Archived Topics / thin film question

mm2424

1. Homework Statement
Two rectangular plates (n = 1.60) are in contact along one edge and separated along the opposite edge (see figure). Light with a wavelength of 600 nm is incident perpendicularly onto the top plate. The air between the plates acts as a thin film. Nine dark fringes and eight bright fringes are observed from above the top plate. If the distance between the two plates along the separated edges is increased by 600 nm, how many dark fringes will there then be across the top plate?

2. Homework Equations

For dark fringes, 2L = mλ/n

3. The Attempt at a Solution

I figured if I plugged in the highest order of dark fringe (m=9) and plugged it into 2L = mλ/n, I would find the initial thickness of air film between the plates. I assumed we should use n = 1, since the film is air.

I then got 2L = 9(600nm)/1
L = 2700 nm

I then added 600 nm to this and got 3300 nm as the new thickness of the air film.

2(3300 nm) = mλ/n
m = 11.

Is this general approach ok? In my professor's answer key, he used the same formula but solved for L initially by plugging in m = 8. This yielded an L of 2400 nm. He then added 600 to this, getting 3000 nm, which he plugged back into the equation to get an m = 10. He reasoned from here (without any explanation why) that the number of dark fringes increased from 9 to 11. Is there some type of logical step here I'm not getting? Can I use my approach, or is there something to his approach of using m = 8 initially?

Thanks!

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julianty

I'm taking a guess, but if you're using Halliday & Resnick Funadmentals of Physics 10th Ed, p1067 tells you why you would use m = 8.

Basically, for thickness L < 0.1λ, the phase difference is only due to phase shifts in reflections (the path length difference is ignored). So if you have one phase shifting reflection (as in this problem) the first dark fringe is m = 0.

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