The Mathematics of Parabolic Mirrors The Mathematics of Parabolic Mirrors by Dr Profidious Vile

An analogy

Curved mirrors, such as those you find at funhouses, scatter the light of images they receive to produce an image which is distorted. I explain this phenomenon of light by comparing this property to the behavior of water waves.

In a water bath, when waves are made on the surface of the liquid, reflected waves travel at the same angle, but opposite direction from the normal to the wall of the water bath compared to the incident waves.

Diagram representing the reflection of water waves off the wall of a water bath

Something similar must occur with light. Like water waves, light travels in the same direction until something happens to cause it to change. In funhouse mirrors, the change occurs as the light hits the mirror surface. As the surface of reflection is distorted, the normals to the tangents of the curved surface are at different angles, thus causing initially parallel rays of light to reflect and divide in different directions.

Describing the gradient of reflected light

We can describe the angle of reflection with the following equation:

a(n) = (a(i)+a(r))/2

Graphically, using coordinates in two dimensions, we can describe this in terms of gradients of lines which we attribute to reflected beams, which can all be described by:

y = mx + c

Where m is the gradient.

As the gradient of a straight line is its rise on the y axis, divided by its run across the x axis, we can convert between gradients and angles from the y axis easily with:

a = arctan(1/m)

And since the gradient of the normal to the curve f(x) is equal to the negative reciprocal of the differential:

a(n) = arctan(-f'(x))

Let's now assume that all incident light is perpendicular to the x axis. This makes the angle of the incident light from the y axis 0, so this can be removed from the equation.

This leaves us with:

arctan(-f'(x)) = arctan(1/m)/2 m = [tan(2(arctan(-f'(x))))]^-1

Where m is the gradient of the reflected light.

We can clearly see that as f'(x) changes, despite identical angles of incident light, the gradient of reflected light changes also. This supports the statement we made earlier, about light being scattered in many directions as it hits a curved plane mirror.

Focusing Light

If we consider this property of light which allows us to control how light reflects off a mirror, it becomes entirely perceiveable that we could focus all parallel light approaching a mirror towards a single, focused point. Why do we want this?

We associate light very closely with warmth. On the surface, when we stand in the bright sun, we feel very warm. When we stand in the shade, we feel relatively cold. Hence it is perfectly valid to suppose that particularly concentrated light will generate a lot of heat at the concentrated point.

Upon considering the ideal mirror shape for focusing light on a single point, the parabola came to mind.

Modeling the parabola and light rays

In its simplest form, a parabola is modelled by:

y = x^2

For the sake of simplicity, we should turn the parabola on its side, making all incident light rays approach parallel to the x axis. Thus, we have:

x = y^2

And to describe the gradient of the reflected light, we can modify the equation from the previous section:

m = tan(2arctan(-dx/dy))

And since dy/dx = 1/2y

m = tan(2arctan(-2y))

m is the gradient of the ray of light reflected from the parabolic mirror. This assumes that all incident rays are modeled by y = k, and thus have gradients of 0.

Reflecting to a common point

If the parabola is a successful model, all reflected rays will meet at a single, common point on the x axis. This will be the point at which the light will focus, or the focal point, if you will. It does not take a genius to understand that the focal point is the solution to the simultaneous equations of two lines representing reflected rays of light.

Let's consider two arbitrary incident light rays.

incident ray: y=1 m=tan(2arctan(-2y)) m=tan(2arctan(-2))=4/3 y=mx+c y^2=x x=1 c=y-mx=1-4/3=-1/3 therefore, y = 4x/3 - 1/3

incident ray: y=-2 m=tan(2arctan(-2y)) m=tan(2arctan(4))=-8/15 y=mx+c y^2=x x=4 c=y-mx=-2+(4*8)/15=2/15 therefore, y = -8x/15 + 2/15

intersect: (Simultaneous equations) y = 4x/3 - 1/3 ...(1) y = -8x/15 + 2/15 ...(2)

4x/3-1/3 = -8x/15+2/15 x = 1/4 substitute into (1) y = 4/3 * 1/4 - 1/3 = 0 Thus, the focal point for the mirror described by y=x^2 is at (1/4, 0)

As this is on the x axis, we have three concordant reflected rays, and have confirmed that the parabola model has a focal point here.

Interpretation

Consider the slightly more complicated parabola:

y = kx^2

A value for k less than one causes the parabola to widen, also causing the focal point to extend further out. We may want this if we'd prefer the focal point to be outside of the parabolic mirror. In exact terms, simple logic tells us that the location of the focal point for a parabolic mirror described by y = kx^2 is:

(0, 1/(4k))

This makes it very easy to know how to construct a parabola to control where the focal point is. Potentially, with a sufficiently large and accurate parabola, one could create focal points many kilometers away. See the next section for applications.

The rest of the pages in this book have been neatly torn out.

On the upper section of the spine of this book, this simple emblem has been stamped: