Quadratic Forms, Elliptical Mirror

Elliptical Mirror

In the last section we discussed the parabolic mirror, which focuses light from a distant object onto a point. This point is called the "focus" of the parabola. In this section we turn to the elliptical mirror, which reflects the light from one focus onto the other. Place a light bulb at one focus, and all the energy from that bulb is directed towards the other focus.

Let the major axis of an ellipse run along the x axis, from -a to +a, and let the minor axis run up the y axis from -b to +b. The foci are at ±f, where f2 = a2-b2.

Imagine a particle moving along the ellipse. Its position, as a function of time, is a parameterization of the ellipse. Let's not worry about the precise parameterization for now. We simply have x(t),y(t), and the particle's velocity is x′(t),y′(t). For notational convenience, let u = x′(t) and v = y′(t). Remember that the velocity vector is also the tangent vector, so the line tangent to the ellipse is defined by the vector u,v.

Stop the particle at a point p, with coordinates x,y and tangent vector u,v. Draw the segments from p to the two foci. If you recall our notation from an earlier section, the distance from p to +f is q, and the distance from p to -f is r. This is summarized below.

q = sqrt((x-f)2 + y2)

r = sqrt((x+f)2 + y2)

If the elliptical mirror reflects light from one focus to the other, then the two lines, from p to the two foci, must make the same angle with the tangent vector u,v. In other words, angle of incidence equals angle of reflection. If we can prove this, then our mirror behaves as described above.

Remember that the angle can be characterized by the dot product. In other words, the two angles are equal iff the dot product of u,v with either segment, divided by the length of that segment, gives the same number.

(u,v) . (x-f,y) over q = (u,v) . (x+f,y) over r

Cross multiply and expand the dot products.

(u×(x-f) + vy) × r = (u×(x+f) + vy) × q

Square both sides and substitute for q2 and r2.

(u×(x-f) + vy)2 × ((x+f)2+y2) = (u×(x+f) + vy)2 × ((x-f)2+y2)

Yes, I used a computer to expand and simplify the above. After clearing out the common factor 4cy, we are left with this.

(v2-u2)xy +uv(x2-y2) - f2uv = 0

Now pick a parameterization. For notational convenience, let c be the cosine of θ and let s be the sine of θ. Thus x = ac and y = bs. Differentiate to find u = -as and y = bc. Substitute for x y u and v in the above. Yes, I used a computer again. Clear out the common factor of abcs and get this.

-a2×(c2+s2) + b2×(c2+s2) + f2 = 0

Remember that c2+s2 = 1, and a2-b2 = f2, hence the above equation is always true. That completes the proof.

A Hall of Mirrors

A funhouse is a house where every wall is a mirror. If you shine a flashlight directly into a wall, any wall, the light beam bounces straight back to you. If you point your flashlight at an angle, the beam of light bounces all around the house faster than you can imagine. In fact it probably makes its way into every nook and cranny.

Assume the fun house is closed to the outside world, so that light never leaves the house. Yet inside, none of the rooms are closed off. In other words, every room has an open door leading to another room. The interior forms a connected set. There is one light bulb in the house. When it is turned on, does light spread throughout the entire house? Or is there a small room somewhere that remains in darkness?

If all the walls consist of straight lines, we're pretty sure the entire house is lit. Nobody has proved this yet, but that's the way it appears. However, if some of your walls are curved, e.g. circular arcs, then some of the rooms might remain dark.

Place the light bulb at the center of your picture and draw a circle of radius 1 around it. Then draw another circle of radius 2 around the light bulb. Beams of light leave the bulb, bounce straight back from the inner circle, and return to the bulb. The outer room, the annulus between the two circles, remains dark, but that's because it is closed off. Cut a small door in the inner circle. Light now streams out through this door, bounces off the outer circle, and returns to the light bulb. As long as you're not standing near the doorway, the outer room remains dark.

This is a simple construction in theory, but it is very hard to build in practice. If a person or object is placed in the inner room, some of the light bounces off that object and leaves the inner room at a different angle. This light does not bounce straight back into the inner room. Instead it bounces away and spreads throughout the annulus. I like the following solution better, because you could actually build it in a science museum, and it would work, even if people were standing inside to watch the action.

Near the bottom of your paper, draw a horizontal line 6 inches long. Then draw four vertical line segments 2 inches high, standing up from your horizontal base at 2 inch intervals. You have built three rooms, each room 2 inches square. Now draw the top half of an ellipse, joining the two outer walls. This closes off the house. Let the two foci of the ellipse be the top end points of the two inner walls. If a beam of light bounces off the end point of either of these inner walls, it is actually leaving one focuse of the ellipse. It bounces off the curve of the ellipes and returns to the other focus, the other inside wall. If a beam of light leaves the middle room in any other way, passing between the two foci, it bounces off the curve and returns to the middle room. The left and right rooms remain dark.

Alternatively, place the light in the left room, and it will bounce over to the right room, and back again, leaving the middle room in darkness. For a great efect, place a red light in the middle room and a green light in the left room. Visitors will see either a green world or a red world, depending on which room they are standing in.