I have the following "Parabolic Dish" z=c(x^2+y^2).

I have to prove that all the reflecting light rays that hit that dish go back through the same point Q in the Z axis, and then I have to find said point Q.

Please help (Speechless)

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- December 14th 2011, 03:23 AMGIPCSmall problem regarding Parabolic Dish z=c(x^2+y^2)
I have the following "Parabolic Dish" z=c(x^2+y^2).

I have to prove that all the reflecting light rays that hit that dish go back through the same point Q in the Z axis, and then I have to find said point Q.

Please help (Speechless) - December 14th 2011, 03:40 AMjensRe: Small problem regarding Parabolic Dish z=c(x^2+y^2)
This is a classic property of parabolic functions. See for example:

Parabola - Wikipedia, the free encyclopedia - December 14th 2011, 03:47 AMGIPCRe: Small problem regarding Parabolic Dish z=c(x^2+y^2)
I know that it's a classic property, that's why our teacher wants us to prove it. I can see it intuitively or when I draw it, but can't seem to think of a concrete proof.

- December 15th 2011, 10:57 AMGIPCRe: Small problem regarding Parabolic Dish z=c(x^2+y^2)
Sorry for the bump.

I'm still stuck with proving it and finding the exact focus point Q

So i've only had to consider rays entering parallel to the z axis...

consider the.cross section where y = 0, it is the parabola z = cx2

a ray parallel to the z axis strikes the mirror at point ( a, ca2 ) at an angle

θ to the normal...

it is then reflected at an angle θ on the other side of the normal

but I couldn't find that θ somehow.

Maybe I should get the gradient of the normal?

I'm not sure how to follow and finish the proof. - December 15th 2011, 03:54 PMjensRe: Small problem regarding Parabolic Dish z=c(x^2+y^2)
I would proceed step by step as follows.

Assume your light ray arrives at the point of coordinates , where is your parabolic function. You know that the tangent at this point has a slope of . So a tangent vector is . Apply a 90° degrees rotation to obtain a vector normal to the tangent. Now, you know that the angle of incidence is equal to the angle of reflection. So you can easily compute the direction of your reflected ray. Say this direction vector is . Now to wrap up, you know that your reflected ray departs from with a direction vector . Where will it cross the axis?

It's pretty simple geometry, you'll manage to solve it by reasoning methodically.