<p>I keep reading about fresnel and there is one thing I don't fully understand.</p>
<h4>My understanding :</h4>
<p>Here is my understanding so far, so please correct me where I am wrong.</p>
<ol>
<li>Fresnel is related to ior (eta) of the surface. Snell's law enable to know where the ray should &quot;go&quot; when going through a surface with different ior. Fresnel is of use to know how much light reflects / is transmitted.</li>
<li>Applying Schlick's approximation enables to know the proportion of the reflected vs the refracted light.</li>
</ol>
<h4>Now, my question:</h4>
<p>When a ray of light touches a refractable surface (glass..), I apply the Schlick's approximation which gives a coefficient.</p>
<p>But does that mean that I take the ray and calculate both its reflection and transmission angles; and then, given the fresnel coefficient, reshoot the ray in <strong>both directions</strong> (reflection and refraction) but with the power of each rays further attenuated with the coefficient.</p>
<p>Or is it <strong>only one</strong> ray with reduced power ?</p>
<p>All my readings don't make that point clear.</p>
<h4>Example</h4>
<p>For an incident ray touching the surface with a power of 1, and a fresnel coefficient of 0.5, both reflected and refracted rays will both be 0.5 before leaving the surface.</p>
<p>Is my understanding correct?</p>
<p>Thanks</p>

CC BY-SA 4.0
2023-03-07T23:37:13.503Z
You have to shoot both reflected and refracted rays. Though if you are doing a progressive raytracer, you can use Monte-Carlo to randomly go either the reflection or refraction route once per iteration so you don't have to do both in one go. If you don't do both you'll obviously miss out the color coming from either path. So if you don't do refraction how will you be able to see through glass?
gallickgunner
CC BY-SA 4.0
2023-03-09T07:24:56.197Z
@gallickgunner: +1 (i will put it when my rep is high enough, or someone else can). Yes, it seems logical, I've already implemented it that way in the meantime.
LeXav
<reflection><refraction><fresnel>
Fresnel and light reflection/refraction
LeXav
# Fresnel and light reflection/refraction

(Asked by: username_0 on 2023-03-06)

I keep reading about fresnel and there is one thing I don't fully understand.

#### My understanding :

Here is my understanding so far, so please correct me where I am wrong.

1. Fresnel is related to ior (eta) of the surface. Snell's law enable to know where the ray should "go" when going through a surface with different ior. Fresnel is of use to know how much light reflects / is transmitted.
2. Applying Schlick's approximation enables to know the proportion of the reflected vs the refracted light.

#### Now, my question:

When a ray of light touches a refractable surface (glass..), I apply the Schlick's approximation which gives a coefficient.

But does that mean that I take the ray and calculate both its reflection and transmission angles; and then, given the fresnel coefficient, reshoot the ray in **both directions** (reflection and refraction) but with the power of each rays further attenuated with the coefficient.

Or is it **only one** ray with reduced power ?

All my readings don't make that point clear.

#### Example

For an incident ray touching the surface with a power of 1, and a fresnel coefficient of 0.5, both reflected and refracted rays will both be 0.5 before leaving the surface.

Is my understanding correct?

Thanks

**Tags:** reflection, refraction, fresnel

> **Comments:**
> **username_1**: You have to shoot both reflected and refracted rays. Though if you are doing a progressive raytracer, you can use Monte-Carlo to randomly go either the reflection or refraction route once per iteration so you don't have to do both in one go. If you don't do both you'll obviously miss out the color coming from either path. So if you don't do refraction how will you be able to see through glass?  
> **username_0**: @username_1: +1 (i will put it when my rep is high enough, or someone else can). Yes, it seems logical, I've already implemented it that way in the meantime.  

---

## 0 answers

