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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 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

2 Answers 2

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 $OQ' = x_Q$. enter image description here

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

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


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

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

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

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