# Law of reflection

1. Aug 26, 2009

### _Andreas

1. The problem statement, all variables and given/known data

An electromagnetic wave is incident upon a planar interface at an oblique angle $$\theta_i$$, where it is reflected. For the wave vector components parallel to the interface, we have $$k_{xi}=k_{xr}$$. Thus, $$\theta_i=\theta_r$$. 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.

2. Relevant equations

$$\cos\theta_i=\cos\theta_r$$

See attached picture.

3. The attempt at a solution

Thus, from the picture, $$\frac{k_{zr}}{k}=\frac{k_{zi}}{k}\Longrightarrow k_{zr}=k_{zi}$$.

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?

#### Attached Files:

• ###### reflection.jpg
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2. Aug 26, 2009

### tiny-tim

Hi _Andreas!

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

3. Aug 27, 2009

### _Andreas

Hi!

It follows from cosθi=kzi/k and cosθr=kzr/k, since θir and k=|ki| = |kr|.

4. Aug 27, 2009

### tiny-tim

Perhaps I'm misunderstanding your picture, but isn't cosθi = -kzi/k ?

5. Aug 27, 2009

### _Andreas

Uhm... can you explain how you get this result?

6. Aug 27, 2009

### tiny-tim

kzi points right?

7. Aug 27, 2009

### _Andreas

Sure, but that's in the positive z direction.

8. Aug 27, 2009

### tiny-tim

ah, then isn't cosθr = -kzr/k ?

9. Aug 27, 2009

### _Andreas

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