Home > 0877 > GR0877 Problem 75

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).

Found a typo? Comment!

Jump to the problem

1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100

Want to get a full PDF with solutions? Read THIS.

The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Advertisements
Categories: 0877 Tags: , , , ,
  1. betelguese05
    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. betelguese05
    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. betelguese05
    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. Jacob
    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?

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: