Double Reflectivity Principle

The "Principle of Double Reflectivity" (not so clearly explained here) arises from the problem of attempting to view a star's angle to the horizon ( to figure out latitude and longitude and "localize" a traveller).

Using Two Mirrors to Tell Angle of Star to horizon.

Using two parallel mirrors as depicted below (black lines are mirrors B and C and the grey lines are imaginary extensions to indicate their parallel geometry used later):

Figure 1: two parallel mirrors

If you rotate the mirror B in order to view a star at point A say as follows:



The star's angle to the horizon (angle 2) is twice the angle of rotation of mirror B(angle 1).




In other words : the total change in the angle of reflection from the rotated mirror is twice the amount of rotation . Specifically speaking : If a body is 70 degrees above the horizon one of the parallel mirrors must be displaced by 35 degrees in order to send its reflected light rays to the eye.

Following is an explanation of why this principle of double reflectivity is true. Suppose you had those mirrors in figure 1 and you were to first draw two "normals" to these mirrors (yellowish lines which are perpendicular to the mirrors) and then rotate the mirror B. We know that these two normals are parallel to each other just like the two mirrors as depicted below.

Now if we were to rotate mirror B above we would see the following configuration where the greyed out line shows the older orientation of mirror B and the faded yellow line shows the old normal of mirror B:

We would have a new orientation of mirror B with a new normal (in orange). We know that the amount of rotation of the mirror is the angle marked 1:

We know that the amount by which the mirror rotates is the same amount by which the normals rotate hence both these angles are marked as equal:

Now we observe that the line EG cuts through the previous yellow normals (two parallel lines) and hence angle EGC is the same as the angle of rotation of the mirror B (opposing interior angles of a parallel line intersected are the same).

So we have shown that angle BGC is the angle of rotation of the mirror B. (I)
Now what we are trying to show is the following :

That the angle of the star at A with respect to the horizon (angle BHC or angle marked 2 in red) is twice the rotation of the mirror B (angle 1 at the mirror B) . We can show this by demonstrating that angle BHC is twice the angle BGC (which is the same as angle 1 at point B or the angle of rotation : see I).

First we look at triangle BCG above.
We notice that
angle (EBC) =  angle (BGC) + angle (BCG) ( since exterior angle is sum of two opposite interior angles of a triangle).

This means that rearranging the above terms gives us:
angle (BGC) = angle (EBC) - angle (BCG)  (II)


Second we look at triangle BCH above.
We notice that
angle (ABC) =  angle (BCH) + angle (BHC) (since exterior angle is sum of two opposite interior angles of a triangle).

This means that re arranging the above terms gives us :
angle (BHC) = angle (ABC) - angle (BCH)  (III)


We also know that in mirrors:
angle of reflection = angle of incidence (law of reflection). (4)

Since the angle of reflection is equal to the angle of incidence in mirrors we see that
angle (ABE)= angle (EBC) (5)
and
angle (BCF)= angle (BCH) (6)

Therefore from (5)
angle (ABC) = 2 * angle(EBC) (7)
and from (6)
angle(BCH) = 2 * angle (BCG) (8)

We plug in (7) and (8) into III above and we get:

angle(BHC) = 2 * angle(EBC) - 2 * angle(BCG)
angle(BHC) = 2 * ( angle(EBC) - angle(BCG) ) (9)

We know from (II) above that angle (BGC) = angle (EBC) - angle (BCG)

So we plug in (II) into (9) to get:
angle(BHC) = 2 *( angle(BGC) )

And since angle(BGC) is equal to the angle of rotation of the mirror (I) we have shown that:
angle(BHC) = 2 *( angle(rotation of the mirror B) )