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## GR0877 Problem 75

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) The shift for the wave reflecting off the top surface of the film is $\Delta_{top} = \lambda/2$ (since the light is reflecting from a higher-$n$ medium). The shift for the wave reflecting off the bottom surface of the film is $2 \,t$, where $t$ is the thickness of the film. Thus, $\Delta = \Delta_{bottom} - \Delta_{top} = 2t- \lambda/2$. For constructive interference one has $2t-\lambda/2 = m \lambda$. But $\lambda$ here is the wavelength in the medium, so we must replace it with $\lambda/n$: $\displaystyle 2t - \frac{\lambda}{2n} = m \frac{\lambda}{n}$ or $4t = \frac{\lambda}{n} (2m +1)$. The film first appears bright for $\lambda$ and $m = 0$ $\implies$ $\displaystyle 4t = \frac{\lambda}{n}$. Then it appears bright for $\lambda'$ and $m = 1$ $\implies$ $\displaystyle 4t = \frac{3 \lambda'}{n}$ and $\lambda' = \lambda/3 = 540/3 = 180$ (nm).

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Categories: 0877 Tags: , , , ,
1. 07.10.2012 at 04:22

Your reasoning, I’m pretty sure, is wrong, but it (happily) gets you the right answer. You’re measuring $\lambda$ in air, so we clearly don’t need to replace $\lambda$ by $\lambda/n$ everywhere. However, the wavelength shift going through the medium is not simply 2t, since the wavelength is longer in this medium. Here, $\lambda$ goes to $\lambda/n$, so you’re covering an effective distance (if $\lambda$ was the same everywhere) of $t n$, so you should in fact replace $2 t$ by $2tn$. Now, if you divide through by $n$ you get your answer, so it’s only the reasoning that is incorrect (not the steps that follow from when you get $2t-\lambda/2n = m \lambda /n$.

• 07.10.2012 at 10:10

NO, EVERYTHING IS OK for one-film interference. When light propagates in the medium with $n>1$ its speed decreases to $v=c/n$ (its frequency is unchanged). We can handle this in two ways: we can say that the optical path $\Delta$ increases by $n$, i.e. $\Delta = \Delta_g n$, where $\Delta_g$ is geometrical path AND treat $\lambda$ as a wavelength _in vacuum_, OR we can say that $\lambda$ decreases by $n$, i.e. $\lambda = \lambda_{vacuum}/n$ and don’t bother with optical path. These are two possible ways to handle the same physical situation.
P.S. I’m NOT measuring $\lambda$ in the air, I’m measuring it INSIDE the MEDIUM, thus $\lambda = \lambda_{air}/n$.

2. 07.10.2012 at 10:23

I don’t think your situation is really equivalent; say that we have two films. Then we would have a thickness for each (call them $d$ and $t$) and we could not enter both into our equations as simply $d$ and $t$ respectively, as one would need to be scaled. So this technique of replacing $\lambda$ by $\lambda/n$ does not generalize, and for this reason I think it just happens to work for a one-film situation (because the equation allows that freedom) but is not true in general.

As a note, fixing $\lambda$ in air (which is how I imagine almost everyone is reading the problem anyway) and then using $n_1 d$ and $n_2 t$ will get the correct solution for the two-film situation, so my way does generalize.
_______________________
Also, am I doing something wrong with the latex? It keeps showing me single dollar signs (although my first post now looks correct).

• 07.10.2012 at 11:37

First, I didn’t say this is true for multiple-film interference, did I? What is more, multiple-film interference is far more complex than you imagined it (see M. Born, E. Wolf Optics, §1.6). For ONE-film interference my reasoning is correct.
Secondly, PLEASE don’t post multiple replies.
Thirdly, for latex support, see this http://en.support.wordpress.com/latex/ Your first post now looks correct because I fixed it.

3. 07.10.2012 at 19:08

Look, there’s no need to get mad here. I appreciate your solutions. But I’m still unconvinced by this problem’s solution, so by replying multiple times we can have a discussion. This isn’t spam or anything.

Final comment: Consider the light reflecting off the top of the film. It never enters the medium. Thus $\lambda$ is not in the medium, so how could you argue that we can replace it by $\lambda/n$? This still doesn’t make any sense. Since $\lambda$ exists both inside and outside of the film, only the pieces inside should get scaled.

4. 09.09.2016 at 00:29

I don’t understand why we know there’s a phase shift of lambda/2 for the upper reflective layer, but *not* one for the bottom layer. We get a 180 degree phase shift if we go from a low n to a higher n, but nothing is explicitly stated about the substance under the film– it could very well have n>1.3, so we could have a 180 degree shift on both waves. But we don’t. How do we know that?