Thin-Film Interference

1. Apr 9, 2006

BoogieL80

I feel really lost in this interference stuff. I'm also having trouble with the following problem:

A mixture of yellow light (wavelength = 584 nm in vacuum) and violet light (wavelength = 408 nm in vacuum) falls perpendicularly on a film of gasoline that is floating on a puddle of water. For both wavelengths, the refractive index of gasoline is n = 1.40 and that of water is n = 1.33. What is the minimum nonzero thickness of the film in a spot that looks the following colors because of destructive interference?

What makes matters worse for this problem is there is even an example of this type of problem in my book. However, I used the formula wavelengthfilm = wavelengthvacuum / n and figured out that the wavelength of yellow in the gasoline is 417 nm and for violet is 291 nm. The problem also said that the color is caused by destructive interference. So in my mind this would mean that 2t + 1/2 wavelengthfilm = conditions for destructive wavelenth. I figured it was 2t since wave 2 traveled extra distance. In the end this would mean that for a miniumum nonzero thickness of film my formula would equal t= 1/2 * wavelength film. However webbassign is saying my answers are incorrect. Any help would be appreciated.

2. Apr 9, 2006

lightgrav

2t = half wavelength IS the condition for destructive interference here.
(there is no extra 1/2 wavelength at reflections from a faster material)

But then t = 1/4 wavelength , for the color that is NOT seen.

Last edited: Apr 9, 2006
3. Apr 9, 2006

BoogieL80

So I guess I'm suppose to assume that all of the other colors show up due to constructive interference?

4. Apr 9, 2006

lightgrav

?? In this problem, there's Yellow light and Violet light.

Sorry, I mis-read the question earlier ...

The TOP reflection (air-gasoline) is from material in which the wave is slower.
So it has an extra half-wavelength (Electric field is flipped on reflection)
But light is faster in the water, so the bottom reflection (gasoline-water)
does NOT have an extra half-wavelength.

The total path difference is therefore 2t - lambda/2 ...
which equals lambda/2 for destructive interference.

Sorry again for my mis-reading of the gasoline n .

5. Apr 10, 2006

BoogieL80

Well that's kind of where I was before. I calculated, for example, the wavelength for yellow to be 417nm (584nm/1.40 = 417nm). I tried diving that wavelength by 2 and got 208nm. However, my webassign is saying that is incorrect....