Sunday, July 1, 2018

Reflection From Two Mirrors


  In Part I, section III of Hamilton's Theory of Systems of Rays he states "...if it be possible to find a mirror, which shall reflect to a given focus the rays of a given system, those rays must be perpendicular to a series of surfaces..." but it is unclear what surfaces he is talking about. Are they surfaces of constant action? Perhaps a simple double reflection problem will help clarify matters.

Let's place a luminous body at the origin and find the positions of reflection from two plane mirrors needed to arrive at some chosen point, r̄3. The planes of reflection can be specified by the position in each plane that is closest to the origin, say, r̄1 and r̄2. The lines connecting these points to the origin will be perpendicular to their respective mirror.


We need the sum of the path lengths to be a minimum subject to the constraints that the reflection points be on the mirror or equivalently that the difference between the reflection points and proximal points is perpendicular to the mirror normals.


Solving for the minimum total path length gives the conditions on the directions of the path segments required for reflection from each mirror.


From the relations connecting the reflected rays with the incident rays we can define two matrix operators, P and Q, associated with the mirrors.


We can verify that the inverse of these operators is the transpose and that their determinant is unity. This tells us the magnitudes of the vectors operated on by these matrices are unchanged. When we look at the action of the operators on vectors perpendicular to the mirror and those parallel to it we see that vectors are reflected through the plane of the mirror.



These operators can be used to find the apparent source of illumination for each mirror, r̄'0 and  r̄''0.


Using a trick at least as old as Hero of Alexandria we can use the apparent sources of light to determine the direction of a line segment and its length in reverse order.


Choosing arbitrary values for the positions of r̄1, r̄2 and r̄3 we can now solve for the unknowns using the method shown above.



Using the basis vectors for the polar coordinates for the position of each mirror we can show that the reflection points are on the mirrors and that the requirements for reflection are satisfied. The fact that the determinant of the matrix containing the direction vectors is not zero tells us that the reflected rays do not all lie in the same plane.

*projections on left should read ρ₁·B₁

Here's a plot of the reflected rays in the y,z-plane.


If the positions of the mirrors, r̄1 and r̄2, are chosen arbitrarily along with the mirror normals one can use êii·r̄i) for the proximal points.

Supplemental (July 1): Instead of finding images for the light source one could determine the mirror images of the target point and then solve for the reflection points in order. Being able to straighten out the bent line segments enables one to solve for the unknowns. The surface of constant action through the target point would have an apparent radius of curvature about the final image of the source.

Edit (July 1): Corrected a minor error. The reflection operator is its own inverse. It is also equal to its transpose.

*edit (Jul 2): typo

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