Society & Culture & Entertainment Hobbies & Science

How Do Flashlights & Car Headlights Utilize Mirrors?

    Reflection

    • Angles in a Plane (Courtesy: library.thinkquest.org)

      Flashlights and cars use mirrors, specifically parabolic mirrors, to reflect backwardly directed light forward in a straight beam. In fact, the backward light is what gives a flashlight or headlight its functionality, since forward light disperses.

      Flashlight mirrors use the property that incoming light is reflected at the same angle. The incident and reflected ray have the same angle to the normal.

    Parabolic mirrors

    • Parabolic Mirror (Credit: Library.Thinkquest.org)

      There exists a shape with the property that light at the shape's focus, if reflected off the shape, would become parallel. This shape, a parabola, reflects light originating from the parabola's focus to parallel lines. Notice the dotted lines in the diagram, showing how the normal lines off the reflecting surface bisect the light paths.

      Rays can go in the opposite direction as well. In flashlights and headlights, the light source is at the focus and rays eventually travel out in parallel. Obversely, light rays entering in parallel are focused to a point. Parabolas therefore are also used in radio antennae and mirror telescopes.

      While the diagram is in two dimensions, the focusing property applies in three dimensions. This is seen by just rotating the ray's plane about the central axis of the mirror.

    Proof of Parallel Reflection

    • Tangent Line and Lengths FP and QP.

      A parabola is the set of all points equally distant from a given point and line. The given point is called the focus, and the line is called the directrix. It can be shown that the algebraic form of a parabola as well as its reflective properties follow from this definition.

      Without loss of generality, orient the directrix and focus so that the vertex of the set of points on the parabola is at the origin and the focus, "F," is located at (0,f). The directrix is horizontal. It is located at y = -f, obviously, since (0,0) is on the parabola and equidistant to F=(0,f) and the closest point on the directrix, (0,-f).

      Equality of distances furthermore means that point P=(x,y) on the parabola is equidistant to F and the directrix. By the Pythagorean theorem, (y+f)^2 = (f-y)^2 + x^2. Reducing gives y = x^2/(4f). It has therefore been proven that parabolas are 2nd-order polynomials.

      To prove the reflective property of parabolas, use the derivative of the algebraic formula. By calculus, the slope of the parabola at x is x/(2f) = 2y/x. A tangent line touching the parabola at P would therefore have an x-intercept at (0,x/2). Why? Because for the tangent line to drop "y," "y = -y. Therefore, by the slope formula "y/"x=2y/x, "x must drop to x/2. So (x/2,0) is the x-intercept of the tangent line of P.

      By similar reasoning, it can be found that the y-intercept of the tangent line of the parabola at P is (0,-f).

      Call the x-intercept G. Call the point (x,-f) Q. Q is on the directrix, y = -f, so P is equidistant from F and Q.

      Q, G and F fall on a line in which G is the midpoint.

      Therefore, the triangles FGP and QPG have sides of the same length and are therefore congruent. This is important because what can be said about the angle QPG can therefore be said about the angle FPG.

      Extending the line QP upward to some arbitrary point T and extending the line GP upward to some point R, the lines GR and QT are crossing lines. Therefore, the angle QPG matches the angle RPT. The angle RPT matches the angle FPG. The line PT is that of an outgoing light ray, traveling vertically. The line FP is the path of an incident light ray from the light source at F to the mirror surface at P. The equality of angles for incident and reflected rays is therefore established for a parabolic surface.

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