Question 1186320: 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 axus. if the paraboloid has a depth of 1.8in and the diameter on its surface is 6in,how far should the light source be placed from the vertex?
Answer by ikleyn(52803) (Show Source):
You can put this solution on YOUR website! .
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 1.8 in and the diameter
on its surface is 6 in, how far should the light source be placed from the vertex?
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If you are not familiar with the general theory and optical properties of parabolas, you may start reading the lesson
Parabola definition, canonical equation, characteristic points and elements
in this site.
We are given that the diameter of the paraboloid of revolution is 6 inches and its depth is 1.8 inch.
It means that if we consider the parabola as the section of the given paraboloid, it has an equation
y = (1)
Indeed, this equation gives y = 1.8 at x = 3.
For such applications, the standard/canonical form of the parabola equation is
y = , (2)
Then the general theory says that the distance from the parabola vertex to its focus, or focal distance,
is .
So, in our case, the parameter "p" in the equation (1) is equal to 5/2 = 2.5.
OK. Surely, we will place the light bulb in the focus of the parabola to use its optical property,
and we need to determine the distance from the focus to the parabola vertex.
According to the general theory, summarized above, the focus is located at the distance from the vertex.
So, the distance under the question is = = 1.25 inches.
ANSWER. The light bulb should be placed at the distance of 1.25 inches from the paraboloid vertex.
Solved.
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