The Mirror Equation

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• This is a geometric/optic problem related to convex spherical mirrors. In the plane, let (M) be an arc of a circle of center O and or radius R, A be the
Message 1 of 2 , Apr 1, 2002
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This is a geometric/optic problem related to convex spherical
mirrors. In the plane, let (M) be an arc of a circle of center O and
or radius R, A be the midpoint of (M), F be the midpoint of OA. Let P
be a point outside the circle, but on the line OA when produced
outside the circle. Trace an incident ray from P to (M), and let's
say this ray hits (M) at a point N. Trace the reflected ray, i.e. the
symmetric of PN with respect to ON. When produced, the reflected ray
will intersect OA at a point I.

Is the point I well-defined? No, because if you take a different
incident ray, you'll get a different point I. However, assuming N is
very close to A (what I really mean is that the angle AON is
negligible), then I becomes well-defined.

Let do=distance(O,P)
di=distance(O,I)
f=distance(F,A)=R/2

Prove that:
(1/do)+(1/di)=(1/f) (the mirror equation)

I proved that by mistake while I was working on a different (but
related) problem :)
• Sea D=[ (x2-x1)^2 + (Y2-Y1)^2]^(1/2) y sea el punto medio entre dos puntos de la hipotenusa de un triangulo que parte en el origen: P Medio = [ (X2-X1)/2 ,
Message 2 of 2 , Apr 1, 2002
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Sea D=[ (x2-x1)^2 + (Y2-Y1)^2]^(1/2)

y sea el punto medio entre dos puntos de la hipotenusa de un
triangulo que parte en el origen:

P Medio = [ (X2-X1)/2 , (Y2-Y1)/2 ]

sin perdida de generalidad, si deplazamos X1, ya no estara en el
origen y su distancia al origen serÃ¡ de X1, por lo que al punto medio
se deberÃ¡ sumar X1:

(X2-X1)/2 + X1

(X2 + X1 )/2

Si hacemos esto mismo para que el eje Y, tenemos:

(Y2+Y1)/2

Por lo que el punto medio de un plano esta dado por:

P Medio = [ (X2+X1)/2 , (Y2+Y1)/2 ]

Jaime
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