Optical Transmission through a thin Film

[kq6up]

Homework Statement

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
film:

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

Homework Equations

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

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 S=\frac{1}{1-r^{2}sin^{2}\theta} since I let replace r with the number that is in the geometric series. That is r^{2}e^{i\theta}. The book's solution is (1+r^{4}-2r^{2}cos\theta)^{-1}. Not sure how they got that.

Thanks,
Chris Maness

 

[tman12321]

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.

 

[kq6up]

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 \frac{r^{2}sin\theta}{(1+r^{4}-2r^{2}cos\theta)}

Follow this link for a JPG of the textbook page:

https://docs.google.com/file/d/0B3e_cmEDZ8qaREgzTTV6WU8zaEI4QW1xOWl2RkwwSVRtNWxn/edit

Thanks,
Chris Maness

 

[tman12321]

Don't multiply by one. Multiply by the complex conjugate. Maybe you should look up what the absolute square of a complex number is.

 

[kq6up]

Ahhhhh! Because I need S^{2} not just S. This makes sense. Thank you.

Regards,
Chris Maness

 

[notnerd]

I didn't get the idea of the solution very well.. can anybody explain it for me please?