Optical Transmission through a thin Film

In summary, the conversation discusses a problem in optics where an expression needs to be evaluated to calculate the intensity of light transmitted through a film after multiple reflections. The expression involves a geometric series and the solution involves finding the sum S and multiplying it by its complex conjugate to get the absolute square, which then simplifies to the book's solution. There are also discussions about only considering the imaginary part of the series and the use of complex numbers.
  • #1
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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:

[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).

Homework 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.

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.

Thanks,
Chris Maness
 
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  • #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.
 
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  • #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:

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

Thanks,
Chris Maness
 
  • #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.
 
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  • #5
Ahhhhh! Because I need [tex]S^{2}[/tex] not just [tex]S[/tex]. This makes sense. Thank you.

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

Related to Optical Transmission through a thin Film

What is optical transmission through a thin film?

Optical transmission through a thin film refers to the process of light passing through a thin layer of material. The material can be transparent, translucent, or opaque, and the light can either be reflected, absorbed, or transmitted through the film.

What factors affect optical transmission through a thin film?

The factors that affect optical transmission through a thin film include the thickness and composition of the film, the angle of incidence of the light, and the refractive index of the film and the surrounding medium. These factors can either enhance or hinder the transmission of light through the film.

How is optical transmission through a thin film measured?

Optical transmission through a thin film is typically measured using a spectrophotometer, which measures the amount of light that passes through the film at different wavelengths. This allows for the determination of the film's transmittance and absorbance properties.

What are the applications of optical transmission through a thin film?

Optical transmission through a thin film has numerous applications in various fields, such as optics, electronics, and photonics. It is used in the production of anti-reflective coatings, optical filters, and photovoltaic cells, among others.

How can the transmission through a thin film be optimized?

The transmission through a thin film can be optimized by adjusting the thickness and composition of the film, as well as the angle of incidence of the light. Additionally, using multiple layers of thin films with different refractive indices can also enhance the transmission of light through the film.

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