trajectory of light as an example of
the shortest path
The minimal problem
Light is known to propagate along straight lines in a uniform medium.
Going from one point to another, it chooses the shortest path.
In this lab, you will answer the question:
What is the trajectory of light that starts at point P and
passes through point Q after being reflected at some unknown point
R on mirror M in between?
To specify the point
R, we apply the principle of the shortest distance. This principle
requires that the sum
PR + RQ
be minimal. This minimum requirement alone will specify the location
of R on the mirror.
We know from daily experience that an observer at Q sees the
reflected light coming from an imiginary reflected source P'
seemingly located behind the mirror. The trajectory of light connecting
P' with Q should be the shortest possible path between these
points, i.e. it should be the straight line P'Q. This line
intersects with the mirror M at point R*. Since
PR* = P'R*, we have
PR* + R*Q =
P'R* + R*Q = P'Q.
This is the shortest distance because P'Q is a straight line.
To prove this, assume the contrary, i.e. that the reflection occurs at
R rather than at R*. For the triangle
P'QR we have the inequality
P'R + RQ > P'Q = P'R* + R*Q.
This inequality shows that the straight line P'Q yields the
shortest possible distance.
An elementary geometric analysis shows that if N*R* is the normal to M at R*, then the angle of incidence
PR*N* is equal to the angle of the reflection N*R*Q
PR*N* = N*R*Q,
and this equality of angles fixes the choice R* of
R. For any other point R on the mirror, the equality of the relevant
angles will not occur
Introduce the function
(T) = PT + TQ
defined as the sum of distances from two fixed points P and Q to a variable point T. Let us fix a
positive number d > PQ and consider the locus of points
T having the same value of (T)
equal to d:
(T) = d.
All such points belong to an ellipse with foci P and Q. If
we choose a number d1 > d, then the equation
(T) = d1 will define a larger ellipse, confocal with the previous one.
If we picked other values of d and plotted the corresponding
ellipses, we would come up with a series of level curves
(T) = const; this chart being a
family of confocal ellipses with focal points P and Q.
Some of these ellipses intersect with the mirror M at two points,
some do not intersect it at
all, and only one ellipse has one common point with the mirror. This selected point is the required
point R* of reflection, and the relevant ellipse specifies the minimum distance PR* + R*Q.
- Let (xP, yP) be the coordinates of
P, (xQ, yQ) be the coordinates
Q, and (x, y) be the coordinates of R.
Without lack of generality, we may assume that the mirror lies along the
x-axis (y = 0). The sum PR + RQ can be
PR + RQ = sqrt((xP -
x)2 + yP2) +
x)2 + yQ2).
This sum (which is a function of x) should be minimal. Let
x* be the value of x minimizing this sum and
R*(x*, 0) be the corresponding
point on the mirror.
Show by direct calculation that
|min(PR + RQ)||=||(PR +
RQ)x = x*|
|=||P'R* + R*Q
(yQ + yP)2)|
- Comparing the first and second solutions, demonstrate the following property of an ellipse: Two straight lines connecting the focal points of an
ellipse with an arbitrary point R on its periphery make equal
angles with the tangent to the ellipse at point R. Your
solution does not need to contain any formulas. A well-worded
description will suffice.
- Based on the second solution, suggest, in words, a procedure for
finding the point
of light reflection from a curvilinear mirror of arbitrary shape.
- Three towns A, B, and C want to build three roads
towards a common transportation center. They want to choose the location
of this center so as to minimize the total length of the roads and, hence,
the construction costs. Formulate this minima problem and solve it
a procedure similar to the one described above in the second solution.
Use the result of Problem 2 to give a complete descriptive
characterization of the desired location R.
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Created by Henry Fink
Last updated: Sunday, September 28, 1997