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Homework Help: Optical Transmission through a thin Film

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data

    In optics, the following expression needs to be evaluated in calculating the intensity
    of light transmitted through a film after multiple reflections at the surfaces of the

    [tex]{\sum _{ n=0 }^{ \infty }{ { r }^{ 2n } } cos\quad n\theta })^{ 2 }+{ \sum _{ n=0 }^{ \infty }{ { r }^{ 2n } } sin\quad n\theta })^{ 2 }[/tex]
    Show that this is equal to [tex]{ \left| \sum _{ n=0 }^{ \infty }{ { r }^{ 2n } } { e }^{ in\theta } \right| }^{ 2 } [/tex] and so evaluate it assuming |r| < 1 (r is
    the fraction of light reflected each time).

    2. Relevant equations

    It looks like geometric series to me, so [tex]S=\frac{a}{1-r}[/tex] where S is the sum of the and r is some decimal number less than one.

    3. The attempt at a solution

    The text says the trick is to use only the imaginary part of the series (which is sign). I get a different answer than the book. I get [tex]S=\frac{1}{1-r^{2}sin^{2}\theta}[/tex] since I let replace r with the number that is in the geometric series. That is [tex]r^{2}e^{i\theta}[/tex]. The book's solution is [tex](1+r^{4}-2r^{2}cos\theta)^{-1}[/tex]. Not sure how they got that.

    Chris Maness
    Last edited: Apr 2, 2014
  2. jcsd
  3. Apr 2, 2014 #2
    The book's solution is right. Find the sum S. S will have a complex exponential in the denominator. Multiply it by its complex conjugate and simplify. There will be a term that you can identify as twice the cosine.
  4. Apr 3, 2014 #3
    Working on it. I see that multiplying the the denominator by the conjugate gives me the book's answer. However, I don't understand why I would only be multiplying the denominator by the conjugate and not the numerator too. If I multiply by 1 using a ratio of the complex conjugate, I then get that term in the numerator, and it is no longer one. According to the text I am actually only interested in the imaginary part of the sum. Then I would expect that answer to be [tex]\frac{r^{2}sin\theta}{(1+r^{4}-2r^{2}cos\theta)}[/tex]

    Follow this link for a JPG of the textbook page:


    Chris Maness
  5. Apr 3, 2014 #4
    Don't multiply by one. Multiply by the complex conjugate. Maybe you should look up what the absolute square of a complex number is.
  6. Apr 3, 2014 #5
    Ahhhhh! Because I need [tex]S^{2}[/tex] not just [tex]S[/tex]. This makes sense. Thank you.

    Chris Maness
  7. Oct 2, 2016 #6
    I didn't get the idea of the solution very well.. can anybody explain it for me please?
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