The flashlight is shaped like a paraboloid

Question

A flashlight is shaped like a paraboloid, so that if its light bulb is placed at the focus, the light rays from the bulb will then bounce off the surface in a focused direction that is parallel to the axis. If the paraboloid has a depth of 4 in and the diameter on its surface is 8 in, how far should the light source be placed from the vertex? (Will you please also help me how does it look like? Thank you very much)

Answer 1

Hi Jang K.,

I'll help you to think a little about the problem, and then you'll get an idea of how to solve it.

Imagine slicing the paraboloid reflector right in half vertically (assuming the beam is "shining" directly upwards). Then look at the curve of the cut edge, as a parabola. You have two points (coordinates in inches): (0,0) (the vertex, at the base) and (4,4) (the first 4 is the depth of the dish, and the second 4 is the radius of the top because the diameter was 8"). Well, you know the general equation for a (centered) parabola: y=kx^2. I think you can determine k from that, and if you don't know where the focus is already, remember that at the vertical position of the focus, light travelling horizontally outwards MUST hit the reflector at a 45 degree angle, so that it can be reflected directly upwards. So you have to consider the equation for the slope of that parabola, and see where the derivative is slope 1 (that's a 45 degree angle, isn't it). That vertical position, but in the middle axis, is where the focus is.

--Cheers, --Mr. d.