# Thin Film Interference Help?

1. Jan 3, 2009

### Nivlac2425

I don't have a particular homework inquiry to ask, just some questions I have about thin-film interference and its related equations. I was told by my instructor to learn these concepts on my own and I am just having a few problems I want to clarify.

1) Equations related to thin-film interference
The only equation given about this concept is one relating wavelengths of light in vacuum and in film; $$\lambda$$ film = ($$\lambda$$ vacuum) / n
There aren't any others explicitly stated, but the examples uses some parts of some equations dealing with the previous section which is about Young's double-slit experiment. I believe the examples do not explain how the equations were manipulated or used. These equations are for constructive and destructive interference and use m, which is the order of the bright fringe, and d, the distance between the slits:
sin$$\theta$$ = m($$\lambda$$/d); m=0,1,2,3....
sin$$\theta$$ = [m+(1/2)] ($$\lambda$$/d); m=0,1,2,3....

In one specific example, a typical thin-film interface is presented where white light reflects off the film and yellow light is observed. In the interface, soap is surrounded by air. Destructive interference is said to occur, eliminating the blue color. A minimum thickness is asked for.
In presenting a solution and after calculating in-film wavelengths, the book states that the condition for destructive interference must be:
(m+(1/2))($$\lambda$$film) = the sum of the extra distance travelled by wave(2) and the net phase change of the waves
where the extra distance travelled by wave(2) is denoted as 2t, t being the thickness; and where the net phase change is (1/2)($$\lambda$$film)

My questions so far are, how was that equation produced and how is the value of m decided, this being destructive interference and when minimum thickness is asked for? (The book chooses m=1 for this purpose)

2) More Thin-Film Interference
Some problems given for exercise dealing with thin-film interference have a mixture of light reflected off the film, and therefore two wavelengths are given. Again, a minimum thickness is asked for where destructive/constructive interference occurs for both wavelengths.
My question is how do I use both wavelengths to find a common thickness at where interference is occuring?
I'm guessing if my questions in 1) were answered, I'd be able to figure 2) out on my own.

So those are my main concerns for now, and I thank all in advance who try their best to help me clarify these points!!!
Thank you!

2. Jan 3, 2009

### jambaugh

The key to the thin film problem is understanding that light will reflect both off of the top and bottom layer of the film. These two reflected waves may destructively or constructively interfere depending on the extra distance the bottom reflected wave must travel. In particular if that distance is an odd half multiple of the wavelength (in the medium) then you get destructive interference and if it is an even half multiple you get constructive interference.

Beware! This assumes the thin film is laying on some higher index medium (and that it has a higher index than air.)

There is another issue of the 180deg phase-shift when light is reflected.

If the above assumption is made then this won't matter since both reflected waves have the same shift upon reflection. However if you are considering say a soap bubble then only one of the two reflected waves will be shifted and the cases of constructive vs. destructive will be reversed.

I think this covers all of the conceptual basics.

Last edited by a moderator: Apr 24, 2017
3. Jan 3, 2009

### Nivlac2425

Yes, I understand those basic concepts, but I am confused on how the equations were produced for the condition of destructive interference. Is it just common sense in relating the phase changes and such, or is there something else?
I believe my confusion arises from the presence of m, and its usage.

4. Jan 4, 2009

### jambaugh

The m is included because of the periodicity of the wave. The thickness of the film can be 28.5 wavelengths and you get the same effect as if it were 0.5 wavelengths.

Yes it is just common sense about the phase. In detail you are adding two sine functions. In the case of destructive interference:
$$A\sin(\phi) + A\sin(\phi + 180^o) = A\sin(\phi)-A\sin(\phi) = 0$$
or possibly
$$A\sin(\phi) + A\sin(\phi + 43\cdot180^o) = A\sin(\phi)-A\sin(\phi) = 0$$
or any other odd half cycle added in. Similarly with even multiple of half cycles i.e. multiples of full cycles you are adding the same numbers instead of negatives of each other and the waves' amplitudes reinforce.

Here $\phi$ is the phase of the wave...(this is not to be confused with the angle off of the double slit which you wrote as $\theta$.

Let's see... $\phi$ will be something like $\phi=\omega t - x/\lambda$.

5. Jan 4, 2009

### Nivlac2425

Great! Thank you for clearing up my confusion! I can't believe I let something as simple as these concepts slow my progress...

So with mixtures of light, the only wavelength of light that would need to be considered would be the one on which interference took place. I believe that answers my second question.

Thanks a lot!