Optics and thin film interference question.

[x12179x]

Homework Statement

The thin film is of two glass with air in the middle.
refractive index of...
air: n=1.0003
glass: n=1.5
lamba=632.8nm

Q1.) Calculate optimum thickness of the film (for 1st order interference)

Q2.) Why does the distance between interference fringe beome longer (increased fringe spacing) when the film thickness is reduced?

Homework Equations

We aren't given any, we have to find them ourselves.
I found:
2t=m(lamba/n)
and 2t=(m+1/2)(lambda/n)
one is constructive one is destructive, not sure which is which for this case.
not sure if they are even relevant here.

The Attempt at a Solution

Q1.) I found: t=m(lambda) / 2n
and used that to get 3.19 x10^-7 m
Is that correct?

Q2.) I drew a diagram that explains it, and I think it has to do with missing fringes in between due to destructive interference, however I don't have any equation that shows that.
I think this is relevant: 2t=m(lamba/n)
I think that's for constructive interference. t is directly proportional to m (# of visible fringes). When t is decreased, the # of fringes decrease. Thus, providing empty spaces between fringes. But how do I prove that it's every other fringe that disappears? Is there any equation that shows thickness is inversely proportional to fringe spacing?

 

[anathema]

Hello there,

Thank you for your post. Let's go through the questions one by one.

Q1. Yes, your calculation for the optimum thickness of the film is correct. The formula you used, t=m(lambda)/2n, is for constructive interference. This means that for the first order interference (m=1), the thickness of the film should be equal to 1/4 of the wavelength of light in the film, which is 3.19 x 10^-7 m.

Q2. Your explanation for why the fringe spacing increases when the film thickness is reduced is correct. As you mentioned, this is due to the destructive interference between the light waves reflected from the top and bottom surfaces of the film. The equation you are looking for is the one you already mentioned, 2t=m(lambda)/n. This is the equation for destructive interference. As you can see, the thickness of the film is directly proportional to the number of fringes, which means that when the thickness is reduced, the number of fringes decreases, resulting in increased fringe spacing.

To prove that it is every other fringe that disappears, you can use the equation for constructive interference, t=(m+1/2)(lambda)/n. This equation tells us that when the thickness of the film is increased by half a wavelength, the number of fringes increases by one. So, for every increase of half a wavelength in film thickness, one fringe appears. This means that for every decrease of half a wavelength in film thickness, one fringe disappears. Since the thickness of the film is directly proportional to the number of fringes, this means that every other fringe will disappear.

I hope this helps. Let me know if you have any further questions.

 

[nlightner]

I would like to commend you for your efforts in finding relevant equations and attempting to explain the phenomenon using a diagram. Your calculation for the optimum thickness of the film appears to be correct, and your reasoning for why the fringe spacing increases with decreasing film thickness is on the right track. However, I would like to clarify a few points and provide some additional information that may help you in your understanding.

Firstly, the equations you have found are indeed relevant to this situation. The equation 2t=m(lambda)/n represents the constructive interference condition, where t is the thickness of the film, m is the order of the interference (1st, 2nd, 3rd, etc.), lambda is the wavelength of the incident light, and n is the refractive index of the film. This equation tells us the thickness of the film required for constructive interference to occur at a given order.

The other equation, 2t=(m+1/2)(lambda)/n, represents the destructive interference condition, where the film thickness is half a wavelength different from the constructive interference condition. This results in destructive interference, causing the fringes to disappear. Therefore, for the 1st order interference, the film thickness would be given by t=(1/2)(lambda)/n.

Now, to address your second question, the reason for the increased fringe spacing with decreasing film thickness has to do with the path difference between the two beams of light that are interfering. When the film thickness is increased, the path difference between the two beams also increases, resulting in a larger fringe spacing. On the other hand, when the film thickness is decreased, the path difference decreases, resulting in a smaller fringe spacing.

To prove that it is every other fringe that disappears, you can use the equation t=(1/2)(lambda)/n to calculate the thickness of the film for the 1st, 2nd, 3rd, etc. orders of interference. You will notice that for every even order (2nd, 4th, 6th, etc.), the thickness of the film will be the same as the thickness for the previous odd order (1st, 3rd, 5th, etc.). This means that the path difference between the two beams of light will be the same for these orders, resulting in destructive interference and the disappearance of fringes.

In summary, the increased fringe spacing with decreasing film thickness is due to the decreasing path