What is the width of the flashlight's opening?

Question

a flashlight is shaped like a paraboloid and the light source is placed at the focus so that the light bounces off parallel to the axis of symmetry; this is done to maximize illumination. A particular flashlight has its light source located 1 cm from the base and is 6 cm deep.

Answer 1

Let's imagine that the parabola shape of the flashlight is centered on the y axis and has a vertex at the origin.

Hopefully you know that if the light bounces off the mirror parallel to the axis of symmetry, then the light source must be located at the focal point. This is a provable fact of life that you learn in optical physics. This means the focal length of our parabola is 1cm. We now need to use formulas from conic sections.

For a vertically oriented parabola, 4py=x^2. In our case (p=1), they want the width at the top, which is 6 cm from the light source, or 7 cm from the vertex (base).

When y=7,

4*1*7=x^2

x=plus or minus sqrt(28)

 

The width is the distance between positive and negative sqrt(28), or 2sqrt(28).

If you want, you can simplify this to 4sqrt(7).