Parabolic Reflector
From Math Images
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Near the bottom of the parabola the perpendicular line is nearly vertical, meaning an incoming beam barely changes its angle after being reflected, allowing it to reach the focus above the bottom part of the parabola. Further up the parabola the perpendicular becomes more horizontal, allowing a light beam to undergo the greater change in angle needed to reach the focus. | Near the bottom of the parabola the perpendicular line is nearly vertical, meaning an incoming beam barely changes its angle after being reflected, allowing it to reach the focus above the bottom part of the parabola. Further up the parabola the perpendicular becomes more horizontal, allowing a light beam to undergo the greater change in angle needed to reach the focus. | ||
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| + | Parabolic reflectors can also work in reverse: if a light emitter is placed at the focus and shined inward towards the parabola, the light that reaches the parabola will be reflected straight out, with the beams of light parallel to each other. Headlights on cars often use this effect to shine light directly forward. | ||
|ImageDesc= | |ImageDesc= | ||
| - | [[Image:Parabdiagram3.JPG|thumb|400px|right| | + | [[Image:Parabdiagram3.JPG|thumb|400px|right|Figure 1]] |
| - | [[Image:Parabdiagram4.JPG|thumb|400px|right|'''A''' represents equal angles: the line normal to the parabola has the same slope relative to the y-axis as the line tangent to the parabola has relative to the x-axis.]] | + | [[Image:Parabdiagram4.JPG|thumb|400px|right|Figure 2: '''A''' represents equal angles: the line normal to the parabola has the same slope relative to the y-axis as the line tangent to the parabola has relative to the x-axis.]] |
The fact that a parabolic reflector can collect light in this way can be proven. A proof follows: | The fact that a parabolic reflector can collect light in this way can be proven. A proof follows: | ||
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::Taking the derivative with respect to x gives the slope of the tangent at any point on the parabola: | ::Taking the derivative with respect to x gives the slope of the tangent at any point on the parabola: | ||
::*<math> \frac{x}{2p} = \frac{dy}{dx} </math> | ::*<math> \frac{x}{2p} = \frac{dy}{dx} </math> | ||
| - | ::The slope of this tangent line is relative to the x-axis | + | ::The slope of this tangent line is relative to the x-axis: when the slope is zero, the tangent line is parallel to the x-axis. The line normal to the parabola has the same slope relative to the y-axis as the line tangent to the parabola has relative to the x-axis, as shown in Figure 2. |
| - | ::We can use this slope to find the angle between the normal and the y-axis, which is the same as the angle between the normal and an incoming beam of light. The desired angle <math> \ | + | ::We can use this slope to find the angle between the normal and the y-axis, which is the same as the angle between the normal and an incoming beam of light. The desired angle <math> \theta</math> can be expressed as: |
::*<math> \tan\theta= \frac{x}{2p} </math> (Equation 1) | ::*<math> \tan\theta= \frac{x}{2p} </math> (Equation 1) | ||
::As mentioned previously, a beam of light is reflected 'over' the normal. This means that the angle the beam of light takes relative to a vertical line is equal to two times the angle the normal makes with the same vertical line. | ::As mentioned previously, a beam of light is reflected 'over' the normal. This means that the angle the beam of light takes relative to a vertical line is equal to two times the angle the normal makes with the same vertical line. | ||
| - | ::We now must show that the direction the light takes after being reflected is exactly the angle needed to hit the focus. | + | ::We now must show that the direction the light takes after being reflected is exactly the angle needed to hit the focus; that is, we want to show that the angle the light travels at after being reflected is the angle needed to hit the focus. |
| + | ::Notice that geometrically, this angle <math> 2\theta</math> satisfies the relationship | ||
::* <math>\tan2\theta= \frac{x}{p-x^2/4p}</math> | ::* <math>\tan2\theta= \frac{x}{p-x^2/4p}</math> | ||
::Which, by a trigonometric identity, also implies | ::Which, by a trigonometric identity, also implies | ||
::* <math>\tan2\theta= \frac{x}{p-x^2/4p} =\frac{2\tan\theta}{1-\tan\theta^2} </math> (Equation 2) | ::* <math>\tan2\theta= \frac{x}{p-x^2/4p} =\frac{2\tan\theta}{1-\tan\theta^2} </math> (Equation 2) | ||
| - | ::We now manipulate | + | ::We now manipulate Equation 1 for <math> \tan\theta</math> to show its equivalence to Equation 2: |
::* <math> 2tan\theta= 2\frac{x}{2p} = \frac{x}{p} </math> , and | ::* <math> 2tan\theta= 2\frac{x}{2p} = \frac{x}{p} </math> , and | ||
::* <math> 1 - \tan\theta^2 = 1 - (\frac{x}{2p})^2 = 1 - \frac{x^2}{4p^2} </math> . Combining these two expressions gives | ::* <math> 1 - \tan\theta^2 = 1 - (\frac{x}{2p})^2 = 1 - \frac{x^2}{4p^2} </math> . Combining these two expressions gives | ||
::* <math> \frac{\frac{x}{p}}{1 - \frac{x^2}{4p^2}} = \frac{x}{p-x^2/4p} </math> | ::* <math> \frac{\frac{x}{p}}{1 - \frac{x^2}{4p^2}} = \frac{x}{p-x^2/4p} </math> | ||
| - | ::Which is the same as the former expression | + | ::Which is the same as the former expression in Equation 2. |
| + | :Therefore, a beam of light will hit the parabola's focus after being reflected. ■ | ||
|AuthorName=Energy Information Administration | |AuthorName=Energy Information Administration | ||
Revision as of 14:18, 17 June 2009
| Parabolic Reflector Dish |
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can be expressed as:
(Equation 1)
satisfies the relationship
(Equation 2)
to show its equivalence to Equation 2:
, and
. Combining these two expressions gives

