# Fermant's principle

1. Feb 3, 2006

### UrbanXrisis

There is an elliptical mirror, and this mirror is defined as $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$. I am supposed to use Fermat's principle to show that the light leaving a point (-c,0) is focused to a point (c,0) where $$c=\sqrt{a^2+b^2}$$

Fermat's principle state that light takes the shortest path. What I dont understand is why is the light not reflected when it hits the mirror? Shouldnt it focus back to (c,0)?

2. Feb 3, 2006

### Tide

Light leaving the first focus (in any direction) does reach the second focus and it does that by reflecting off the elliptical mirror. Likewise, if there is nothing to stop the original light rays then they will continue on to reflect off the mirror again and return to the original focus.

3. Feb 5, 2006

### UrbanXrisis

how does this exemplify Fermant's principle?
are you saying that $$c=\sqrt{a^2+b^2}$$ is the focus?

4. Feb 5, 2006

### Tide

I was answering your question regarding reflection. Fermat says that the path taken in going from point A to point B is such that the time of flight is minimized (shortest path).

5. Feb 5, 2006

### UrbanXrisis

so how would this example NOT follow Fermant's principle? I mean, why does Fermant's principle tell me that 1. the ray is not reflected (is it because it is an elliptical mirror? cause a flat mirror would reflect it back?) and 2. if Fermant's principle was not obeyed, would the ray hit the point (c+/-x, 0) where x is some arbitrary number?

6. Feb 5, 2006

### Tide

By the way - it is Fermat (no "n").

The essential result from Fermat relating to your problem is that, at the surface of the mirror, the angle of reflection of a light ray equals its angle of incidence. In other words, can you solve the geometry problem of showing that a light ray from one focus AND reflecting off the elliptical mirror (anywhere!) must necessarily pass through the other focus?

7. Feb 6, 2006

### UrbanXrisis

so the distance that light travles from the first focus to the mirror is $$d= \sqrt{c^2+b^2}$$ while the distance from the mirror to the other focus is $$d= \sqrt{(-c)^2+b^2}$$.

The total distance is then $$d= 2 \sqrt{c^2+b^2}$$

there is another path that the light can take, which would be 2c+2a, which would travel from the focus to a point on a, then back to the focus, crossing (c,0).

now, I know my original equation needs to add 2c to it for it to give the the right answer, but I dont know how to do it. Why must there be another 2c added to the original equation?

that is... $$d= 2\sqrt{(c)^2+b^2}+2c=2c+2a$$ ??? why must there be a +2c for the $$d= 2 \sqrt{c^2+b^2}$$ ?

8. Feb 7, 2006

### Tide

I would approach the problem differently.

First, I'd pick a point P(x, y) on the ellipse and find a unit vector normal to the tangent:

$$\hat n = \frac {b^2 x \hat i + a^2 y \hat j}{\sqrt {b^4 x^2 + a^4 y^2}}$$

Then I would attempt to show that this vector makes the same angle with respect to displacement vectors from each of the foci to P(x, y).