c 2000 Society for Industrial and Applied Mathematics
SIAM REVIEW
Vol. 42, No. 4, pp. 727–735
Exploring Reflection: Designing
Light Reflectors for Uniform
Illumination∗
Gary W. De Young†
Abstract. Many students first learn of the law of reflection when they are told about the reflective
properties of conics. Typically students are exposed to a curve or surface, and then its
reflective properties are derived or verified. In this paper we explore reflection by starting
from a desired property of an unknown reflector and proceeding to find a reflector with
those properties. This is done in the context of finding a reflector that produces uniform
illumination of nearby objects. The reflectors considered are those whose analysis can be
reduced to studying curves in two dimensions.
Key words. reflector design, uniform illumination, differential equations, optics
AMS subject classifications. 34-01, 78-01, 78A99
PII. S0036144599358857
1. Introduction. Reflective surfaces are widespread in our society, so much so
that we tend to take them for granted. A child who has made a periscope from
cardboard, tape, and two small mirrors implicitly understands the application of the
basic principle of reflectors. The basic principle is, of course, the law of reflection: the
normal to the mirror bisects the angle formed by the incident ray and the reflected
ray (Figure 1.1).
Calculus students often meet the law of reflection while verifying the reflective
property of conic sections. Frequently, the reflective property of the parabola is stated
(Figure 1.1) and then verified.
Rather than starting with a curve and finding its reflective property, we shall
begin with the reflective property and find a family of curves with that property.
That is, we will work the inverse problem. We shall prescribe a property and ask
what curves satisfy it. The understanding that one gains from verifying or finding
the reflective property of a parabola (or other conics, for that matter) fails to provide a method that can be exploited to develop reflectors with a desired reflective
property.
We develop a general method to solve this problem when the analysis of the
reflecting surface can be reduced to a two-dimensional curve with a point light source
emitting light uniformly in all directions. This includes infinite-length reflectors that
are generalized right cylinders, or “trough” reflectors with a light source that is a
line parallel to the axis of the cylindrical reflector, emitting light uniformly along its
∗ Received by the editors July 7, 1999; accepted for publication (in revised form) May 2, 2000;
published electronically October 30, 2000.
http://www.siam.org/journals/sirev/42-4/35885.html
† Mathematics Department, The King’s University College, 9125 50th Street, Edmonton, Alberta,
Canada T6B 2H3 ([email protected]).
727
728
GARY W. DE YOUNG
N
β α
Fig. 1.1 The law of reflection states that the normal N to the mirror bisects the angle formed by
the incident ray and the reflected ray (α = β). The reflective property of a parabola is that
light rays emanating from the focus are reflected parallel to the axis of symmetry.
length and in all directions. We can consider this an idealized model for designing a
reflector for a long thin straight fluorescent light source.
Our goal will be to design a reflector that uniformly illuminates a flat surface
parallel to a fluorescent bulb. The flat surface may be thought of as the diffusing
panel of a typical fluorescent hallway fixture or a sidewalk below a light source. The
method used can be generalized to include surfaces that can be represented in polar
coordinates.
The approach that is taken relies on vector calculus and is accessible to students
studying third-semester calculus or differential equations. The broad thrust of this
paper is to lead students away from finding solutions to equations, toward finding
equations that describe solutions.
2. From Reflective Properties to Curves. To begin, we introduce coordinates
that place the light source at (0, 0), and we assume that the cross section perpendicular
to the axis of the cylindrical reflector is an unknown curve r(t) = (x(t), y(t)), where
the functions x(t) and y(t) are differentiable and never simultaneously zero. (It would
not work well to have the reflector pass through the light source.)
The vector r(t) describes not only the reflecting curve, but also the direction of
the light ray incident at the point (x(t), y(t)). The reflected direction of the light ray
incident at r(t) will be indicated by the unit vector v(r). We shall assume that the
incident ray, r, never has the same direction as the reflected ray, v, since in that case
the incident ray is not reflected.
The desired reflecting property must be incorporated into v(r); examples and
exercises will indicate how this is done. For the present we assume that v(r) is
known.
The unit vectors r/r and v form the sides of a rhombus (see Figure 2.1). Since
the diagonals of a rhombus bisect the angles of the rhombus, the vector
n(r) = v(r) −
r
r
bisects the angle formed by −r/r and v and is normal to the reflector (see Figure 2.1).
If n(r) = (n1 , n2 ), then n⊥ (r) = (−n2 , n1 ) is perpendicular to n and, more
importantly, tangent to the reflector at r.
EXPLORING REFLECTION
729
dr/dt
v
r
r
n
v
A
Fig. 2.1 An illustration of n = v − r/r. The point A is the light source.
By our assumption that v and r never have the same direction, n(r) and n⊥ (r) are
never zero. Thus, n⊥ (r) is everywhere parallel to r . Then, just as any differentiable
curve can be reparameterized to make its tangent vector r of unit length (by using
arc length as the parameter rather than t), our reflector curve could in principle
be reparameterized to make r everywhere equal to the vector n⊥ (r). Denote this
parameterization by r(τ ). Our reflector curve will satisfy the autonomous differential
equation
(2.1)
dr
= n⊥ (r).
dτ
An initial condition for (2.1) fixes a point through which the reflector must pass.
If we can solve the initial value problem (2.1), the solution will provide a parametric representation of the reflector’s cross section that passes through the point
given by the initial condition and produces the desired reflected rays v = v(r) for all
incident rays r.
It is often difficult, if not impossible, to solve the resulting differential equation
in closed form, but numerical methods can always be used to find an approximate
solution that can be plotted and then used as a template for the reflector.
Consider the simplest case where v = v(r) is a constant, say, v = (1, 0). As we
know, the reflector should turn out to be a parabola opening to the right with focus
at the origin. If r = (x, y) is a general point on the reflector, then
r
x
−y
n=v−
= 1− ,
r
x2 + y 2
x2 + y 2
and (2.1) becomes
(2.2)
dx
y
,
=
2
dτ
x + y2
dy
x
.
=1− 2
dτ
x + y2
This system of differential equations describes the set of all curves that reflect light
emanating from (0, 0) parallel to v = (1, 0).
730
GARY W. DE YOUNG
System (2.2) is certainly not the typical set of differential equations for parabolas.
To see that parabolas are solutions, consider the slope of the reflector at (x, y):
dx
y
dx/dτ
.
=
=
2
dy/dτ
dy
x + y2 − x
Substitution of x = y 2 /(4a) − a verifies that the solution is a parabola with focus at
the origin. (See [6] for an alternative derivation of this differential equation and its
solution by traditional means.)
The important point to note is that we started with the desired reflective property
and found equations that defined a set of curves with that property. In a similar manner differential equations for other conic sections can be derived from their reflective
property.
Exercise: Find (2.1) for a reflector that reflects all light rays emanating from the
origin to the point (5, 2). This is the reflective property of an ellipse with foci at (0, 0)
and (5, 2). (Hint: If (x, y) is a point on the reflector, then v is the unit vector that
gives the direction from (x, y) to (5, 2).)
Exercise: Find the equations that describe a reflector that exactly doubles the
illumination on the line x = 1. This can be done by “redirecting” the light that would
have gone to the point (−1, s) to (1, s). To find v, consider the light ray to (−1, s). It
hits the reflector at some point r = (x, y) and is sent to (1, s). Using similar triangles
it is possible to find (2.1) by finding s in terms of x and y and then computing
v(r).
3. Uniform Illumination. Let’s consider the problem posed at the beginning:
Design a reflector that uniformly illuminates a flat surface parallel to a long fluorescent
bulb. We use the distance from the light source to the flat surface as a unit distance;
thus the flat surface may be idealized as the line x = 1.
Both direct and reflected light must be considered. The reflected light must
compensate for the variation of the direct illumination. Typically in reflector design
either the direct light component is ignored [1, 3, 5, 4] or the reflected light component
is ignored [2]. Sometimes the nonuniformity of the direct light is exploited by adding
several light sources over a surface hoping to compensate for the uneven illumination
of a single light [2].
3.1. Direct Illumination Levels. When considering a linear light source (represented by the origin under our simplifying assumption) that emits light uniformly
in all directions, illumination is measured in radians per unit arc length along the
illuminated surface (see Figure 3.1, right panel).
Suppose the illuminated curve has polar coordinates R = R(θ). Then the average
illumination over a segment is proportional to |∆θ/∆s|, where ∆s is the length of
the arc spanned by ∆θ. We shall assume for simplicity that the constant of proportionality, which has to do with the intensity of the light source, can be normalized
to 1.
Direct
illumination at a point, Idir , is found by taking the limit as ∆s → 0, so
Idir = dθ
ds . Thus, using the formula for arc length in polar coordinates, we have
(3.1)
dθ 1
Idir = = .
2
ds
(R (θ)) + (R(θ))2
731
EXPLORING REFLECTION
∆s
R(θ)
∆θ
s
θ
1
1
Fig. 3.1 Left panel: The polar representation of the illuminated surface, R(θ) = sec(θ). Right
panel: The average illumination over a segment of a curve is given by ∆θ/∆s, where ∆s
is the length of the curve spanned by ∆θ. Illumination at a point is found by considering
lim∆s→0 ∆θ/∆s, which is dθ/ds.
The total direct light falling on an interval is, as it should be,
s2
s2
dθ
(3.2)
ds = θ(s) = θ(s2 ) − θ(s1 ),
s1 ds
s1
which is the angle subtended by the portion of the curve between the points s = s1
and s = s2 .
The direct illumination on the line x = 1 is easily computed from its polar form
R(θ) = sec(θ). Equation (3.1) becomes
Idir = 1
sec2 (θ) tan2 (θ)
+
sec2 (θ)
= cos2 (θ).
The signed arc length s along the line x = 1 starting at (1, 0) is simply the y coordinate, but we shall continue to denote it by s since we want to save x and y for later
use as coordinates of the reflector.
From the right triangle in the left panel of Figure
√
3.1, we see that cos(θ) = 1/ s2 + 1, so
Idir =
1
.
s2 + 1
As expected, the maximum direct illumination on x = 1 occurs at the point closest
to the light source (1, 0) and falls off rapidly and symmetrically on either side.
3.2. Reflected Illumination. The reflected light must compensate for the rapid
fall off in the direct illumination. To produce uniform illumination level K on a line
segment of x = 1, we must have a reflected illumination, Iref , of
(3.3)
Iref = K − Idir = K −
1
s2 + 1
over the line segment. Since illumination levels are additive, the maximum of Idir
on the line segment is the minimum possible uniform illumination. The maximum of
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GARY W. DE YOUNG
1
Iref
−3
Idir
−2
−1
1
2 s 3
Fig. 3.2 To find the length of the line segment that can be uniformly illuminated, the area enclosed
in the box must be 2π. Shown is the maximal range for minimal uniform illumination on
x = 1 symmetric with respect to the brightest location.
1/(s2 + 1) is 1 at s = 0; thus if the point at s = 0 is included in the line segment that
is to be uniformly illuminated, we must have K ≥ 1.
The total illumination that can emanate from our light source is 2π, thus
s2
s2
K ds = (s2 − s1 )K
(Idir + Iref ) ds =
2π =
s1
s1
or (s2 − s1 ) = 2π/K. So the length of the line segment of x = 1 that can be uniformly
illuminated is inversely proportional to the intensity of the uniform illumination. If
the line segment includes s = 0, then K ≥ 1 and the maximum length that can be
uniformly illuminated is 2π. A brighter light source will not uniformly illuminate a
longer segment! Figure 3.2 makes this visually clear for the case of minimal level
illumination (K = 1).
3.3. Reflector Curves. The level of reflected illumination is found in the same
way as direct illumination, thus Iref = |dβ/ds|, where β is the angle along which
the light ray is emitted from the origin. (This assumes that the mirror reflects light
perfectly.) The difference between direct illumination and reflected illumination is
that the angle at which the light ray is emitted from the origin is not directly toward
the final point of illumination.
For the reflected light rays to produce uniform illumination we must have
dβ 1
Iref = = K − 2
(3.4)
.
ds
s +1
Integrating (3.4) leads to an equation for β:
β(s) = ±(Ks − tan−1 (s)) + C.
Note that dβ/ds is of one sign and nonzero (except possibly at isolated points;
e.g., s = 0 when K = 1), so β(s) is an invertible function s(β) = β −1 (s). The reflector
can be described as the curve that sends the light ray emitted at angle β to the point
(1, s(β)). The constant C determines which light ray is sent to the point s = 0, that
is, (1, 0). The sign on the derivative drastically effects the final curve. Although both
curves will have the desired property, one curve generally will result in a reflector that
is undesirable.
In our example it works best to have β move clockwise while s increases, thus
we shall choose dβ/ds = −(K − 1/(1 + s2 )). Once the sign and C are chosen, the
function s(β) is uniquely determined.
EXPLORING REFLECTION
733
In order to find the equations that describe the reflecting curve we need to prescribe v(r). If r = (x, y) is an arbitrary point on the reflecting curve, then v is the
unit vector that gives the direction from (x, y) to (1, s(β)), where β = tan−1 (y/x).
Thus,
n=v−
(x, y)
r
(1 − x, s(β) − y)
.
−
=
r
(1 − x)2 + (s(β) − y)2
x2 + y 2
Applying (2.1) and separating the components gives the following system of differential equations:
(3.5)
(3.6)
dx
y − s(β)
y
=
+
,
2
2
2
dτ
(1 − x) + (s(β) + y)
x + y2
x
dy
1−x
−
.
=
2
2
2
dτ
(1 − x) + (s(β) + y)
x + y2
Since β = tan−1 (y/x), the system is autonomous and the solution of the system gives
the family of curves that have the desired reflective property and can be numerically
computed.
There is one slight problem in trying to compute the solution numerically: s(β) is
a function that is known only as the inverse of β(s). There are two possible solutions
to this problem; the first is to numerically invert β(s) while computing the solution
to the system (3.5)–(3.6), and the second is to add another equation for ds/dτ .
We will take the second approach. Even though the equation is messy, it admits
a number of advantages: (1) it allows use of standard numerical integration packages,
(2) it avoids problems with β = tan−1 (y/x) being a discontinuous function of x,
(3) the image point (1, s(τ )) of the light ray reflected by (x(τ ), y(τ )) is computed
at the same time as the curve is computed, and finally (4) in many cases dβ/ds
cannot be integrated to find β(s) (as we have done), so s(β) can only be found from
ds/dβ = (dβ/ds)−1 . There is one disadvantage to this approach. It may introduce
singularities where dβ/ds = 0. In the case of minimal illumination (K = 1), the
equation for ds/dβ introduces a singularity at s = 0.
Using the inverse function theorem rule for derivatives and implicit differentiation,
ds/dτ can be found:
ds
ds
dβ
=
dτ
dβ
dτ
−1 y d( x )
1
1
=− K−
(3.7)
2
2
1+s
1 + (y/x)
dτ
−1 1
dy
1
dx
=− K−
−x
.
y
1 + s2
x2 + y 2
dτ
dτ
Since dx/dτ and dy/dτ are known to be functions of x, y, and s, the equations (3.5),
(3.6), and (3.7) form an autonomous system. The initial conditions determine a point
on the reflector and location to which that point reflects its light.
We have used (3.5), (3.6), and (3.7) together with Maple to compute two curves
that produce uniform illumination on segments of the line x = 1 and to compute a
template for a reflector that produces uniform illumination on a path.
In Figures 3.3 and 3.4 there are pairs of curves.The left curves in Figures 3.3 and
3.4 are the reflector (x(τ ), y(τ )) and the right curves are graphs of (s(τ ), y(τ )) for
734
GARY W. DE YOUNG
1.5
1.5
y(τ )
y(τ )
(1, s)
1.0
1.0
−1.5
0.5
1.0
x(τ )
0
(s, 0.5)
0.5
0.5
−1.0
−0.5 0
−0.5
−0.5
−1.0
−1.0
−1.5
−1.5
0.5
1.0
1.5
s(τ )
Fig. 3.3 This reflector (left curve (x(τ ), y(τ ))) produces uniform illumination on the line segment
between (1, −π/2) and (1, π/2) (shown). The right curve is (s(τ ), y(τ )), which can be used
to find the final point of the reflected light. For example, the line y = 0.5 intersects both
curves. The intersection point on the reflector (x, 0.5) reflects light to the point (1, s),
where s is determined by the intersection point (s, 0.5) on the right curve. Parameters for
generation are K = 2, x(0) = −1/4, y(0) = 0, and s(0) = 0.
6
y(τ )
6
y(τ )
5
5
4
4
3
3
2
2
1
1
x(τ )
−1
1
s(τ )
1
2
3
4
5
6
Fig. 3.4 This reflector produces minimal uniform illumination on the line segment shown. The left
curve is the reflector (x(τ ), y(τ )). The right curve is (s(τ ), y(τ )), which can be used to find
the final point of the reflected light (see caption of Figure 3.3). Parameters for generation
are K = 1, x(0) = 0.99, y(0) = −0.01, and s(0) = 0.02.
the same range of τ . The abscissa of the right graphs correspond to the final point
of illumination on the line x = 1. Thus, the projection onto the abscissa gives the
interval of uniform illumination on the line x = 1.
Figure 3.5 provides a template of a reflector that uniformly illuminates a path 10
times brighter than the direct illumination directly below the light. The path is 4 feet
735
EXPLORING REFLECTION
1
4
2
2
4
1
2
10
3
4
5
6
2.5
4
Fig. 3.5 A template for a reflector (right, scale 1 unit = 1 in) to uniformly illuminate a path 4
feet wide and 2.5 feet away from a 10-foot wall with the light tube 0.5 feet away from the
wall (left illustration). The illumination on the path is 10 times greater than the direct
illumination of the point directly below the light. Over 70% of the total available light falls
on the path.
wide and 2.5 feet away from a 10-foot wall, and the light tube is 0.5 feet away from
the wall.
REFERENCES
[1] W. B. Elmer, The Optical Design of Reflectors, TLA Lighting Consultants, Salem, MA, 1989.
[2] M. A. Gennert, N. Wittels, and G. L. Leatherman, Uniform frontal illumination of planar
surfaces: Where to place the lamps, Optical Engineering, 32 (1993), pp. 1261–1271.
[3] J. M. Gordon and P. Kashin, Achieving uniform efficient illumination with multiple asymmetric compound parabolic luminaires, Optical Engineering, 33 (1994), pp. 267–271.
[4] J. M. Gordon, P. Kashin, and A. Rabl, Nonimaging reflectors for efficient uniform illumination, Applied Optics, 31 (1992), pp. 6027–6035.
[5] J. M. Gordon and A. Rabl, Uniform and efficient illumination from pair-overlap flux maps
using nonimaging reflectors, Optical Engineering, 32 (1993), pp. 1957–1962.
[6] B. Mueller and R. Thompson, Discovering differential equations in optics, The College Mathematics Journal, 28 (1997), pp. 217–223.