[abonnement gratuit]
the author
Emmanuel Bigler is a professor (now
retired) in optics and microtechnology at ENSMM, Besançon, France, an
engineering college (École Nationale Supérieure d'Ingénieurs) in
mechanical engineering and microtechnology.
He got
his Ph.D. degree from Institut d'Optique, Orsay (France).
E. Bigler
uses an ArcaSwiss 6X9 FC view camera.
Mail

Download the article
in pdf format
Version Française


Scheimpflug's rule :
a simple raytracing
for high
school ?
(Version Française)
Abstract:
We show
here how Scheimpflug’s
rule can be found by an elementary ray tracing procedure
based on common rules of image formation in a thin positive
lens element.
Introduction
Photographic textbooks on Large Format cameras present most of the times
Scheimpflug’s rule without
any proof, even if elementary ray tracing procedures for image formation by
a positive lens (position, magnification) are always presented and sometimes
illustrated in detail.
A good reason is that anybody attempting to restart from Mr.
Scheimpflug’s complex
original patent [1]
would have hard times to use this wellknow large format photographic rule
in practice.
Elegant proofs exist (see Bob
Wheeler’s documents [2]
as well as Q.Tuan Luong’s
web site [3])
however they require a substantial knowledge of highlevel geometry and
mathematics. Considering the minimum knowledge of geometrical optics taught
in high school in France up to the 1960s, I realized that a student of such
a high school, following basic rules of image formation and ray tracing,
could easily find how the image of a slanted object is in fact located in
another slanted direction obeying Scheimpflug’s rule, i.e. the 2 planes intersecting in the lens plane.
So let us, for few minutes, go back to school and apply the basic rules of
geometrical optics.
Scheimpflug’s rule as a consequence of a simple ray tracing
In elementary courses, one of the first ray tracing procedures that students
learn is how to find the image A′B′
of an object AB through a thin
positive lens of focal distance f.
As a starting point this object is always taken perpendicular to the optical
axis. The following rules apply (see figure
1), they are nothing more than what is usually taught in elementary
classes.

any incident ray, parallel to the optical axis e.g.
BH crosses on exit
the optical axis in the image focal point
F′, following the
path HF′B′,

any incident ray crossing the lens through its optical
center, e.g. BO, is
not deflected on output and follows the path
OB′,

all rays emitted by a single (object) light point
A cross each other
in the image A′,
same for B and
B′,

if
AB is an object
perpendicular to the optical axis, its image
A′B′
is also perpendicular to the axis.
Figure 1: Basic objectimage ray tracing according to high school’s rules

Surprisingly, professors hardly ever ask the students to find where the
image of a slanted object like CD
is located (fig. 2).
It is not so difficult, however, simply following the above mentioned rules,
to find CIC′ (rule #1) and
COC′ (rule #2), then
DKD′ et
DOD′; nothing spectacular or
difficult here, however the relationship that may exist between the planes
BCD et
B′C′D′
is still totally unclear.
Figure 2: Scheimpflug’s rule as a consequence of an elementary ray tracing.

One should in fact keep in mind an additional rule, always presented by
professors
rule #5 : in all raytracing of geometrical
optics, one may expand the vertical scale by any factor without changing
anything to the previous rules, as if the lens was unlimited in the
direction perpendicular to its axis.
Thus, a welldisciplined student will plot without any question the
following rays: BLB′, then
BMB′ and
BSB′ and even
BNB′ in total compliance with all
the above mentioned rules, since all rays emitted by the object
B must intersect in the image point
B′.
Now this is just what we need to say:
Let us forget (rule #5) that the lens is
actually limited, and let us consider a ray like
BCDS;
this ray must go through all successive images
B′,
C′
and D′
of light sources B,
C,
and D
according to rule #3, with the special case of
S
being identical to its image.
The conclusion is that the image of the slanted object plane
BCDS is the slanted image plane
SB′C′D′,
in other words nothing but Scheimpflug’s rule[1].
Final remarks

A first difficulty arises from the fact that, in practice,
the fstop actually limits the usable part of the lens to
such a small diameter that it may seem absurd to consider an
“imaginary” ray like BCDSB′.
But when, in practice, starting from a very small aperture
(rays in the vicinity of BOB′) the fstop is gradually opened to allow other
rays like BLB′ to
enter the lens, one does nothing else but obey all basic
rules, those rules being still valid even if we extrapolate
according to rule #5 up to the “imaginary” ray
BCDSB′.

A second difficulty comes from the fact that a photographic
lens is never a single element but a “thick” compound
optical system. In fact, to get a good idea of what happens
in a real lens, one should simply take a pair of scissors
and cut fig.2
along the lens plane HON
and separate both sides parallel to the optical axis by a
distance HH′
(positive for a wideangle “retrofocus” lens or even
negative for a telephoto) equal to the distance between
principal planes HH′
of the system (or nodal planes, they are identical in air).
The “optical thickness“ HH′
of a photographic lens is in the range of a few centimeters.
In practical terms hardly anything will change with respect
to what has been derived for a single thin lens element.
Simply the planes BCDS
and SB′C′D′
will intersect somewhere between the principal planes
H and
H′.

A last difficulty is that we have only considered in this
derivation rays propagating in a plane containing the
optical axis (technically : meridian rays) but what
about a whole 3D object plane? Then again, ray
tracing (fig.3)
will solve the problem. Consider a family of parallel rays
propagating along a rectangular grid plotted in the slanted
object plane, all those rays being parallel to the ray
BCDS. According to
another wellknow rule, namely that parallel rays on input
will all cross on output as a single point
E′ located in the
focal plane, we can easily see that after refraction by the
lens, the locus of all emerging
Scheimpflug’s
rays coming from those parallel incident rays is actually
another slanted plane in 3D space, this plane
intersecting with the plane of fig.2
as the meridian ray SB′.
As a byproduct we can see also that a rectangular grid will
be distorted and rendered as a bunch of converging rays, the
original rectangles being distorted as a trapezoidal shapes.
Figure 3: 3D ray tracing for Scheimpflug’s conjugate planes

References
See other articles on
www.galeriephoto.com
Questions, comments?
