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I am doing ray tracing and I do the refraction of the ray using the following relation (I got it from PDF "Reflections and Refractions in Ray Tracing"):

$$\mathbf{t} = \frac{\eta_1}{\eta_2}\mathbf{i} + \left(\frac{\eta_1}{\eta_2}\cos{\theta_i} - \sqrt{1 - \sin^2{\theta_t}}\right)\mathbf{n} \tag{22}$$

$$\sin^2{\theta_t} = \left(\frac{\eta_1}{\eta_2}\right)^2\sin^2{\theta_i} = \left(\frac{\eta_1}{\eta_2}\right)^2 (1 - \cos^2{\theta_i}) \tag{23}$$

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But I have seen it in another PDF as follows:

$$\vec{t} = \frac{\eta_1}{\eta_2}\vec{i} + \left(\frac{\eta_1}{\eta_2}\cos{\theta_i} - \sqrt{1 - \sin^2{\theta_t}}\right)\vec{n} \tag{28}$$

$$\sin^2{\theta_t} = \left(\frac{\eta_1}{\eta_2}\right)^2\sin^2{\theta_i} = \left(\frac{\eta_1}{\eta_2}\right)^2 (1 - \cos^2{\theta_i}) \tag{29}$$

Could you please explain for me why?

And how can I reassure that my refraction vector that I calculated is correct?

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  • $\begingroup$ Presumably your question concerns the difference in sign conventions in the equation for the transmitted vector magnitude? Do you know the direction of the normal used in the second case? $\endgroup$ Commented Oct 23, 2013 at 18:39
  • $\begingroup$ I don't know the direction of the norm vector in the second one but even if the normal is in the other direction of the first one that should be (-n) and that should lead to the same equation of the first one but the (-) sign should be out of the brackets and that isn't the case in the second equation. $\endgroup$
    – Jack_111
    Commented Oct 23, 2013 at 18:44

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I've added the homework tag even though it likely isn't. I think the best way to make sure of equations like this is simply to derive them yourself - that way you won't miss any quirky or unwonted conventions the author may be using.

So here we have the basic facts:

  1. $\vec{i}$, $\vec{n}$ and $\vec{t}$ are all in the same plane;
  2. We are free to make these vectors unit vectors if we like (they only define directions);
  3. Snell's law

So, translate these into equations: (1) says that the three vectors are linearly dependent or:

$$\vec{t} = \alpha\,\vec{i} + \beta\,\vec{n};\;\alpha,\,\beta\in \mathbb{R}$$

(3) says that $n_1\, \vec{i} \wedge \vec{n} = n_2\, \vec{t} \wedge \vec{n}$ which simplifies to $n_1 = n_2 \alpha$. Now all we need is $|\vec{t}|^2 = \vec{t}\cdot \vec{t} = \alpha^2 + \beta^2 - 2 \alpha \beta \cos\theta_1 = 1$ and we can solve these equations to give the right expressions.

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  • $\begingroup$ Changed $|\vec{t}|^2 = \vec{t}\cdot \vec{t} = \alpha^2 + \beta^2 + 2 \alpha \beta \cos\theta_1 = 1$ to $|\vec{t}|^2 = \vec{t}\cdot \vec{t} = \alpha^2 + \beta^2 - 2 \alpha \beta \cos\theta_1 = 1$ as this holds for the way that the OP has drawn the vectors $\endgroup$ Commented Oct 24, 2013 at 7:40

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