Do incident and reflected wave normal components have opposite signs?

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_Andreas
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Homework Statement



An electromagnetic wave is incident upon a planar interface at an oblique angle [tex]\theta_i[/tex], where it is reflected. For the wave vector components parallel to the interface, we have [tex]k_{xi}=k_{xr}[/tex]. Thus, [tex]\theta_i=\theta_r[/tex]. The wave numbers for the incident and reflected waves are equal. Find the relation between the wave vector components normal to the interface for the incident and reflected waves.


Homework Equations



[tex]\cos\theta_i=\cos\theta_r[/tex]

See attached picture.


The Attempt at a Solution



Thus, from the picture, [tex]\frac{k_{zr}}{k}=\frac{k_{zi}}{k}\Longrightarrow k_{zr}=k_{zi}[/tex].

To me this seems to imply that both normal components point in the same direction, in addition to being of the same magnitude. But shouldn't the normal components have opposite signs, since the incident and reflected waves travel in opposite normal directions?
 

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Hi _Andreas! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)
_Andreas said:
Thus, from the picture, [tex]\frac{k_{zr}}{k}=\frac{k_{zi}}{k}[/tex]

I don't understand how you get kzr/k = kzi/k :confused:
 
Hi!

tiny-tim said:
Hi _Andreas! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)I don't understand how you get kzr/k = kzi/k :confused:


It follows from cosθi=kzi/k and cosθr=kzr/k, since θir and k=|ki| = |kr|.
 
Perhaps I'm misunderstanding your picture, but isn't cosθi = -kzi/k ? :confused:
 
tiny-tim said:
Perhaps I'm misunderstanding your picture, but isn't cosθi = -kzi/k ? :confused:

Uhm... can you explain how you get this result?
 
tiny-tim said:
kzi points right? :confused:

Sure, but that's in the positive z direction.
 
ah, then isn't cosθr = -kzr/k ? :confused:
 
tiny-tim said:
ah, then isn't cosθr = -kzr/k ? :confused:

I guess so, and there's my problem. I tend to think of kzr without the sign as the z component of kr.