Reflections in Ellipse

Choose two points T1, T2 on an ellipse. Imagine a light source at T1 sending a ray of light in the direction of T2. Assuming the inner surface of the ellipse fully reflective, the ray will bounce off at T2 and reach the ellipse again at T3 where it will bounce again and then again at point T4 and so on.

 

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There are three possibilities:

 
  1. T1T2 passes through one of the foci. Then T2T3, T3T4, ... will pass alternately through the two foci of the ellipse. (Due to a limited accuracy and also because of the difficulty in positioning the points accurately, this case is only partially validated by the applet.)

  2. All the pieces TkTk+1 remain outside the segment F1F2 of the two foci. In which case there is an ellipse confocal with the given one for which TkTk+1's serve as tangents.

  3. All the pieces TkTk+1 cross the segment F1F2 of the two foci. In which case there is a hyperbola confocal with the given ellipse for which TkTk+1's serve as tangents.

In the latter two cases, the straight lines TkTk+1 are said to form an envelope of the ellipse (case 2) or of the hyperbola (case 3.)

Proof

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Copyright © 1996-2018 Alexander Bogomolny

Reflections in Ellipse

The question is, What happens to a ray of light that hops off an elliptic mirror with a light source inside the ellipse?

One case, i.e. where the light source is located at a focus of the ellipse has been discussed from the first principles elsewhere. In this case, after every reflection the ray will proceed from one focus to the other indefinitely. This answer clearly applies to the case where the first the light source is not at a focus but emits a ray that passes through one.

Only other two possibilities are possible: either every piece of the light trajectory crosses between the foci of the ellipse or none does. In the latter case, the pieces envelope an ellipse, in the former a hyperbola. Bellow, we shall look into the case of ellipse.

 

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Let the foci of the ellipse be F1 and F2, and assume that the light source located at T1 on the ellipse emits a ray that reflects of the ellipse at point T2 and reaches it again at T3. If T1T2 does not cross F1F2, there is an ellipse confocal with the given one that touches T1T2. This is because the whole plane is split into the level curves of the function f(M) = |F1M| + |F2M|, each of these curves being an ellipse.

We are going to prove that T2T3 is tangent to the same ellipse.

Find F'1, the reflection of F1 in T1T2, and F'2, the reflection of F2 in T2T3. By the optical property of ellipse, the angles between the tangent at T2 and F1T2 and F2T2 are equal. By the construction, so are the angles between the tangent and T1T2 and T3T2. It follows that the angles F1T2T1 and F2T2T3 are equal and so are their doubles:

  angleF1T2F'1 = angleF2T2F'2,

implying that isosceles triangles F1T2F'1 and F2T2F'2 are similar. It then follows that triangles F1T2F'2 and F'1T2F2 are equal. In particular,

  F2F'1 = F1F'2.

With respect to the constructed ellipse, the locus of points F'1 is a circle with center F2 and radius, say, R = |F1T| + |F2T|, where T is the point of tangency if the ellipse with T1T2. But then also F1F'2 = R; so that F'2 lies on the circle with center F1 and radius |F1T| + |F2T|, which is the locus of reflections of F2 in all the tangents to the constructed circle. This proves that T2T3 is one of these tangents.

References

  1. V. Gutenmacher, N. Vasilyev, Lines and Curves: A Practical Geometry Handbook , Birkhauser, 1 edition (July 23, 2004)

Ellipse

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Copyright © 1996-2018 Alexander Bogomolny
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