In summary: B terms).In summary, the equation (1) is valid for all y if equation (2) is true. However, equation (2) is only true if y is measured relative to the interface between the media.
  • #1
EmilyRuck
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6
Hello!
This post is strictly related to my previous one. Let's consider the same context and the same image. Regarding the oblique incidence of a wave upon an interface between two dielectric, all the texts and all the lectures write an equation like the following:

[itex]e^{-j k_1 y \sin \theta_i} + \Gamma e^{-j k_1 y \sin \theta_r} = Te^{-j k_2 y \sin \theta_t}[/itex]
(1)

where [itex]k_1[/itex] is the wavenumber in the medium 1 (left) and [itex]k_2[/itex] is the wavenumber in the medium 2 (right).
The texts also say: if this relation has to be valid for all [itex]y[/itex], then the [itex]y[/itex] variation must be the same on all the terms and so

[itex]k_1 \sin \theta_i = k_1 \sin \theta_r = k_2 \sin \theta_t[/itex]
(2)

I know that this relation is true (it's Snell's law!) and it can be experimentally proved, but I have a doubt: is equation (2) really the only way to verify the equation (1) for all [itex]y[/itex]? I can't understand - neither physically nor mathematically - why.

Why is this solution the only acceptable one and not just a trivial one?
 
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  • #2
It is based on the orthogonality of the sine function.
## x r sin(\theta) = x t \sin(\phi) ## for all x, implies that t = r, since that is your magnitude, and ##\phi = \theta + k 2\pi ## where k is an integer. Other solutions might have t = -r, and phi an odd number of pis offset from theta, but since k_1, k_2 are restricted to positive numbers and theta_i, theta_r, theta_t restricted to [0,2pi], there is only one solution.
For more on orthogonality of sine ... check out this page on Fourier Series.
ref: http://mathworld.wolfram.com/FourierSeries.html
 
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  • #3
First of all thank you for your answer.
In the link you suggested, the othogonality seems to be just for sines of the form [itex]\sin (mx)[/itex] and [itex]\sin (nx)[/itex] with [itex]n \neq m[/itex] and so this situation is quite different; anyway, the sine functions [itex]k_1 y \sin \theta_i[/itex] and [itex]k_1 y \sin \theta_r[/itex] are actually equal only in the cases you listed.
What is still weird for me is that here, from a single equation (equation (1) of my post), multiple equations arise. Is it due to the features of the sine function, that imposes equality for both its amplitude and its phase, or what else?
 
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  • #4
Sorry I did not reply to this sooner. Let's look at this question in more detail.
You have:
##\forall y, \quad Ae^{-j k_1 y \sin \theta_i } + Be^{-j k_1 y \sin \theta_r}=Ce^{-j k_2 y \sin \theta_t} ##
Then, it must be true for y = 0.
So A + B = C. Which gives:
##\forall y, \quad Ae^{-j k_1 y \sin \theta_i } + Be^{-j k_1 y \sin \theta_r}=(A+B)e^{-j k_2 y \sin \theta_t} ##
Now, by enforcing equality of the imaginary parts when y is not equal to zero, you will see that the imaginary term of the right side is zero when
## y = \frac{ n \pi }{ k_2 \sin \theta_t}##
Then you must also have the imaginary parts on the left be zero, so
##k_1 \frac{ n \pi }{ k_2 \sin \theta_t} \sin \theta_i = m \pi ##
and
##k_1 \frac{ n \pi }{ k_2 \sin \theta_t} \sin \theta_r = m \pi ##
In order for this to be true, you will need
##\frac{ k_1 \sin \theta_i}{ k_2 \sin \theta_t} \in \mathbb{Z}## and ##\frac{ k_1 \sin \theta_r}{ k_2 \sin \theta_t} \in \mathbb{Z}##
Noting that the two sides have to be zero at the same times, there can be no zeros on the left side for y between
##(\frac{ n \pi }{ k_2 \sin \theta_t}, \frac{ (n+1) \pi }{ k_2 \sin \theta_t})##
This imposes the condition that:
##\frac{ k_1 \sin \theta_i}{ k_2 \sin \theta_t} \in \{1, -1\}## and ##\frac{ k_1 \sin \theta_r}{ k_2 \sin \theta_t} \in \{1, -1\}##
Generally, there is an assumption that your theta values are measured relative to the interface, so ##k\sin\theta## is always positive. This reduces our solution set to simply:
##\frac{ k_1 \sin \theta_i}{ k_2 \sin \theta_t} =1## and ##\frac{ k_1 \sin \theta_r}{ k_2 \sin \theta_t} =1##
which can be written as:
##k_1 \sin \theta_i = k_1 \sin \theta_r = k_2 \sin \theta_t ##
If you didn't have the assumption of positive signs, it would look more like:
##k_1 \sin \theta_i = \pm k_1 \sin \theta_r = \pm k_2 \sin \theta_t ##
 
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  • #5
Don't worry, I am late this time more than you. Thank you instead for all your detailed observations.
I followed all your steps and I agree with them. But maybe there is another solution and I would like to ask you for your opinion.
Consider your equation:

[itex]A e^{-j k_1 y \sin \theta_i} + B e^{-j k_1 y \sin \theta_r} = A e^{-j k_2 y \sin \theta_t} + B e^{-j k_2 y \sin \theta_t}[/itex]

and the following system

[itex]\begin{cases} e^{-j k_1 y \sin \theta_i} = e^{-j k_2 y \sin \theta_t} \\ e^{-j k_1 y \sin \theta_r} = e^{-j k_2 y \sin \theta_t} \end{cases}[/itex]

(It is obviously obtained by separately equating the [itex]A[/itex] and [itex]B[/itex] coefficients in the first equation).
Is this system a necessary and sufficient condition in order for the first equation to be satisfied?
The conditions

[itex]\begin{cases} k_1 \sin \theta_i = k_2 \sin \theta_t \\ k_1 \sin \theta_i = k_1 \sin \theta_r \end{cases}[/itex]

would immediately follow.
Can it be a valid solution too or does it lack some details?
 
  • #6
Yes that is a straightforward and valid way to approach the explanation. If you were unsure of whether any of those assumptions had logical holes in them, I feel that my explanation above should remove those doubts.
 
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  • #7
Yes, it does, because it is very detailed. Thank you!
 

FAQ: Wave reflection and refraction, relations between angles

1. What is wave reflection?

Wave reflection is the phenomenon where a wave bounces off an object or surface and changes direction. This can occur with any type of wave, such as light, sound, or water waves.

2. How does the angle of incidence relate to the angle of reflection?

According to the law of reflection, the angle of incidence (incoming wave) is equal to the angle of reflection (outgoing wave). This means that the angle at which a wave hits an object or surface is the same as the angle at which it bounces off.

3. What is the difference between reflection and refraction?

Reflection is when a wave bounces off an object or surface, while refraction is when a wave passes through a medium and changes direction due to a change in speed. Reflection results in an equal angle of incidence and reflection, while refraction results in a change in direction and speed of the wave.

4. How do the properties of a medium affect wave refraction?

The properties of a medium, such as density and temperature, can affect the speed of a wave passing through it. This change in speed causes the wave to refract, or change direction. For example, light waves refract when passing through different materials, such as water or glass, due to differences in density.

5. What is the difference between normal, incident, and refracted angles?

The normal angle is an imaginary line perpendicular to the surface where the wave is incident (incoming). The incident angle is the angle at which the wave hits the surface, and the refracted angle is the angle at which the wave changes direction after passing through the surface. All three angles are related by the law of refraction, which states that the ratio of the sine of the incident angle to the sine of the refracted angle is equal to the ratio of the speeds of the wave in the two media.

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