![]() My diagram is not fully general as drawn (in terms of the angles), but the argument below is fully general, provided that all angle measures are directed-that is, counterclockwise angles are positive and clockwise angles are negative. Note that regardless of what line the angles are measured from, there is a parallel line through the point of reflection on the mirror for which the angle measures will be the same. So we are increasing (or decreasing, for counterclockwise rotation) angle x by d, but angle z (vector angle) by 2d. Likewise when d < 0 for clockwise rotation. Assuming counterclockwise rotation, d > 0, this "pushes" the reflected line by 2d, one d for a direct push to maintain the angle fo reflection, and another d because the angle of incidence also increases by d, so the reflected light must rotate that much more to keep the two angles of the light equal. Now rotate line X by d degrees around the point of reflection, either way, leaving the original light ray (angle y) fixed and letting the reflected ray (angle z) rotate around the point of reflection to maintain the angle of reflection equal to the angle of incidence. The it's clear that the vector angle, z, of the reflected light is the negative of the vector angle, y, of the light itself: z = -y. Start with the mirror line X horizontal, that is, its angle x = 0. Ok, here's another way to look at it, using vector directions and kinda intuitive, but I think it is pretty close to a real proof. (Haven't checked the degenerate cases, but no reason to expect them to fail). It should be $\pi$ z, so the true formula, in vector direction is z = 2x - y, and that works for all cases. If you do that, then the angle labeled z in my diagram is not the correct vector directional angle. So we really need vectors for the light rays, not undirected lines, and/or the original problem is not altogether well-formed, that is the notion "angle" of a light ray needs to be defined as the direction of its vector. ![]() And then the geometry yields z = 2x - y, or the prior formula if you take the angle of the reflected light as the supplement of z. But when z falls to the left of x, it's because the line of the reflected light intersects the x axis "backwards". When z falls to the right of y, the same formula follows because the roles of z and y are interchanged. Also angle x could > $\frac$, but since the lines are undirected, we may assume that x < $\pi$. OK, this is actually far from a general analysis - angles x, y and z could occur in any order on the x axis, but we may assume that y is to the right of x without losing generality, so there are only two other cases. Then, using the two triangles with the x axis as base and the fact that an exterior angle of a triangle is the sum of the other interior angles, you getĪnd solving those for z in terms of x and y gives So here's the picture, with z the angle you are seeking, and a the angle of reflection. ![]() Since you've stated all three angles in similar terms, and want a formula that works in all cases, lets use the angle with the x axis, in 360 degree terms, which accomplishes both purposes and is good for computation. ![]()
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