# Parabolic Reflector

(Difference between revisions)
 Revision as of 11:41, 17 June 2009 (edit)← Previous diff Revision as of 11:48, 17 June 2009 (edit) (undo)Next diff → Line 18: Line 18: ::*$x^2=4py$ ::*$x^2=4py$ ::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: - ::*$\frac{x}{2p} = \frac{dy}{dx}$ + ::*$\frac{x}{2p} = \frac{dy}{dx}$ ::This slope of the 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 the diagram. ::This slope of the 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 the diagram. ::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 as shown in the diagram. The desired angle $\omega$ can be expressed as: ::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 as shown in the diagram. The desired angle $\omega$ can be expressed as: - ::*$\tan\omega = \frac{x}{2p}$ + ::*$\tan\omega = \frac{x}{2p}$ (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. Notice that geometrically, this angle $2\omega$ satisfies the relationship ::We now must show that the direction the light takes after being reflected is exactly the angle needed to hit the focus. Notice that geometrically, this angle $2\omega$ satisfies the relationship ::* $\tan2\omega = \frac{x}{p-x^2/4p}$ ::* $\tan2\omega = \frac{x}{p-x^2/4p}$ ::Which, by a trigonometric identity, also implies ::Which, by a trigonometric identity, also implies - ::* $\tan2\omega = \frac{x}{p-x^2/4p} =\frac{2\tan\omega}{1-\tan\omega^2}$ + ::* $\tan2\omega = \frac{x}{p-x^2/4p} =\frac{2\tan\omega}{1-\tan\omega^2}$ (Equation 2) - ::We now manipulate our former expression for $\tan\omega$ to show its equivalence to the latter form: + ::We now manipulate our Equation 1 for $\tan\omega$ to show its equivalence to Equation 2: ::* $2tan\omega = 2\frac{x}{2p} = \frac{x}{p}$ , and ::* $2tan\omega = 2\frac{x}{2p} = \frac{x}{p}$ , and ::* $1 - \tan\omega^2 = 1 - (\frac{x}{2p})^2 = 1 - \frac{x^2}{4p^2}$ . Combining these two expressions gives ::* $1 - \tan\omega^2 = 1 - (\frac{x}{2p})^2 = 1 - \frac{x^2}{4p^2}$ . Combining these two expressions gives - ::* $\frac{\frac{x}{p}}{1 - \frac{x^2}{4p^2}} = \frac{x}{p-x^2/4p} + ::* [itex] \frac{\frac{x}{p}}{1 - \frac{x^2}{4p^2}} = \frac{x}{p-x^2/4p}$ ::Which is the same as the former expression for ::Which is the same as the former expression for

## Revision as of 11:48, 17 June 2009

Parabolic Reflector Dish
Solar Dishes such as the one shown use a paraboloid shape to focus the incoming light into a single collector.

# Basic Description

Incoming beams of light perpendicular to the directrix bounce off the dish directly towards the focus.
Note that incoming beams reflect 'over' the line perpendicular to the parabola at the point of contact.

The geometry of a parabola makes this shape useful for solar dishes. If the dish is facing the sun, beams of light coming from the sun are essentially parallel to each other when they hit the dish. Upon hitting the surface of the dish, the beams are reflected directly towards the focus of the parabola, where a device to absorb the sun's energy would be located.

We can see why beams of light hitting the parabola-shaped dish will reflect towards the same point. A beam of light reflects 'over' the line perpendicular to the parabola at the point of contact. In other words, the angle the light beam makes with the perpendicular when it hits the parabola is equal to the angle it makes with same perpendicular after being reflected.

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.

# A More Mathematical Explanation

[[Image:Parabdiagram.JPG|thumb|400px|right|Diagram for proof; note that each label refers to the angl [...]

Diagram for proof; note that each label refers to the angle between the two closest lines on either side of the label, and the three angles are not necessarily equal.

The fact that a parabolic reflector can collect light in this way can be proven. A rough proof follows:

Begin with the equation of a parabola with focus at (0,p):
• $x^2=4py$
Taking the derivative with respect to x gives the slope of the tangent at any point on the parabola:
• $\frac{x}{2p} = \frac{dy}{dx}$
This slope of the 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 the diagram.
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 as shown in the diagram. The desired angle $\omega$ can be expressed as:
• $\tan\omega = \frac{x}{2p}$ (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.
We now must show that the direction the light takes after being reflected is exactly the angle needed to hit the focus. Notice that geometrically, this angle $2\omega$ satisfies the relationship
• $\tan2\omega = \frac{x}{p-x^2/4p}$
Which, by a trigonometric identity, also implies
• $\tan2\omega = \frac{x}{p-x^2/4p} =\frac{2\tan\omega}{1-\tan\omega^2}$ (Equation 2)
We now manipulate our Equation 1 for $\tan\omega$ to show its equivalence to Equation 2:
• $2tan\omega = 2\frac{x}{2p} = \frac{x}{p}$ , and
• $1 - \tan\omega^2 = 1 - (\frac{x}{2p})^2 = 1 - \frac{x^2}{4p^2}$ . Combining these two expressions gives
• $\frac{\frac{x}{p}}{1 - \frac{x^2}{4p^2}} = \frac{x}{p-x^2/4p}$
Which is the same as the former expression for