

The question asks us what light is reflected, meaning it is specifically related to constructive interference. So we assume that Δa = 0 + 2NPi where N is an integer.

Equating these two expressions for Δa, we arrive at:
2tn_f(2Pi/λ)=2NPi
2tn_f=(N-½)λ

But now what do I do? I can sub the values in for t and n, but how do i choose the value for N?

 

[revacious]

just realized that edited posts are not bumped. Sorry for this double post!

 

[Doc Al]

But now what do I do? I can sub the values in for t and n, but how do i choose the value for N?

Just start with the lowest value and crank them out one by one. You're looking for something in the visible part of the spectrum.

 

[revacious]

Ok so N=0 churns out a negative wavelength which is clearly wrong
N=1 churns out 1702 nm
N=2 churns out 567nm (the answer is 570 so this is right!)
N=3 churns out 340nm which is apparently in the ultraviolet region
...

I will choose the one which is in the visible spectrum, but will there ever be two in the visible spectrum? If this is the case, would the colour appear to us as a mixture of wavelengths?

What is the significance of the fact that N can take infinite values? Like does this imply that there is also an ultraviolet component reflected at me as well as ... gamma rays, etc?


thanks for your guidance :)

 

[Doc Al]

Ok so N=0 churns out a negative wavelength which is clearly wrong
N=1 churns out 1702 nm
N=2 churns out 567nm (the answer is 570 so this is right!)
N=3 churns out 340nm which is apparently in the ultraviolet region
...

Good! (N=1 is the lowest value allowable.)

I will choose the one which is in the visible spectrum, but will there ever be two in the visible spectrum? If this is the case, would the colour appear to us as a mixture of wavelengths?

I don't see any reason why there couldn't be two visible wavelengths under certain conditions. And yes, in that case the observed color would be a mixture of the two.

What is the significance of the fact that N can take infinite values? Like does this imply that there is also an ultraviolet component reflected at me as well as ... gamma rays, etc?

Don't take all those values of N too seriously. The given value for the index of refraction would only be relevant for the visible spectrum, most likely. And when you're talking about gamma rays, lots of other issues come into play. So just stick with the visible for this exercise.

 

[revacious]

thanks very much for your help. hopefully this comes up in my exam tomorrow lol.