<p>We have <span class="math-container">$t_{12} = t_{21}$</span>, by <a href="https://computergraphics.stackexchange.com/questions/5286/the-reciprocity-of-brdf">the law of reciprocity</a>. Also, <span class="math-container">$$r_{12} + t_{12} = 1\\r_{21} + t_{21} = 1$$</span>
It follows that <span class="math-container">$r_{12} = r_{21}$</span> as well.</p>

CC BY-SA 4.0
2019-01-20T05:03:54.850Z
Ahh..I don't understand why t_12 = t_21...
TIANLUN ZHU
CC BY-SA 4.0
2019-01-21T05:38:29.620Z
@TIANLUNZHU It's because of the law of reciprocity, a fundamental principle of optics. It says if you swap the incoming and outgoing rays, the BSDF (reflectance or transmittance) doesn't change. Note from the diagram that the $t_{12}$ and $t_{21}$ cases are exactly symmetrical with each other, only with the rays going in the opposite directions.
Nathan Reed
CC BY-SA 4.0
2019-01-20T04:10:57.577Z
2019-01-21T05:35:53.683Z
2019-01-21T05:35:53.683Z
Nathan Reed
<p>I am reading a paper about layered materials. However, I am confused on the implementation of this paper. The following picture describe the layered material. <a href="https://i.sstatic.net/CTUZR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CTUZR.png" alt="enter image description here"></a></p>

<p>r_ab is a complex reflection coeffcient when going from medium a to medium b. </p>

<p>And in the code, there is:</p>

<pre><code>        float2 R12, phi12;
        fresnelDielectric(LdotH, 1.0, eta_2, R12, phi12);
        float2 R21 = R12;
</code></pre>

<p>I don't understand why is r_12 equal to r_21?</p>

<pbr><material><fresnel>
Confused on reflection coeffcient of Fresnel equations
TIANLUN ZHU
# Confused on reflection coeffcient of Fresnel equations

(Asked by: username_0 on 2019-01-20)

I am reading a paper about layered materials. However, I am confused on the implementation of this paper. The following picture describe the layered material. [![enter image description here](https://i.sstatic.net/CTUZR.png)](https://i.sstatic.net/CTUZR.png)

r\_ab is a complex reflection coeffcient when going from medium a to medium b.

And in the code, there is:

```
        float2 R12, phi12;
        fresnelDielectric(LdotH, 1.0, eta_2, R12, phi12);
        float2 R21 = R12;

```

I don't understand why is r\_12 equal to r\_21?

**Tags:** pbr, material, fresnel


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## 1 answers

### Answer by username_1 on 2019-01-20. Score: 1 | Accepted answer

We have $t\_{12} = t\_{21}$, by [the law of reciprocity](https://computergraphics.stackexchange.com/questions/5286/the-reciprocity-of-brdf). Also, $$r\_{12} + t\_{12} = 1\\r\_{21} + t\_{21} = 1$$
It follows that $r\_{12} = r\_{21}$ as well.

> **Comments:**
> **username_0**: Ahh..I don't understand why t_12 = t_21...  
> **username_1**: @username_0 It's because of the law of reciprocity, a fundamental principle of optics. It says if you swap the incoming and outgoing rays, the BSDF (reflectance or transmittance) doesn't change. Note from the diagram that the $t_{12}$ and $t_{21}$ cases are exactly symmetrical with each other, only with the rays going in the opposite directions.  

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