|
Question 1164605: A flashlight shaped like paraboloid, so that if the bulb is focused, the light rays from the bulb is faced 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 1.8 and the diameter is 6 in, how far should the light source be placed from the vertex?
Found 2 solutions by KMST, greenestamps:
Answer by KMST(5345)   (Show Source):
You can put this solution on YOUR website! A paraboloid is a bowl-shaped surface, like the surface swept by a parabola as it rotates around its axis of symmetry.
A cross section "cutting" through a plane containing the axis is a parabola.
If we places the axes with the axis of the paraboloid along the y-axis and the origin at the apex of the paraboloid,
the cross section would be a parabola that we can represent by the familiar equation  , and a graph like the one below:
A paraboloid (and a parabola) go on widening forever, but the reflecting surface of the flashlight is just a finite piece of a paraboloid with a depth of 1.8 inches and a diameter of 6 inches at the open end.
The corresponding graph would follow the parabola, but end at the points (-3,1.8) and (3,1.8).
The endpoints with  and  would satisfy , so
 ,  and 
From there, if we like to remember formulas, we could use the one that says that
the focal distance of a parabola with the equation
is to calculate 
that is the the distance from the focus of the parabola to its vertex,
from the focus of the paraboloid to its apex,
and the distance we need between the light source to the center of the bottom (apex) of the paraboloid-shaped reflective "bowl".
Answer by greenestamps(13327)  (Show Source):
|
|
|
| |
