Effective refractive index of a stratified medium

  • #1
Hello PF,
I'm reading a paper for a project. In the paper they derive an equation for the effective refractive index ##n=\sqrt{\epsilon^{e} \mu^{e}}## of two stacked layers ##(n_1^2 = \epsilon_1 \mu_1, a)## and ##(n_2^2 = \epsilon_2 \mu_2, b)## where ##a,b## are the lengths and in my case ##\mu_1=\mu_2=1##. One material is silica glass the other is air.

So using Maxwell's equations they derive this equation for ##n##: $$\frac{\alpha_2}{\mu_2} \tan\left(\frac{b\alpha_2}{2}\right)=\frac{\alpha_1}{\mu_1} \tan\left(\frac{a\alpha_1}{2}\right)$$
where ##\alpha_i=k\sqrt{n_i^2-n^2}=k\sqrt{\epsilon_i-\epsilon^{e}}##.
(I have attached a picture showing the setting used for the derivation.)

But since it can not be solved analytically they do the expansion ##tan(x) \approx x + \frac{1}{3}x^3## and get this: $$ \epsilon^e = \frac{a\epsilon_1+b\epsilon_2}{a+b} + \frac{k^2b^2a^2}{12(a+b)^2}(\epsilon_1-\epsilon_2)^2$$

Homework Statement


I'm trying to retrace the steps to get the equation for ##\epsilon^{e}## to then be able to expand it to third order. ##(\tan(x) \approx x+\frac{1}{3} x^3 + \frac{2}{15}x^5)##

Homework Equations


##\epsilon^{e}_{1,2} = \frac{-a_1\pm\sqrt D}{2a_0}, D=a_1^2-4a_0a_2##

The Attempt at a Solution


When I do the second order expansion, as in the paper, I get the coefficients for the equation ##a_0x^2+a_1x+a_2=0## to be:
##a_0 = \frac{k^2 (b^3+a^3)}{12}##
##a_1 = -(b+a) - \frac{1}{12} 2k^2 (\epsilon_2 b^3+\epsilon_1 a^3)##
## a_2 = \epsilon_1 a + \epsilon_2 b + \frac{1}{12} k^2 (\epsilon_1^2 a^3 + \epsilon_2^2 b^3)##

But when I put in the values from the experiment:
## a = 92\cdot10^{-6} m , b = 58\cdot10^{-6} m, 1400 GHz, \epsilon_1 = 1 F/m, \epsilon_2 = 3.75 F/m## and solve the equation for ##\epsilon^{e}_{1,2}##.
I am not getting the same result as if I would use the equation derived in the paper. And I can't seem to figure out how to reduce the expression, I get when substituting the coefficients into ##\epsilon^{e}_{1,2} = \frac{-a_1\pm\sqrt D}{2a_0}, D=a_1^2-4a_0a_2##, to the equation given in the paper.

I have attached the paper as well in case there is something missing in my post.
To summarize: How is the equation for ## \epsilon^e ## derived? Because I seem to be getting wrong results :olduhh:
 

Attachments

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,737
279
Hello PF,
I'm reading a paper for a project. In the paper they derive an equation for the effective refractive index ##n=\sqrt{\epsilon^{e} \mu^{e}}## of two stacked layers ##(n_1^2 = \epsilon_1 \mu_1, a)## and ##(n_2^2 = \epsilon_2 \mu_2, b)## where ##a,b## are the lengths and in my case ##\mu_1=\mu_2=1##. One material is silica glass the other is air.

So using Maxwell's equations they derive this equation for ##n##: $$\frac{\alpha_2}{\mu_2} \tan\left(\frac{b\alpha_2}{2}\right)=\frac{\alpha_1}{\mu_1} \tan\left(\frac{a\alpha_1}{2}\right)$$
where ##\alpha_i=k\sqrt{n_i^2-n^2}=k\sqrt{\epsilon_i-\epsilon^{e}}##.
(I have attached a picture showing the setting used for the derivation.)

But since it can not be solved analytically they do the expansion ##tan(x) \approx x + \frac{1}{3}x^3## and get this: $$ \epsilon^e = \frac{a\epsilon_1+b\epsilon_2}{a+b} + \frac{k^2b^2a^2}{12(a+b)^2}(\epsilon_1-\epsilon_2)^2$$

Homework Statement


I'm trying to retrace the steps to get the equation for ##\epsilon^{e}## to then be able to expand it to third order. ##(\tan(x) \approx x+\frac{1}{3} x^3 + \frac{2}{15}x^5)##

Homework Equations


##\epsilon^{e}_{1,2} = \frac{-a_1\pm\sqrt D}{2a_0}, D=a_1^2-4a_0a_2##

The Attempt at a Solution


When I do the second order expansion, as in the paper, I get the coefficients for the equation ##a_0x^2+a_1x+a_2=0## to be:
##a_0 = \frac{k^2 (b^3+a^3)}{12}##
##a_1 = -(b+a) - \frac{1}{12} 2k^2 (\epsilon_2 b^3+\epsilon_1 a^3)##
## a_2 = \epsilon_1 a + \epsilon_2 b + \frac{1}{12} k^2 (\epsilon_1^2 a^3 + \epsilon_2^2 b^3)##
I am trying to follow your calculation but am having a hard time. What does your x stand for? You talk about Taylor expanding the tangent but the arguments on both tangent are different so I don't know what your x stands for here.
 
  • Like
Likes AwesomeTrains
  • #3
I am trying to follow your calculation but am having a hard time. What does your x stand for? You talk about Taylor expanding the tangent but the arguments on both tangent are different so I don't know what your x stands for here.
Here's my derivation: (I missed the minus sign in the original post)
##
\frac{\alpha_2}{\mu_2} \tan\left(\frac{b\alpha_2}{2}\right)=-\frac{\alpha_1}{\mu_1} \tan\left(\frac{a\alpha_1}{2}\right)
##
First I expand the tangents on both sides:
##
\frac{\alpha_2}{\mu_2}\left(\frac{b\alpha_2}{2}+\frac{1}{3}\left(\frac{b\alpha_2}{2}\right)^3\right)=-\frac{\alpha_1}{\mu_1}\left(\frac{a\alpha_1}{2}+\frac{1}{3}\left(\frac{a\alpha_1}{2}\right)^3\right)
##
Using the definition of ##\alpha_i=k\sqrt{\epsilon_i-n^2}=k\sqrt{\epsilon_i-\epsilon^{e}}## from the paper and setting ##\mu_1=\mu_2=1##:
##
b(\epsilon_2 -n^2) + \frac{1}{12} k^2b^3(\epsilon_2-n^2)^2=-a(\epsilon_1-n^2)-\frac{1}{12}k^2(\epsilon_1-n^2)^2a^3
##
Since ##n^2=\epsilon^e## and after ordering the terms according to the exponent of ##\epsilon^e## I get:
##
\frac{1}{12}k^2 (b^3+a^3){\epsilon^e }^2
-(b+a + \frac{1}{12} 2k^2 (\epsilon_2 b^3+\epsilon_1 a^3) )\epsilon^e +
(\epsilon_1 a + \epsilon_2 b + \frac{1}{12} k^2 (\epsilon_1^2 a^3 + \epsilon_2^2 b^3)) = 0
##

According to the section I attached a screenshot of from the paper this should be how the derivation is done.

But when I then substitute the values given in the OP and solve for ##\epsilon^e## I am not getting the same result as if I substitute them into ##
\epsilon^e = \frac{a\epsilon_1+b\epsilon_2}{a+b} + \frac{k^2b^2a^2}{12(a+b)^2}(\epsilon_1-\epsilon_2)^2
##
 

Attachments

  • #4
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,975
3,130
I don't think you should expand only the tangent. You have expressions that look like ##\alpha_i \tan\left(\frac{b\alpha_i}{2}\right)##. You should be expanding ##x\tan(cx)##. Note that this is an even function, so there are no odd order terms. It might be easier to do it this way and go up to sixth order for what you want to do.

On edit: I redid the algebra and I agree with your coefficients ##a_0,~a_1~##and ##a_2##. Now as you know these are the coefficients of a quadratic. The solution that is presented is probably an additional approximation to the exact solution of the quadratic because the quadratic does not look like a perfect square, so where is the radical? However, I am not sure I understand the physics of the situation well enough to figure out what the appropriate approximation is.
 
Last edited:
  • Like
Likes AwesomeTrains
  • #5
TSny
Homework Helper
Gold Member
12,960
3,315
##
\frac{1}{12}k^2 (b^3+a^3){\epsilon^e }^2
-(b+a + \frac{1}{12} 2k^2 (\epsilon_2 b^3+\epsilon_1 a^3) )\epsilon^e +
(\epsilon_1 a + \epsilon_2 b + \frac{1}{12} k^2 (\epsilon_1^2 a^3 + \epsilon_2^2 b^3)) = 0
##

##
\epsilon^e = \frac{a\epsilon_1+b\epsilon_2}{a+b} + \frac{k^2b^2a^2}{12(a+b)^2}(\epsilon_1-\epsilon_2)^2
##
These equations are approximations that hold only if ##|\alpha_1 a| \ll 1## and ##|\alpha_2 b| \ll 1##. See the comments just above equation (20) in the paper. But your data doesn't appear to meet these conditions.
 
  • Like
Likes AwesomeTrains
  • #6
I ended up solving the equation numerically which gave me a pretty good fit to the measured values. The numerical solution works until approximately 500 GHz which is where the conditions from post #5 aren't holding anymore, so that is probably the explanation and then expanding to third order would probably not improve the result.
Anyways thanks for the help :approve:
 

Related Threads on Effective refractive index of a stratified medium

  • Last Post
Replies
2
Views
3K
Replies
4
Views
6K
Replies
1
Views
2K
  • Last Post
Replies
9
Views
8K
Replies
8
Views
2K
Replies
4
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
1
Views
603
  • Last Post
Replies
3
Views
5K
Top