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the author
Emmanuel Bigler is a professor 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.
ENSMM, 26 chemin de l'Épitaphe 25030 Besançon cedex, France
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Scheimpflug's rule : a simple raytracing for high
school ?
(Version Française)
Emmanuel BIGLER
Résumé : 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.
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 '60s, 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.
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
Références
Emmanuel Bigler 16 novembre 2002
