# Law of refraction in vector form: how is this derived? [duplicate]

I am told that the law of refraction can be stated in the following two parts:

1. $$\mu_1 \sin(\theta_i) = \mu_2 \sin(\theta_r)$$
2. The incident ray, the normal to the surface separating the media at the point of incidence, and the reflected ray lie in one plane.

I am then told that both parts of the law can be put into vector form as

$$\vec{n}_2 = \vec{n}_1 - (\vec{n}_1 \cdot \vec{s}) \vec{s} + \left( \sqrt{(\mu_2)^2 - (\mu_1)^2 + (\vec{n}_1 \cdot \vec{s})^2} \right) \vec{s}, \tag{1.2}$$

where $$\vec{n}_1$$, $$\vec{n}_2$$, and $$\vec{s}$$ are the unit vectors of the incident ray, the reflected ray, and the normal to the surface, respectively, and $$\theta_i$$ and $$\theta_r$$ are the angles of incidence and refraction, respectively. The refractive indices of the two media are $$\mu_1$$ and $$\mu_2$$, and the incident ray is in the medium of refractive index $$\mu_1$$.

I don't understand exactly how (1.2) was derived. How is this derived from the two parts of the law of refraction? Please show this carefully, so that I can follow.

## EDIT

• You have: vector = vector-vector+scalar, so that equation doesn't work.
– JEB
Jul 10 at 15:48
• @JEB Hmm, I've copied it exactly – I just checked. Jul 10 at 16:11
• Does this answer your question? Snell's law in vector form Jul 10 at 18:20
• Jul 10 at 18:36
• Related question on reflection by the OP: physics.stackexchange.com/q/619594 .. almost in the same format. Jul 10 at 19:42

Given two vectors $$\vec a$$ and $$\vec b$$, the vector projection of $$\vec a$$ onto $$\vec b$$ is:

$$\vec a_{\parallel \vec b}= \frac{\vec a \cdot \vec b}{\vec b \cdot \vec b} \vec b$$

The component of $$\vec a$$ that is perpendicular to this is called the vector rejection on $$\hat b$$:

$$\vec a_{\perp \vec b}=\vec a - \vec a_{\parallel \vec b}=\vec a - \frac{\vec a \cdot \vec b}{\vec b \cdot \vec b}\vec b$$

If $$\vec b=\hat b$$ is a unit vector:

$$\vec a_{\perp \vec b}=\vec a -(\vec a\cdot \hat b)\hat b$$

Armed with that geometric knowledge we can address the question. We can capture the index of refraction and wave direction with the wave vector:

$$\vec k_{\alpha}=\mu_{\alpha}\hat n_{\alpha}$$

(with $$\alpha\in(i, 1)$$ or $$\in(r, 2)$$).

The fact that the phases of the incident and refracted rays need to match along the boundary (perpendicular to $$\vec s$$) means that wave vector rejection onto $$\hat s$$ is continuous across the boundary:

$$\vec k_{r,\perp \hat s} = \vec k_{i,\perp \hat s}$$

or

$$\mu_2 \vec n_{r,\perp \hat s} = \mu_1 \vec n_{i,\perp \hat s}$$

which is Snell's Law.

From here, you go in two directions. First:

$$\vec n_{r,\perp \hat s} = \frac{\mu_1}{\mu_2} \vec n_{i,\perp \hat s}= \frac{\mu_1}{\mu_2}[\hat n_i-(\hat n_i\cdot \hat s)\hat s]$$

solves for the transverse component of the unit vector of the refracted wave $$\hat n_r$$, which can be written:

$$\hat n_r = \left(\begin{array}{c} n_{r,\perp \hat s} \\ n_{r,\parallel \hat s}\end{array}\right)$$

Here, the 3rd dimension (along $$\hat n \times \hat s$$) has been suppressed. The remaining basis vectors are in the $$n-s$$ plane, with $$\left(\begin{array}{c}0\\1\end{array}\right)=\hat{s}$$ and $$\left(\begin{array}{c}1\\0\end{array}\right)$$ along $$(\hat n\times\hat s)\times \hat s$$.

Sin $$\hat n_r$$ is a unit vector:

$$||\hat n_r||^2 = n_{r,\perp \hat s}^2 + n_{r,\parallel \hat s}^2=1$$

so that:

$$\hat n_r = \left(\begin{array}{c} n_{r,\perp \hat s} \\ \sqrt{1-n_{r,\perp \hat s}^2}\end{array}\right)$$

We have a vector expression for the transverse component. Note that:

$$\left(\begin{array}{c}0\\1\end{array}\right)=\hat s$$

so the vector expression for the longitudinal component is:

$$\vec n_{r,\parallel \hat s}= \sqrt{1-n^2_{r, \perp\hat s}}\hat s$$

$$\hat n_r = \frac{\mu_1}{\mu_2}[\hat n_i-(\hat n_i\cdot \hat s)\hat s] + \sqrt{1-n^2_{r, \perp\hat s}}\hat s$$

and substitute:

$$\hat n_r = \frac{\mu_1}{\mu_2}[\hat n_i-(\hat n_i\cdot \hat s)\hat s] + \sqrt{1- \left\Vert\frac{\mu_1}{\mu_2}[\hat n_i-(\hat n_i\cdot \hat s)\hat s] \right\Vert^2}\hat s$$

and turn the crank.

Comment on notation: Note that all $$\hat n_{\alpha}$$ are unit vectors. The projection or rejection of that onto another vector $$\vec v$$ is not necessarily a unit vector, and hence, does't have a hat. So the arrow over the $$n$$ in $$\vec n_{r,\parallel \vec v}$$ refers to the resultant vector after vector projection onto $$\vec v$$, even though the initial vector, $$\hat n_r$$, was a unit vector.

Meanwhile, the operation of scalar projection/rejection, has neither, e.g. $$a=n_{r,\parallel \vec v}$$ is a scalar which can converted into a vector a la:

$$\vec n_{r,\parallel \vec v} = a\hat v$$

• But where is (1.2) in this? It isn't clear to me that this is addressing my question. Jul 11 at 3:01
• How do you get that $\vec a -(\vec a\cdot \hat b)\hat b = \vec a - \frac{\vec a \cdot \vec b}{\vec b \cdot \vec b}$? If we say that $\vec b=\hat b$, then we have $\vec a - \frac{\vec a \cdot \hat b}{\hat b \cdot \hat b}$, but I don't see how that is $\vec a -(\vec a\cdot \hat b)\hat b$. Jul 11 at 7:20
• @Frobenius I don't think $\vec{n}_i$ is a unit vector here, so how can it be that $\Vert\vec{n}_i \Vert \boldsymbol{=} 1$? Jul 11 at 16:09
• Ok, you've got a typo in there. Should be \begin{align} (\vec a\cdot \hat b)\hat b&=\left(\vec a\cdot \frac{\vec b}{|\vec b|}\right)\frac{\vec b}{|\vec b|}\\\\ &=\left(\frac{\vec a\cdot \vec b}{|\vec b|^2}\right) \vec b\\\\ &=\left(\frac{\vec a\cdot \vec b}{\vec b\cdot \vec b}\right) \vec b\\\\ \end{align} Jul 11 at 18:01
• @ThePointer n is a unit vector. It's all derived from the fact that phase needs to match along the boundary
– JEB
Jul 12 at 1:42 Denote by $$\mathrm{A}$$ the point of departure, $$\mathrm{B}$$ the point of arrival of the ray, and $$\mathrm{I}$$ the point of contact with the refracting surface. The optical path is $$L = \mu_1 \mathrm{AI} + \mu_2 \mathrm{IB}$$.

Fermat’s principle tells us that this path is extremal with respect to a small displacement $$\overrightarrow{d\mathrm{OI}}$$ tangent to the surface of the refracting surface.
At the first order in $$\overrightarrow{d\mathrm{OI}}$$, it is not difficult to show (see below!) that $$d(\mathrm{AI}) = \vec{n_1} \cdot \overrightarrow{d\mathrm{I}}$$ and $$d(\mathrm{IB}) = -\overrightarrow{n_2} \cdot \vec{d\mathrm{OI}}$$ with $$\vec{n_1}$$ and $$\vec{n_2}$$ globally from $$\mathrm{A}$$ to $$\mathrm{B}$$. So, to the first order:

$$dL = {(\mu}_1\vec{n_1} - \mu_2\vec{n_2}) \cdot \overrightarrow{d \mathrm{OI}} = 0$$

As $$\overrightarrow{d \mathrm{OI}}$$ is any displacement in the tangent plane, this means that $${(\mu}_1 \vec{n_1} - \mu_2 \vec{n_2}) = \alpha\vec{s}$$.

We then write $$\mu_2 \vec{n_2} = -\alpha \vec{s} + \mu_1 \vec{n_1}$$ and we square it, the vectors being unitary: $${\mu_2}^2 = \alpha^2 - 2\alpha \mu_1(\vec{n_1} \cdot \vec{s}) + {\mu_1}^2$$.

It is a quadratic equation whose solutions are: $$\alpha = \mu_1(\vec{n_1} \cdot \vec{s}) \pm \sqrt{{\mu_1}^2{(\vec{n_1} \cdot \vec{s})}^2 - ({\mu_1}^2 - {\mu_2}^2)}$$.

If $$\mu_1 = \mu_2$$, we must find $$\vec{n_1} = \vec{n_2}$$ and therefore $$\alpha = 0$$. So we keep: $$\alpha = \mu_1(\vec{n_1} \cdot \vec{s}) - \sqrt{{\mu_1}^2{(\vec{n_1} \cdot \vec{s})}^2 - ({\mu_1}^2 - {\mu_2}^2)}$$. Hence finally:

$$\mu_2\vec{n_2} = \mu_1\vec{n_1} - \mu_1 \left( (\vec{n_1} \cdot \vec{s}) - \sqrt{{\mu_1}^2{(\vec{n_1} \cdot \vec{s})}^2 - ({\mu_1}^2 - {\mu_2}^2)}\vec{s} \right)$$

I can't find the same relationship as you? But by examining when the term under the root vanishes (limiting angle), I find a correct relation and therefore it is possible that your formula is incomplete?

Complement: To prove $$d(\mathrm{AI}) = \vec{n_1} \cdot \vec{d \mathrm{OI}}$$, it suffices to write $$\mathrm{AI} = \sqrt{{\vec{\mathrm{AI}}}^2}$$ and differentiate: $$d(\mathrm{AI}) = \dfrac{\vec{\mathrm{AI}} \cdot d\vec{\mathrm{AI}}}{\sqrt{{\vec{\mathrm{AI}}}^2}} = \vec{n_1} \cdot \vec{d \mathrm{OI}}$$ with $$\vec{\mathrm{AI}} = \vec{\mathrm{OI}} - \vec{\mathrm{OA}}$$ and $$\mathrm{A}$$ fixed.

Sorry for my poor english! English is not my native language.

• You say that I is the point of contact. What do you mean by "dI" later on? And with vector sign on top.
– nasu
Jul 10 at 15:13
• The ray is AIB, from A to B. I is on the the surface of separation (dioptre, and not diopter ! I hope dioptre is the correct word ?). Using Fermat's principle, you have to imagine a small displacement $\vec{dI}$ of the point I. The correct point is such that the optical path is extremal. Jul 10 at 15:22
• By your definition "I" is just a label for a point. If you want to look at the position of the point you should have a variable which represents the position of that point in respect to some coordinate system. Of course, you could choose to call this variable "I" again but it would be confusing, unless you disclose this explicitly.
– nasu
Jul 10 at 15:27
• To be more clear, I could have written $\vec{dOI}$ for the small displacement. (with O the origin). i have changed the notation. Jul 10 at 15:29
• \bullet is normally used for unordered lists, while dot product is usually denoted with \cdot. Jul 10 at 15:34