Sketch created with limnu iphone app from limnu.com |

**N**pointing to the side with refractive index

*n*. One thing that means is that in the diagram

**N**might be pointing "down" along the dotted line.

First what about reflection? It is usually the classic formula:

**S**=

**D**- 2*

**N***dot(

**D**,

**N**)

Does that formula still work if

**N**points along the dotted line? Happily, yes-- try substituting -

**N**in there and you get a "-" on the

**N**and the dot product and they cancel. Also,

**D**need not be a unit vector. I rarely require that in my ray tracers so I can use instancing if I want.

**N**however does need to be a unit vector.

Now what about refraction? It obeys Snell's law:

*n*sin(

*theta*) =

*n*' sin(

*theta*')

We can use the same trick as in reflection where we use

**N**and

**D**as a 2D basis. First let's make life easy on ourselves and assume

**D**is a unit vector and

**N**points "up" like in the picture, and make an orthogonal basis (let's change the sin() and cos() with c, c', s, s')

**D**= -c

**N**+ s

**Q**

Here

**Q**is to the right and perpendicular to

**N**. We don't know what is is yet, but it we know what it is by rearranging the terms above to be get

**Q**= (

**D**+ c

**N**) / s

We also know that

**T**= -c'

**N**+ s'*

**Q**

We know c = -dot(

**D**,

**N**), and s = sqrt(1 - c^2)

Further we know that s' = (

*n*/

*n*') s

Right there is a place we can get into trouble-- n can be high (over 2 for diamond) so if the more dense medium is where

**D**is (the top medium), we could have s > 1. In that case we get all reflection and no refraction (total internal reflection). Here's a nice image of that:

The water acts as a perfect mirror due to total internal reflection (from wikipedia) |

**T**= -c'

**N**+ s'

**Q**

**T =**-c'

**N**+ s'(

**D**+ c

**N**) / s

We know s'/s = n'/n.

We also know c' = sqrt(1-s'^2) = sqrt(1 - (n/n')^2 (1-c^2) ).

**T =**(n'/n)

**D +(**c(n'/n)

**-**sqrt(1 - (n/n')^2 (1-c^2) ))

**N**

putting back in c = -dot(

**D,**

**N**) we get

**T =**(n'/n)**D +(**-dot(**D,****N**)(n'/n)**-**sqrt(1 - (n/n')^2 (1-dot(**D,****N**)^2) ))

__N__Ok! That agrees with the ray tracing news formula. Now what happens when

**N**points down? Only the last term is different:

**T =**(n'/n)

**D +(**-dot(

**D,**

**N**)(n'/n) + sqrt(1 - (n/n')^2 (1-dot(

**D,**

**N**)^2) ))

**N**

Well isn't that annoying? That sign difference is just sgn(dot(

**D,**

**N**)). I never know how to best code that sort of thing, but when in doubt, be very readable!

if (dot(D,N) > 0) temp_normal = -N // now proceed with original formula

## 2 comments:

Heh, I of course like seeing that the Ray Tracing News is a source for this bit of info. That's both great and sad - I would have hoped by now there would have been some grad student who got possessed by the ray tracing bug and scrawled all these sorts of bits (how to form a camera matrix, what the center of a pixel is labeled, reflection and refraction formulae, etc.) and put them on a nice web page somewhere.

I was surprised Phil Dutre's old document http://people.cs.kuleuven.be/~philip.dutre/GI/TotalCompendium.pdf doesn't have this vector formula. Morgan McGuire's Graphics Codex might well have it, http://graphicscodex.com/, but drat he's making dozens of dollars off it and flying around in his sky-yacht with the earnings.

Of course, now we need a new equation for metamaterials, which can have a negative index of refraction for some wavelengths: https://en.wikipedia.org/wiki/Negative_refraction

I was deriving the equations on my own and I think, I saw a small mistake in part of your equations, namely:

T =

(n'/n)D +(c(n'/n)- sqrt(1 - (n/n')^2 (1-c^2) )) NT =

(n'/n)D +(-dot(D, N)(n'/n)- sqrt(1 - (n/n')^2 (1-dot(D, N)^2) )) Nand

T =

(n'/n)D +(-dot(D, N)(n'/n)+ sqrt(1 - (n/n')^2 (1-dot(D, N)^2) )) NShouldn't it be: n/n', because of n*sin(theta) = n'*sin(theta'), sin(theta') = (n/n')sin(theta)?

That aside, I want to really thank you for bringing this precious blog as well as the numerous books you've published on different topics of computer graphics. As a CS student, who is fascinated by the subject, I found them invaluable!

Wish you all the best,

Harry!

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