# Geometrical construction for Snell's law?

Snell's law from geometrical optics states that the ratio of the angles of incidence $\theta_1$ and of the angle of refraction $\theta_2$ as shown in figure1, is the same as the opposite ratio of the indices of refraction $n_1$ and $n_2$.

$$\frac{\sin\theta_1}{\sin \theta_2} = \frac{n_2}{n_1}$$

(figure originally from wikimedia)

Now let $P$ be a point in one medium (with refraction index $n_1$) and $Q$ a point in the other one as in the figure. My question is, is there is a nice geometrical construction (at best using only ruler and compass) to find the point $O$ in the figure such that Snell's law is satisfied. (Suppose you know the interface and $n_2/n_1$)?

Edit A long time ago user17762 announced to post a construction. However until now no simple construction was given by anybody. So, does anybody know how to do this?

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@user3445: Do you know the interface? (I think if you know just the position of $P$, $Q$ and the ratio $\frac{n_2}{n_1}$, there can be many solutions.) – user17762 Feb 20 '11 at 17:25
Yes, I know the interface. – student Feb 20 '11 at 17:33
Assume that the interface is the $x$ axis. The algebraic equations that characterize the point $O$ is a quartic polynomial with coefficients given by the coordinate values of $P$, $Q$, and $n_2/n_1$. So the ruler-and-compass constructibility can probably be considered in the usual way. – Willie Wong Feb 20 '11 at 18:51
@WillieWong So the ruler-and-compass constructibility probably is impossible. – vesszabo Jan 6 '13 at 13:12
Solving a quartic equation like I've found as: $$x^2(B^2+(d-x)^2)=(n_2/n_1)^2(d-x)^2(A^2+x^2)$$ seems to be prohibitive; it's a mess even with a computer algebra system. – Han de Bruijn Mar 2 '15 at 11:20

If you know the interface, then drop perpendiculars from $P$ and $Q$ to the interface. Let the points of intersection be $P'$ and $Q'$. Let $PP' = y_P$ and $QQ' = y_Q$.

Now consider the line segment $P'Q'=x$. You need to find a point $O$ inside $P'Q'$ such that $OP' + OQ' = x$.

Let $OP' = x_P$ and $OQ' = x_Q$.

We now have two equations to solve for $\theta_1$ and $\theta_2$.

$x_P + x_Q = x$ i.e. $$y_P \tan(\theta_1) + y_Q \tan(\theta_2) = x$$

and

$$\frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{n_2}{n_1}$$.

So the problem is well-defined and hence solving for $\theta_1$ and $\theta_2$ gives $x_1$ and $x_2$.

I shall post the geometric construction later.

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It would be nice if you could post the geometric construction in detail. – student Feb 26 '11 at 15:09
I am still interested in the geometric construction. So it would be great if you could post it in detail. – student Dec 30 '12 at 14:59

To expand on Han de Bruijn's comment: Assume $P=(x,y_1)$ and $Q=(x',-y_2)$ with $x'-x=:d>0$. Then we have to solve the system \eqalign{x_1+x_2&=d\cr {n_1x_1\over\sqrt{x_1^2+y_1^2}}&={n_2x_2\over\sqrt{x_2^2+y_2^2}}\cr} for $x_1$ and $x_2$. Introducing $x_2:=d-x_1$ into the squared second equation leads to an equation of degree $4$ for $x_1$ having no obvious solutions in terms of second degree equations. From this we may conclude that there is no ruler and compass construction of the path in question, given the ratio ${n_1\over n_2}$.

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This might be a little trivial, but...

$n_1,n_2$ are the radius of the circles. The horizontal red lines are equidistant from the horizontal axis.

Interactive graph in GeoAlgebra, here: http://tube.geogebra.org/student/m771807

You can play by moving the point $B$ horizontally (hence changing $n_1/n_2$) and moving the point $C$ over the circumference (changing the incidence angle).

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This doesn't construct the point $A$ on the interface when the points $P$ and $Q$ are given. – student Mar 8 '15 at 20:09
@student : you're right – leonbloy Mar 8 '15 at 21:40

EDIT 3: (EDIT1 / EDIT2 are withdrawn due to an error found.)

As queried in my comments options so far not replied by OP, I assume that points P and Q are given as fixed, and the interface direction is given,but it is not required to pass through any given point. Accordingly, given temporary interface direction straight line L can be parallelly displaced to pass through a point of incidence I, yet to be found out.

Label I is used instead of O to avoid imagining point of incidence as fixed at origin.

Draw parallel and perpendicular directions to L passing through P and Q so that they intersect at C.Divide PC in the ratio $n_2 : n_1$ in the usual way ( Draw $n_2 , n_1$ inches or centimeters from a corner and draw parallels for such proportionate division of line L into two segments). Draw line NN parallel to QC. Let the perpendicular bisector of PQ intersect NN at I, the point being sought.Draw line XX parallel to PC. Now XX is the interface straight line.Since $PI= QI$,

Special Case when P and Q are given with respect a given fixed horizontal Interface (slope L = 0):

It is easier. Divide horizontal projection of PQ in the ratio $n_2:n_1$ to draw vertical line NN. The perpendicular bisector of PQ cuts NN at desired point I.

$$\dfrac{\sin i}{\sin r}= \dfrac{\sin \theta_1}{\sin \theta_2} = \dfrac {PN/PI}{QN/QI} = \dfrac{PN}{QN} = \dfrac{n_2}{n_1} = \mu.$$

The interface cannot be a fixed line. I had earlier answered the question in German newsgroup de.sci.mathematik.

The one and only way for a ray of light to travel between two given fixed points is to have an interface curve shaped as one among curved confocal Cartesian Ovals. This is a direct geometric consequence of Fermat's Law.

In other words perpendiculars onto correctly positioned interface normal should be in ratio $n_2/n_1.$

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The distance $PX$, $QX$ of $P$ and $Q$ from the interface are given in advance and cannot be the result of the construction. – Christian Blatter Mar 5 '15 at 17:32
@Christian Blatter: The case of fixed L direction with zero slope is a special simple case shown. – Narasimham Mar 6 '15 at 10:09
Thanks for your link to the german math newsgroup. There seems to be an answer which proves that it is not possible to construct it with ruler and compass only. groups.google.com/forum/#!msg/de.sci.mathematik/BPoRZK2_dvY/… – student Mar 8 '15 at 20:16
I just did an analytic construction here: tube.geogebra.org/material/show/id/800821 where a fixed interface works. So I don't understand the part about the cartesian ovals. (You probably need to download the ggb file to make it work) – student Mar 8 '15 at 20:40

Since I cannot draw a picture I will describe the proof with words, and using the image of the OP. The proof requires only euclidean geometry and the result

Lemma: light moves in a straight line in a given medium.

Consider a fictitious problem where the source on the side 1 has been desplaced vertically so far up that it will lie on the continuation of the line $QO$ into medium 1, call this source $P'$. In such fictional setting the source $P'$ lies on the straight line $P'OQ$ and light follows a straight path, because now the fictional medium has the same refraction index $n_2$. However the new path $P'O$ accounts for the different speeds on different media, in particular it must be $n$ times longer than the original path (assumming $n_1=1$, $n_2= n$ with no loss of generality). By construction of $P'$ it holds:

$$P'O \sin(\theta_1) = OP \sin (\theta_2)$$ Now the kinematical condition tells us that the new fictional path $P'O$ must be $n$ times longer than $PO$ because light is now $n$ times slower and paths $PO$ and $P'O$ take the same time to travel through. Than means:

$$P'O = n PO$$ from which the relation follows. We have transformed the problem into a trivial one of a light ray travelling from $P'$ to $Q$, and which takes the same time as the original.

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Yes. I guess there is.

Draw perpendicular from point P to interface say PP'. Draw perpendicular from point Q to interface say QQ'.

Divide segment P'Q' in ratio n_2/n_1.

O is the point that divides P'Q' in n_2/n_1.

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This is not right as if you just make Q twice as far from the interface along the same ray, your point O gets changed. If PP' and QQ' are the same length and the angles are small so $\sin \theta=\theta$ this works – Ross Millikan Feb 20 '11 at 17:40

here is the geometric argument advanced by daniel pedoe. this is from his paper 'a geometric proof of the equivalence of fermata principle and shells law' in the selected papers on precal by maa.

his proof use the ptolemys theorem that state that in any quadrilateral $ABCD, \,AB\cdot CD +BC \cdot DA \ge AC \cdot BD$ and the equality holds iff $ABCD$ is cyclic quadrilateral.

let $l$ be the interface. construct the circus circle of $POQ$ of diameter $d$ where $P$ is in medium $1, Q$ in medium $2$ and $O$ on $l$ so that $POQ$ is the fermat path so that $$\frac{v_1}{\sin \theta_1} = \frac{v_2}{\sin \theta_2}$$

draw a line through $O$ perpendicular to $l.$ let this line cut the circumcircle at $R.$ by the law of sine, we have $$QR = d \sin \theta_2, PR = d \sin \theta_1.$$

now, ptolemys theorem applied to the cyclic quadrilateral $POQR$ gives $$PO \cdot QR + OQ \cdot RP = PQ \cdot OR \tag 1$$ and for any other point $O'$ on the interface $l$ we have $$PO' \cdot QR + O'Q \cdot RP \ge PQ \cdot O'R\tag2$$

substituting for $Qr, PR$ in $(1), (2)$ gives $$\frac{PO}{\sin \theta_1} + \frac{OQ}{\sin \theta_2} = \frac{PQ \cdot OR}{d \sin \theta_1 \sin \theta_2}, \frac{PO'}{\sin \theta_1} + \frac{O'Q}{\sin \theta_2} \ge \frac{PQ \cdot O'R}{d \sin \theta_1 \sin \theta_2} \ge \frac{PQ \cdot OR}{d \sin \theta_1 \sin \theta_2}$$ we used the fact that $OO'R$ is a right angled triangle and $O'R \ge OR.$ replacing $\sin \theta_1, \sin \theta_2$ by $v_1, v_2$ we have $$\frac{PO}{v_1} + \frac{OQ}{v_2} \le \frac{PO'}{v_1} + \frac{O'Q}{v_2} \text{ for all O' on the interface. }$$

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I don't really understand how to get from this a concrete construction. – student Mar 8 '15 at 20:18
@student, this is a geometric proof of snell's condition. but i will think of geometric construction of finding the point on the interface given two points on either side of it. – abel Mar 8 '15 at 20:34