# Homework Help: Optical Transmission through a thin Film

1. Apr 2, 2014

### kq6up

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
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).

2. Relevant 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.

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 $$S=\frac{1}{1-r^{2}sin^{2}\theta}$$ 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

Last edited: Apr 2, 2014
2. Apr 2, 2014

### 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.

3. Apr 3, 2014

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

Thanks,
Chris Maness

4. Apr 3, 2014

### 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.

5. Apr 3, 2014

### kq6up

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

Regards,
Chris Maness

6. Oct 2, 2016

### notnerd

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