**L7**** Light**

**Double refraction**** (**Bartolinus** ****1669****)**

**Refraction of light by Iceland
spar **

We return now to the beginning
of our study of optics which led us to refraction of light. Let
the wall refracting the light be isotropic. Refraction through
this wall expresses itself in the law of conservation of the
plane of incidence and the constancy of the ratio of sines .
We now replace the plate of glass by one of made out of a
crystal, in fact - in order to make the phenomena under
consideration especially obvious - by a plate of Iceland spar; if
it has not been processed, it is, as a rule, like a cube (Fig.
677). We cut the plate parallel to one of the crystal's natural
bounding planes, fit them into the shutter in an otherwise dark
room and let from the outside fall a parallel bundle of light *B* on to the
plate. If it were glass, we would see the rays emerge from it and
cause a spot of light on the opposite wall; However, *two *bundles of light come out of the Iceland
spar and we see on the opposite wall two spots *o* and*
ao*. This phenomenon is referred to as the *double refraction of light*. All crystal systems, except for the
cubic ones, manifest this,* **Iceland spar* by far most clearly.

Experiments show: The one ray -
*o* - obeys the law of refraction of Snellius (it passes, perpendicularly incident, unrefracted), in
general, the other - *ao* - does not; it does *not *lie in the plane, determined by the
incident ray and the perpendicular and the ratio (*sin of angle of incidence)/(sin of
angle of refraction)*
does *not* have, in general, for *every *angle of incidence the same value; for
example, in Iceland spar, for every ray *o *at *every** *angle of incidence, *n*_{D
}= 1.658; for the ray *ao*, it lies in
dependence on the angle of incidence between 1,486 and 1. 658.
Hence you say that the first ray has been refracted *ordinarily**,* the other *extraordinarily**. *(You see that, for the
extraordinary ray, *n* is occasionally equal to that for
the ordinary ray [1.658] and both rays are then refracted in the
same way.) How differently from Snellius' law the
two rays behave is shown, for example, by turning the Iceland
spar plate in the shutter about the perpendicular of incidence
(like a wheel about its axle); this does not change at all the
ordinarily refracted ray and its corresponding spot of light on
the wall. In contrast, the extraordinarily refracted ray rotates
about the ordinary ray (*also* its plane of refraction) and the
corresponding sort of light on the wall circulates about that
corresponding to the ordinary ray. If we imagine that the crystal plate has been inscribed like
the face of a clock, you would see: For the ordinary ray, the
position of the plate is not important - whether it is the
position in Fig. 679a or b, but this is not true for the
extraordinarily refracted ray. What changes as the plate is turned from the one
position into the second position and then, continuing to turn it
in the same direction, back into the first position? You are dealing with a *crystal plate**. *If you cut from an isotropic
block of *glass** *a plate, the direction of
cutting is unimportant for its refraction of light. However, for
a crystal - unless it happens to belong to the cubic system - the
angle between that of the cut and the *optical axis* is critical. What is understood by the optical axis
of a crystal? In a doubly
refracting crystal exists a *certain** *direction (in crystals of some
systems *even two*) along which light is only *simply*, that is, *ordinarily**
*refracted. *This** *direction is that of the
optical axis of the crystal. If you rotate the plate from *a *to
*b *(Fig. 679), the incidence of the light changes with
respect to the optical axis of the crystal, whence the spot of
light corresponding to the extraordinary rays circulates about
the other one.

The refraction of light on
optically bi-axial crystals is very complicated, whence we will
only deal with mono-axial ones, which include Iceland spar. What is understood by the optical axis
of Iceland spar? The
crystals of Iceland spar are rhombo-hedra (Fig. 677), that is,
they have six faces bounded by rhombi and have the appearance of
warped cubes. If you make out of equally long bars a model of a
cube, the corners and sides of which have hinges, and deform it
by pressing on the corners *a *and *b*, you obtain
a rhombo-hedron. At *a* and *b **only** *obtuse angles meet (at the
other corners two sharp angles and one obtuse angle). The
straight line through *a* and *b*, also one
parallel to it, is called the *principal axis** *- it is simultaneouly the *optical** *axis - *of the crystal*; a cut through the crystal which
contains the principal axis and every cut parallel to it is a *principal cut*.

**Refraction by an
axi-perpendicular Iceland spar plate**

You cut from the crystal a
plate with parallel faces *perpendicularly
to the optical axis**
*and fit it
into the shutter (Fig. 680). The axis then lies horizontally. The
bundle of light is to fall at arbitrary angles on to the plate;
let the plane of incidence be horizontal. There appear then two
bright spots horizontally side by side on the line of
intersection of the plane of incidence with the opposite wall. If
we rotate the plate about the perpendicular, this action does not
affect either of the bright spots - this is a sign that also the
extraordinary ray lies in the plane and therefore obeys the the
law of Snellius. But it obeys it only in *this* respect. If you change the angle of
incidence and measure every time its magnitude and that of the
angle of refraction, you will find for the *one* ray *always** n*_{ao}*=*1.658,
but not for the other ray; the smaller (larger) is the angle of
incidence - it is the angle between the ray and the *axis*, for the axis is perpendicular to the
plate (cf. above), that is, parallel to the perpendicular - the
larger (smaller) is *n*_{ao}. For the
90º angle of incidence, *n*_{ao }=
1.486. As the angle of incidence (ray/axis) becomes smaller, *n*_{ao
}becomes larger, the extraordinary ray then approaches
the ordinary one and at the *angle
of incidence* 0º it
coincides with it and *n*_{ao}=1.658.

For the -crystal plate,
you have the following geometrical characteristic: Its crystal
axis *always *lies in the plane of incidence, or, in
other words, the *plane
of incidence coincides all the time with a principal cut of the
crystal*. Whenever this
happens - you should remember in the following *whenever**!* - *on**e part* of the law of Snellius applies: The one concerning the plane of incidence, but
*not the second*:* *If the angle ray/axis
changes, also *n* changes - the ratio of the sines.

**Refraction by an Iceland spar
plate with parallel faces**

You now cut from the crystal a plate with
parallel faces *parallel** *to the optical axis and use it, without any other change,
instead of the * *plate, to start
with, in the position when the crystal axis is horizontal, that
is, it lies in the (previously assumed to be horizontal) plane of
incidence. The plane of incidence then coincides with a principal
cut. Their appear *again** *two spots of light *o *and
*ao *on the wall, horizontally side by side. However, if
you rotate *this** *plate about the perpendicular,
one spot wanders - the spot *ao *further away from the
perpendicular and caused by the extraordinary ray - in the sense
of rotation of the plate and, if the plate has been turned so far
that the axis is vertical, that is, has been turned by 90º, it
arrives again horizontally beside the other, but further away
from the perpendicular than at the start of its motion. This
means that the extraordinary ray, to which the moving spot
belongs, is refracted more *strongly*, when the axis lies in the plane of
incidence and less strongly when the* **plane of incidence is perpendicular** *to the axis.

If the two spots lie side by
side horizontally, then both rays lie in the horizontal plane of
incidence. Hence, in this respect (maintenance of plane of
incidence), the two positions - plane of incidence to the axis and to the axis - do not
differ, however, they do with respect to the ratio of the sines.
The experiment tells you: In the first position, the axis lies in
the plane of incidence. We *know** *already from the * *plate (Fig. 566 "whenever"), that then, depending on the magnitude of the angle
(between 90º and 0º) between the incident ray and the axis, for
the extraordinary ray 1.486 < *n* < 1.658. In the
second position, the axis is to the plane of incidence. The experiment tells us
that in this position also for the extraordinarily refracted ray *sin* *angle of incidence/ sin
angle of refraction**
*is* **always** *1.486 for every angle between
0º and 90º. Thus, in this position of the plane of incidence
with respect to the axis of the crystal, the extraordinary ray
obeys completely the law of Snellius. If the angle of incidence is 0º, that
is, the incident ray is to the crystal plate, then the angle of
refraction for the ordinary as well as the extraordinary ray is
the angle of refraction is 0, as demanded by the Snellius law; that
is, both rays leave the plate in the direction of the incident
ray, that is, perpendicularly to the plate and therefore
coincide, that is, only *one*
spot of light appears on the wall. Exactly
the same occurs when the incident ray meets the * *plate perpendicularly - at least *subjectively** *- as we see, that is, it is
the same. However, it is not *objectively** *the same, because in the second case the
refractive index for the extraordinary ray is *equal** *to that for the ordinary ray,
that is, there exist actually only *a single refracted** *ray. - However, in the first case, when it is
for the ordinary ray as always 1.658, it is for the extraordinary
ray 1.486. Actually, there arise *two* rays,
one of which - the extraordinary one - propagates faster than the
other, but the *eye **cannot
distinguish between them*, whence they *seem** *to be a single ray.

In summary: The ray,
extraordinarily refracted by optically mono-axial crystals,
obeys, *in general*, neither of the laws, which the
Snellius law demands for ordinarily refracted rays; however, *under special conditions *- relating to the position of the plane
of incidence with respect to the axis of the crystal - it obeys *both** *or at least *one of them*: It obeys *both*
(maintenance of the plane of incidence and constancy of the ratio
of sines), when the plane of incidence is perpendicular to the
axis, *one *(maintenance of plane of incidence) when
the axis lies in the plane of incidence.

**Positive and negative mono-axial
crystals**

Iceland spar refracts the
ordinary ray more strongly than the extraordinary ray: *n*_{o}
= 1.6585, *n*_{ao} = 1.4864. They same
occurs with tourmaline, corundum, sapphire, emerald, which are
said to be *negatively
mono-axial*. Other
crystals such as rock crystal, zircon and ice refract the
extraordinary ray more strongly than the ordinary one; they are
said to be *positively
non-axial**. *For
rock crystals - except calcite which is a crystal most frequently
employed in Optics - you have *n*_{o} =
1.5442, *n*_{ao} = 1.5533. (Measurements,
first performed by Rudberg 1800-1839 1828 with prisms; the
refracting edge is cut parallel to the optical axis and during
the measurement is placed perpendicularly to the plane of
incidence of the light, because then both rays obey the Snellius law. The refractive index of such a crystal prism for
both rays is measured in the same way as that of a glass prism.)

The
difference between the
two refractive indices is for calcspar (0.1721) much larger than
for *all other** *crystals, whence it allows
most readily observations of double refraction (Fig. 681). You
can see it immediately as you replace the wall in Fig. 678 by the
retina of your eye. For example, if you place the crystal - it
must be very transparent - on the cross (Fig. 682) and then looks
through it as perpendicularly as possible, you will, as a rule,
see the cross doubled; if you rotate it in the process about the
line of viewing, you see that only one cross stays in place,
while the other is displaced, indeed in such manner that the one
cross stays put while the other circulates. In effect, your
observe here that same thing which you have really employed in
Fig. 678 for the description of double refraction.

**Characteristic properties of
doubly refracted light**

If double refraction in all *non-isotropic* substances were so strong that you
could see it as readily as in the case of Iceland spar, one could
detect immediately whether a substance is isotropic or
anisotropic. However, as a rule its is very weak and demands
special tools for its detection. Nevertheless, it is a clear
indicator regarding the presence of isotropy, because the light
after passing through a doubly refracting substance has special
properties, which distinguish it fundamentally from ordinary
light: It is *polarized*. If you send *polarized** *light through a substance, you
observe certain optical phenomena, which indicate whether it
refracts doubly or not (*examination
under polarized light*).
For example, you obtain *polarized** *light when you pass ordinary
light through a calcspar crystal. However, since two bundles of light emerge and
the simultaneous presence of two disturbs, you remove one of
them. The prism of William
Nicol 1768-1851 1839 serves
this purpose and is *one
of the most important optical tools *(Figs. 683/4).

This four-sided prism,
comprising two three-sided prisms, only allows the
extraordinarily refracted ray *ao *to pass and deflects
the ordinarily refracted ray *o* by total reflection at
the plane of contact *bc* of the two three-sided prisms in
such a manner sidewards that it cannot exit; it is absorbed by
black paint covering the prism's side faces. You manufacture the
prism of Fig. 684 out of a natural calcspar crystal. You split
off from it a piece, which is about three times as long as it is
broad, and give it to start with a somewhat different shape. In
the natural crystal, the end faces form with the edges at *a*
and *b *angles of 71º. You reduce them by grinding to
68º and bisect this four-sided prism (along *bc*) by a
cut, which is perpendicular to the main cut (here the plane of
the drawing) and simultaneously perpendicular to the ground faces
*ae* and *gd*; you then polish the cut faces *bc*,
stick at them the two halves of the prism together with *Canada balsam* and obtain Nicol's prism.
Fig. 684 displays the course of the rays in it. For yellow light,
the refraction index of calcspar (ordinary ray) is 1.658, that of
Canada balsam 1.536. Hence the ordinary ray transits at the
balsam layer *bc* from an optically denser into an
optically thinner material. The angle of total reflection is
68º. Hence all rays are totally reflected at the balsam layer,
the angle of incidence is larger than 68º. A computation
confirms: If the rays enter parallel to the prism's edge (or in a
direction which does not deviate too much from that direction),
the angle of incidence *i* at the Canada balsam layer is
larger than 68º. The refractive index of the *extraordinary *rays is smaller than that of the balsam
layer, whence they are allowed to pass. If you place Nicol's
prism instead of the earlier employed plate into the shutter, so
that the edge *k* is horizontal and perpendicular to the
shutter, and let a bundle of light *mn* meet it parallel
to *k*, then there forms on the wall only *one* spot which belongs to the *extraordinary* ray. - We will return to Nicol's prism
below when dealing with the polarization phenomena of light.

We cannot deal further with the
complicated conditions of double refraction. It becomes a more
mathematical than physical problem. It is sufficient to point out
that you can find geometrically the two refracted rays on the Fresnel *wave surface* for every crystal - single or double axial - for every
given incident ray of light. - While the discussion of double
refraction follows logically to that of simple refraction, it is *only** *encountered again during that
of polarization of light. Until then, we will only be confronted
with *simple** *refraction.

**Refraction of light by spherical
surface**

**Geometrical relations between
distances of objects and images from the refracting surfaces**

The refracting wall which light
encounters during
its spreading and which affects its
further progress was to be *plane*. How does a *curved* wall affect
its progress (Fig. 685a)? This
question takes us to the task of *forming an image** *like with a camera or projector and of *supporting** *the eye by auxiliary tools
like spectacles, a magnifying glass, a microscope and a
telescope. In these instruments, spherical surfaces are employed
most often and we will confine our study to them.

We will resume the
approach,
which we adopted when we were dealing with the reflection of
light by spherical surfaces. Let *L *on the line *LC *through
the centre *C* of the surface be a point source of rays -
the object point. We follow the ray which encounters at *P *the
spherical border *PS *between glass and air. The radius *CP
*is the perpendicular at *P*, the angle of incidence the angle *i*.
The ray remains after refraction in the plane of the drawing; it
is refracted *towards *the axis by the angle of refraction *r*
and intersects it at *L'. *If you imagine that the drawing
has been rotated about *LC *as axis, you see that all rays
on the conical surface (with axis angle *PLS)*, generated
by the rotation, pass on refraction through *L'*, which
you call the *image
point* of *L. *You
say that *L *and* L*' are *conjugate points*. Since *PS *is a spherical
surface and *n* sin *i *= *n*'·sin *r*,
where *n* and *n*' are the *absolute* refraction indices of air and glass, we
can find the position of *L*' with respect to *S*.
- In the triangle *PCL*:* LC*/*l* = sin *i*/sin*j*, in the triangle *PCL*': *L*'*C*/*g*
= sin *r*/sin*j*,
whence

(*LC*/*L*'*C*)·(*g*/*l*)
= *n*'*/n* (since sin *i*/sin* r = n*'*/n*).

Now assume that *P *is so
close to *S* that we can substitute *l *for *LS *and*
g* for *L*'*S* and obtain

(*LC*/*L*'*C*).(*L'S*/*L*S)
= *n*'*/n*. (1)

If we can consider *SP** *to be small, the opening of
the bundle of rays is so small that *all *its rays are
approximately perpendicular to the spherical surface (like
in the case of a reflecting surface). Equation (1) applies subject to this
restrictive assumption. It holds, when light encounters a concave
refracting surface, as well as when it enters air from glass
(hitherto glass from air), that is, *n* corresponds to a *more strongly *refracting substance and *n*' to
a less strongly refracting medium. This is demonstrated by Figs.
685 a,b,c,d, where *L *is the source of light, *C*
the centre of the sphere, *S *the vertex of the refracting
spherical surface, *LP *the incident ray, *CP *the
perpendicular for the incident ray, the arrow *towards P* the ray, the arrow *starting *at *P *the ray after refraction
and *L*' the *image
point** *- the
intersection of the refracted ray with the axis through* L *and
*L*'. If the ray is refracted *towards the axis *(*a* and *d*), it itself crosses the axis *behind *the refracting surface (real), if it is
refracted *away from the
axis *(*b* and *c*),
its backwards extension intersects the axis, and indeed *in front of** *the refracting surface
(virtual). In all four cases, *n *is the refractive index
of the substance from which *comes
*the light, *n*'
that of the substance into which it *enters*.
In Figs. 865 a,c, *n*' > *n* (path of light from
air into glass), in Figs. 865 b,d, *n*' < *n *(
path from glass into air). In all four cases, *i* is the
angle of incidence, *r *the refractive angle, that is,
Equation (1) applies. Hence you find the distance *L*'*S*
of the image point from the vertex from the other quantities in
all cases *from the same
equation*. But the
image point *L*' lies either on the right or on the left
hand side from the vertex. In order to be able to predict where
it lies with respect to the vertex, you must know *at the same time** *whether the distance is to be
computed towards the left or the right of the vertex. (Similarly,
it is not sufficient to know: *A temperature lies at a certain distance from zero*; you must know simultaneously whether
it lies above or below zero.) We will denote again the contrast
between **+** and **-**,
employ again the *vertex
**S* as
reference point and compute distances, starting from it, *as negative** *to the left hand side and *as positive** *to the right hand side. If we
denote the distances *L*'*S* by *a*', *LS
*by *a*, *CS* by *r*, then, for the case of Fig. 685a, (-*a *+ *r*)/(+*a*' - *r*)·(+*a*'/*-a) = n*'*/n*
and similarly for *every** *case of Fig. 685, provided you
attach to each distance the appropriate sign (+/-) relative to *S*
(as, for example, is shown by the case of Fig. 685d). An
elementary algebraic conversion of the left hand side of the
formula yields for *each* case: (*n*'*/a*'*) - *(*n*/*a*)
= (*n*' - *n*)/*r.*

**Application of the formula for
refraction by spherical surfaces**

For the sake of
simplicity, replace the absolute refractive indices by their
relative values: If *n *is the absolute refractive index
of air, *n*' that of glass, then the refractive number from air into glass is *N = n'/n*. Denoting the
radius by *r*, you have

*N*/*a*' - 1/*a = *(*N*
- 1)/*r*.

Assume for *N* the value 3/2. If, as in
Fig. 685a, the spherical surface is convex towards the left hand
side, that is, the centre is *on the
right hand side** *of *S, *and the
radius is 50 cm, then *r* = + 50. If the object point *L*
is on the *left hand side *of
*S *at a distance of 200 cm, then *a *= -200. You
have now for the image point of *L*:

(3/2)/*a*' - 1/-200 = [(3/2) - 1]/(+50)
or 3/2*a*' + 1/200 = 1/100,

whence *a*' = +300, that is, *L*'
cm on the right hand side of the vertex. If *L *is only 50
cm from *S*, then

3/2*a*' + 1/50 = 1/100, whence *a*'
=*-*150,

that is, *L*' also lies, like *L*,on
the left hand side of *S*. Since it lies on the same side
of the vertex as the object point, it is virtual; not the
refracted rays intersect themselves, but their backwards
extensions. If the spherical surface is concave to the left, then
*r = -50 *and *a* = -200, whence

(3/2)/*a*' - 1/-200 = [(3/2) - 1]/(-50),
whence *a*' = -150,

that is, the image point lies also on the left
hand side of *S *and, as in the last numerical example, is
virtual.

It is often
convenient to rewrite the formula (*n*'*/a*'*) - *(*n*/*a*)
= (*n*' - *n*)/*r *and introduce focal lengths. If the
object point *L *lies infinitely far away from the
refracting surface, that is, *a *= , then the rays meeting it are parallel to
each other and the

axis. The image
point at which these rays (parallel before refraction) intersect
after their refraction - *F' *in Fig 686 - on the left
hand side is called the *rear
focal point *of the
refracting surface. Its distance *a*' from the refracting
surface - denoted by *s* '-
follows from *n*'/*a' - n*/*a = *(*n' - n*)/*r.*
Since *a *= ,
*n*/*a = *0, that is *a*' = *n*'·*r*/(*n'
- n*) *s*'.

In contrast, if the *image *point
forms on the axis at an infinite distance from the surface, that
is, if *a*' = , the rays *after
*passage through the
refracting surface are parallel to each other and the axis. The
object point from which these rays - parallel after refraction -
come prior to their refraction - *F* in Fig. 686 right -
is called the *front
focal point* of the
refracting surface. Its distance *a* from the refracting
surface - denoted by *s *-
is, since *a*' = , that is, *n'*/*a' *= 0,

*a = -n*·*r*/(*n'
- n*) *s*.

You employ the *focal lengths **s* and *s*' to rewrite the equation *n*'/*a' - n*/*a
= *(*n' - n*)/*r.* After multiplication of both
sides by *r*/(*n' - n*), you obtain *s*'/*a*' + *s*/*a* = 1.

The distances of
the focal points from the refracting surface depend only its
radius of curvature *r* and its refractive index. If you
employ again instead of the absolute refractive numbers *n *and
*n'* their relative values, you find *s*' = *N*·*r*/(*N - *1)
and

*s *=* -r*/(*N - *1). For
example, letting *N* = 1.5 and *r* = 3 cm, so that
the refracting surface is convex (Fig. 686), then *s*'=1.5·3/0.5 = 9 cm, *s* =-3/0.5 = - 6cm. If the refracting
surface is concave, that is, *r* = -3cm, then *s*' = 1.5·(-3)/0.5 = -9 cm, that is, the
rear focal point lies on the same side of the refracting surface
as the infinitely far away object point: It is *virtual*, the backwards extensions of the
refracted rays intersect at the focal point (Fig. 687 left).
Moreover, *a *= +6 cm, that is, in order to become
parallel after refraction, the rays must must converge to a point
lying behind the refracting surface (Fig. 687 right)

* Hence the *rear** *focal
point of the *concave *refracting surface lies (in
agreement with the definition) in the sense of the motion of the
light *ahead** *of , the *front** *focal point *behind *the surface.

**Elementary bundles of rays of
light**

The formula *n*'/*a' - n*/*a = *(*n'
- n*)/*r *interrelates the positions of the object and
image points relative to the refracting plane. For its
derivation, we have assumed that the angle between the ray *LP *and
the axis is so small, that the rays arrive almost perpendicularly
to the spherical surface. O*nly*
then is this formula valid.
However, if *it is true**,* it applies to every ray of
the cone of rays starting from *L, *because the rays
internal to the cone form yet smaller angles with the axis. - A
bundle of rays with so small an opening will be called an *elementary bundle*. If, as in the present case, its *principal ray** - *the ray *LC - *which
you can conceive to be a centroidal axis, meets the sphere
perpendicularly, the entire bundle is almost perpendicular to the
sphere (*normal** *incident elementary bundle).

We now understand:
All the rays of a normally incident elementary bundle of rays
pass after refraction through a common point on the principal
ray, that is, the bundle is also after refraction *homo-centric*; the point *L *is *mapped** *into the point *L*'.

**Mapping of infinitely small
objects**

What is true for
the line *LC* through the centre of the sphere, is also
true for *every** *other line through *C*:
Every point which lies on such a line at a distance *a* in front of the
refracting surface, is mapped into a point behind the refracting
surface at a distance *a*' from the corresponding vertex.
Hence, if all points *a*_{1}*, a*_{2, }···
*a*_{n} are at the same distance* a*
from the refracting surface (Fig. 688), their images *a*'_{1}*,
a*'_{2, }··· *a*'_{n} also
have the same distance from the corresponding vertex. Hence, if
the *object points** *are located on a spherical
surface, which is concentric to the refracting sphere, also their
images *a*'_{1}*, a*'_{2, }··· *a*'_{n}
lie on such a sphere. If we consider only points in a plane -
those lying in the plane of the drawing - we understand that
points, which lie on a *circular* arc, are again mapped onto a *circular* arc. However, a piece of a spherical
surface which is very small, can be viewed to be a plane, a piece
of a very short circular arc a straight line, whence we conclude:
The element *oo* *of
a plane**, *perpendicular
to the axis *LC, *is mapped into a perpendicular element *o*'*o*'*
*of a plane. We see in the drawing only the *two lines* one of which is the image of the other;
you must imagine the drawing being rotated once about the axis *LC*,
in order to survey the process in *space*.

Briefly speaking:
Under the known condition regarding the opening of the incident
bundle of rays and the expansion of the object to be mapped -
understanding by object an infinitely small line or area element
- every *point* which lies on the axis (through the
sphere's centre) is mapped into a *point*,
which also lies on the axis, and a *line* or *plane** *perpendicular to the axis into
a *line** *or *plane** *perpendicular to the axis.

**Refraction through a centred
sequence of surfaces**

The formula *n*'/*a' - n*/*a = *(*n'
- n*)/*r *only relates to refraction in *one *plane. In reality, you deal almost always with two - a *lens *bounded by two spherical surfaces. What happens to a bundle of rays which
after passing through the first spherical surface encounters a
second, third, etc. surface? We will employ again the absolute refraction index and
denote it for the first substance by *n*, for the second
by *n*', for third by *n*'', etc. (Fig. 689).Later
applications (optical instruments) allows us to assume that the
centres of the spherical surfaces lie *on a straight line **NN*. Such a *refracting system *is said to be *centred*. Moreover, we will assume that we are
dealing again with an elementary bundle which starts from a point
*L*_{1} of the axis. This bundle remains also
homo-centred after refraction at the surface * 1*
and yields the image point

A *lens *is formed by a light refracting
substance bounded by any pair of spherical surfaces. We are here
only concerned with *glass*
lenses, used in optical
instruments. There are six different shapes for which pairs of a
convex and concave surface and a plane bound the substance; the
plane is conceived as a spherical surface with infinite radius.
The terminology is shown in Fig. 690.

Let a point *L* on the axis (Fig. 689)
emit an elementary bundle to the *lens*.Where lies the image point *L*',
conjugate to *L*? It
also lies on the axis. The general lens formula tells us
whether it lies to the right or to the left hand side of the
lens. Assume to start with that the lens is so thin that its
thickness can be neglected in comparison with other distances,
that is, it is *infinitely
*thin. "This is
admissible for a preliminary approximate estimate of the action
of an optical system" (Siegfrid
Czapski 1861-1907).
Accordingly, we measure from the vertex of the first bounding
surface *also** *the radius, the image and
object distances which relate to the *second *surface.
Denote by *a* the distance of the object point from the
first refracting surface, *a*' the distance of the image,
which arises due to refraction, and *r*_{1}* *its
radius; then

*n*'/*a*'*-
n*/*a = *(*n*'* - n*)/*r*_{1}.
(1)

This image at the
distance *a*' from the vertex *S *is now the *object** *for the second surface. Since,
by assumption, the vertices of these two surfaces coincide, the *object** *point has from the second
surface the distance *a*'. Let *b *be the distance
of the* **image**, *which arises through
refraction at the second surface, and *r*_{2} the
radius. The light leaves the substance with the refractive index *n*'
into the substance with the refractive index *n*",
whence *n*"/*b- n*'/*a*'* = *(*n*"*
- n*')/*r*_{2}. Since *n" = n - *the
lens is bordered on both sides by air! - this equation becomes

*n*/*b-
n*'/*a*'* = *(*n*'* - n*)/*r*_{2}.
(2)

Addition of equations (1) and (2) yields

*n*/*b-
n*/*a = *(*n*'* - n*){1/*r*_{1}
- 1/*r*_{2}}, that is, 1/*b- *1/*a = *{(*n*'*
- n*)/*n*}{1/*r*_{1} - 1/*r*_{2}}

and with *n*'/*n
*= *N*, the *general
lens formula** *becomes:

1/*b- *1/*a
= *(*N - *1){1/*r*_{1} - 1/*r*_{2}}.

This formula
becomes simpler by introduction of the focal points of the lens -
note: of the lens! The same argument and notation as before for
the surface now yield the focal points of the *lens*. If the source of light is infinitely
far away - *a* = - that is, the rays reach the lens parallel to each
other and the axis, its image arises at the *rear focal point of the lens**. *All the rays then pass behind
the lens (for diverging lenses extended backwards!) through that
point on the axis, the distance *f* of which (since 1/*a*=0)
is given by 1/*f=*(*N-*1){1/*r*_{1}-1/*r*_{2}},
whence you can rewrite the general formula in the form:

1/*b* -
1/*a = *1/*f*.

The front focal
point of the lens is the point *F* on the axis from which
the rays must emit (for concave lenses: to which the rays must
aim!) so that they only intersect behind the lens at an infinite
distance from it (Fig. 691). The planes perpendicular to the axis
through the focal points are called the *focal planes*. The distance of the front (back) focal
point of the lens from the first (second) *principal plane** *is called the frontal* *(rear)
focal *width*. For infinitely thin lenses, the focal
width and distance of the focal point are the same.

The general lens
formula yields directly the equations for each of the six types
of lenses (Fig. 690), you must only be sure to take the signs into account.
If the *object point **lies to the left of the vertex* on
the axis, *a* is given a negative sign. If the lens is
biconvex, *r*_{1}*
*is positive*, r*_{2}
is negative, if it is biconcave, *r*_{1}* *is negative, *r*_{2} is positive, whence
for infinitely thin. biconvex/biconcave
lenses

1/*b *+ 1/*a = *(*N* - 1){1/*r*_{1} + 1/*r*_{2}},

which becomes after introduction of the focal length

1/*b *+ 1/*a = f.*

**Principal planes and principal
points****
**(Gauss)**. ****Nodal
points ****(**Johann Benedict Listing 1808-1882 1845**)**

You obtain also
for lenses of finite thickness a clearly set out formula, if you
do not measure the the distances (of object, image, focal points)
from the *vertex* of the refracting surfaces, but from
the *principal points** *of Gauss - the two points at which the two *principal planes** *intersect the axis. Fig. 690 shows for each of the six types of
lenses the locations of their principal planes and principal
points. The corresponding formulae will not be presented here,
but the geometrically special position of the principal planes
will be discussed, as it assists to follow rays of light through
a lens and to find for a given object its image.

Consider a biconvex lens with focal
points *F *and *F*' (Fig. 692). A ray of light *S*,
which meets parallel to the axis the first surface, passes after
refraction at the second surface through *F*', that is,
the ray *S *and the ray through *F*' are *conjugate*. The point of their intersection is at *K*'*.*
Moreover, imagine there is a second ray *S*' in the same
plane of incidence as *S *(the plane of the drawing),
whicharrives parallel to the axis from the opposite direction at
the same height over the axis, that is, in the extension of *S*.
It passes after refraction through *F*. The two rays
intersect at *K. *Now apply to the ray *S*' the
principle of the invertibility of the paths of rays, that is,
imagine it to start at *F*; this does not alter the
geometry. (It has then the direction of the double arrow.) Consider now the points *K *and *K*'.
*K *is the point of intersection of *S*' with the
ray through *F*. However, it is also - and this is now *important **-* the point of intersection of *S*
with the ray through *F*. Similarly important is that *K*'
is the intersection of *S*' with the ray through *F*'.
However, we have already seen that the ray *S* is
conjugate to the ray through *F*' and that *S*' is
conjugate to the ray through *F*. Hence *K*' is the
point of intersection of two rays which are conjugate to those
coming from *K*. This characterizes *K*' and *K *as
the conjugate points, that is, when *K* is an object
point, then *K*' is its image point and inversely. The *image *and *object points lie here on a line parallel to the
axis**. *We have
selected arbitrarily the distance from the axis, at which the
rays *S* and *S*' run parallel to the axis, and
have considered only the case of the plane of incidence, which is
identical with the plane of the drawing. However, what applies to
the points *K *and* K*' is also valid for each
point of the two lines, which you draw from them perpendicularly
to the axis. Moreover, what applies to the plane of incidence -
identical with the plane of the drawing - applies to every plane
of incidence *AA*' through the optical axis. If you rotate
the drawing once completely about the axis, the lines through *K
*and *K*' describe two planes in which pairs of points
correspond like *K *and *K*'. This means that every
point of the one plane has its image opposite to it in the other
plane, and indeed at the same distance from the axis on the line
through it, which is parallel to the axis. - These planes are
called *principal planes**, *and the points *H* and
*H*', at which they are intersected by the axis, *principal points*. It can be shown that every optical
system has only two principal planes.

The focal points
and the principal points are important for the geometry and
computation of an optical system, whence they are called *cardinal points*. Given the focal points of the lens and the position of
an object, Listing's *nodal
points also** *assist
with the discovery of the geometrical orientation of an image. *N
*and *N*' are the nodal points in the biconvex lens
(Fig. 693). They are characterized by the following geometrical
properties: If a ray is aimed before refraction at the point *N
*on the first refracting surface, it continues *after *refraction at the second surface of the
lens parallel to the direction of incidence, and, indeed, as if
it was coming from the second nodal point. - The *nodal* and *principal points *coincide,
when the first penetrated substance and the last such substance
are the same, *what is
usually the case for the lenses in optical instruments*, which are on *both sides *surrounded by air. For this reason, we
will not give details of the nodal points. The eye is an
exception, since for it air is the first, but not the last
substance.

**Images constructed geometrically
with the aid of the cardinal points**

The following figures demonstrate the
significance of the principal planes and nodal points combined
with the focal points as geometrical auxiliary tools. Fig. 694a
shows a convex lens, *E* and *E*' its principal
planes, *K *and *K*' simultaneously their node
points, *B*** **and *B*' their focal
points and *OP *an object. In order to construct the image
of *OP* - to understand geometrically
where it lies *in the case that it is formed *- for example, to find the image point belonging
to the object point *O, *you can employ three of the rays
coming from *O *(two are sufficient!), the paths of which
you can construct geometrically from the object to the image with
the aid of the six cardinal points*. *The three rays are:

ray |
before refraction |
property |
||||

1. | 1 |
" | parallel to axis | |||

2 | 2 |
" | through front focal point | |||

3. | 3 |
" | towards the first nodal point |

About the ray * 1*, we know, firstly,
that it passes behind the lens through the focal point

The construction becomes
simplest when the lens can be considered to be *infinitely
thin**. *The
thinner it becomes, the closer approach the principal points each
other and the optical centre *C*. If it is infinitely
thin, these points coincide and the main principal planes
collapse into the symmetry plane *MM' *through the optical
centre *C, *perpendicular to the optical axis. For such a
lens (Fig. 694b), if you want to construct an image point for a
given object point, you must only keep in mind (the notation is
the same as in Fig. 694a):

The ray *I' *passes
behind the back focal point and indeed from the *same* point of the plane *MM*,
at which the ray *I *has intercepted it - the same,
because the two principal planes if Fig. 694a coincide in Fig.
694b. The ray *2'* proceeds after the lens parallel to the
axis and indeed from the same point of the plane *MM* at
which the ray *2* has intersected it - for the same
reason. The ray *3'* is the extension of the ray *3*,
that is, *3 *continues through the optical centre *C*,
because *C *joins both nodal points. The formerly* **parallel* displaced rays *3*
and *3'* become now one continuous ray through the single
remaining nodal point.

Figs. 695 display the construction of the image which
disregards the thickness of the lens, *a* and *b *for
the* *biconvex*, c* for the biconcave lens, and
illustrate the formulae
for the two lenses.

For the biconvex lens (Fig. 695a), you find: If the object
lies in front infinitely far away from it (*a* = ), its image is located
at the posterior focal point *F'* of the lens, is *real**,** inverted *and* **smalle**r *than the object.
As the object approaches the lens, its image distances itself
from it and grows in size. If the object has approached the lens
so that its is only twice the focal distance from it (*a*
= 2*f*), then also *b *=* *2*f* and
the image is* **also**
*two focal distances from the lens and as large as the object
(Fig. 695a: Object 5 and image 5'). When the object reaches the
frontal focal point *F*, its image is infinitely far away
from the lens. If* *the object comes* *closer to
the lens beyond the focal point *F* (Fig. 695b), no *real* image whatsoever arises, but
a virtual one* *which is *upright**,*
*enlarged *and lies on the
side of the object. The biconvex lens then *enlarges*
(magnifying glass).

You observe for the biconvex lens (Fig. 695c): If the object
is infinitely far away from the lens (*a* = ), the image is at the
back focal point of the lens - on the side of the object - *virtual**, **upright *and *smaller
*than the object; as it approaches the lens, also the
image comes closer to the lens and grows in size, but remains
virtual, upright and smaller than the object. For the biconvex
lens, the *front *focal point
lies behind the lens, the rear focal point in front of the lens.
(Fig. 687).

However, these are only drawings for the
construction, which tell you where and how the images lie and
their relationship to their objects, *if *an image can be
realized, however, they do not give information regarding their
realization. In fact, we have all along only spoken of the
mapping of *infinitely
small *objects and of *infinitely narrow** *bundles of rays, facts which have not been
taken into consideration. We will return to this aspect below.

Everyone knows these lenses;
the biconvex lens as burning glass, magnifying glass, also as the
large lens of an theatre binocular, the biconcave lens as the
small lens of the ordinary theatre binocular. Fig. 695a indicates
that the biconvex lens causes the rays, emanating from objects
within reach to generate a real image, beyond the lens - *on the side of the image* (in contrast to *on the side of the object*) unless, like in Fig. 695b, the object
lies closer to the lens than the focal point; they *converge *to *real** *points of intersection. In
contrast, the biconcave lens causes the backwards rays, that is,
the prolongations towards the object of the rays which originate
from objects within reach, to intersect. Also the refracted rays *appear** *therefore to come from a point
*on the side of the
object*; they *diverge *from *virtual**
*points of intersection. Hence you can intercept with a
biconvex lens the image on a screen (ground glass plate) and
display it like an image; in contrast, through a biconvex lens,
if you place your eye in the *direction of the rays*, you can see the image (as in Fig.
695c), but you *cannot** *intercept it on a screen.

In order to see how differently
the two lenses behave towards incident rays, let the sun's rays
fall on a sheet of paper and place in their path *before they reach the paper *one time a biconvex lens, another time a
biconcave lens. If you place the former one at a suitable
distance from the paper, you will see: **1. **The
lens, although *transparent*, creates a circular, strong *shadow** *and **2. **its
centre (strictly speaking a very small circular disk about its
centre) is intensely bright. Strong heat develops there and the
paper ignites (focal point). The lens creates a shadow because
all rays of light meeting it are directed towards the focal
point, whence the points behind the lens do not receive anything.
The bright small circular disk is an image of the sun. - If you
place the biconcave lens between the sun and the paper, the
brightness of the paper is somewhat reduced, because the lens
scatters the rays over a larger surface. It is impossible to
create an image of the sun! However, if you place the lens in
front of your eye, you see the *virtual*
image of the sun on the same side of the lens as the sun (on the
side of the object).

Summarizing, we may say:
Biconvex lenses refract parallel rays towards each other,
biconcave lenses disperse them - the first *collect**,* the second *disperse** *rays. This explains the
principal difference in the generation of images. If we look
through one or the other at a *very far away* object, of which you can discern details, you see
everything inverted through the biconvex lens - up and down,
right and left interchanged. You see someone on a horse, who
rides from the right hand side towards the left hand side, move
from the left hand side towards the right hand side and the
horse's feet up on top, the rider's head at the bottom, and all
of it reduced in size. You also see through the biconcave lens
the size of everything reduced, but everything in its original
relative position. Also this difference originates from the fact
that the rays refracted by a biconvex lens *intersect behind* the lens; the real intersection of the
rays causes the inversion, shown by the drawing of the
construction.

The focal number *N* in
the lens formula changes with the colour of the light, whence the
focal width of a biconvex lens *f *= *r*/2(*N
- *1) has a different length for red from that for yellow or
violet. With

N_{red}
= 1.527 |
f_{red}
= 0.949·r |
|||

N_{violet} = 1.527 |
f_{violet}=
0.922·r |

you find that
(Fig. 696): If you place a screen into the focal point *v*
of the violet rays, the image becomes less sharp by a faded
external *red* coloured band. It arises by the screen also being
intercepted by the red rays which *aim* at their focal point
at *r *as well as from those rays the focal points of
which lie between *r* and *v*. In contrast, if you
move the screen to *r*, the image losses sharpness by a
faded external *violet* band. It arises from the rays with different colour
which have already passed their focal points between *r*
and *v*. The rays, the focal points of which lie between *r*
and *v*, generate coloured *dispersion* circles on
the screen - as each individual cone of rays is intercepted by
the plane of the screen in a circle of different diameter - at *v*
the red circles are largest, at *r* the violet circles.
The concentric superpositions of the different dispersion circles
generate the *coloured** *edges*.* Hence the lenses of optical
instruments (cameras, microscopes, etc.) would yield in
non-homogeneous light images with coloured edges unless it were
not possible to remove these errors due to colour dispersion.
This removal is called *achromatization* (decolourization), the result of this action achromacy (removal
of colours).

Complete achromacy demands union of the foci of
the rays of all colours at a *single** *point,
which cannot be done technically, and is also unnecessary - the
brightness of the different sections of the spectrum differs so
much that it becomes sufficient to make the foci of the *brightest**
*rays as close together as possible. In order to understand
how approximate achromacy is attained, we return to the colour dispersion of white light by a prism.

A prism (Fig. 651)
creates a number of coloured images of the opening *o* -
the red one least, the violet one most deflected from the
position *a*, at which the white light would have met the
wall opposite the opening. In order to let the violet image
coincide with the red one, you need only connect with the prism * 2*
another one, which is like the first and is placed like

But this experiment suggests the
direction in which the solution of the task can be found: You
must simply produce a prism, *which
disperses the colours equally strongly *as the first
prism, which however *refracts less**
*and therefore when linked to the first *reduces**
*its dispersion, but does not *cancel**
*it.* *For this purpose, you make the first prism out
of crown glass, the second out of flint glass*. This difference of the kinds of glass (Fig. 698a)
results, for example, in a flint prism with a refracting angle of
only 37º in an equally long spectrum, that is, it separates the
colours *equally
strongly** *as, all else being
equal, a crown prism of 60º. In this process, the flint prism
deflects the green-yellow rays minimally only by 25º48', the
crown prism by 40º. When the two prisms are linked, a deflection
of 14º12' remains.

* Crown glass has no lead in it and is in the first place of glass for sheets. Flint glass has much lead; it was initially produced from pulverized flint.

Fig. 698b shows the action of such an achromatic
system. White light enters through a small opening in the
direction of the arrow. If it were not refracted, it would yield
the spot of light *W*. If you place the prism *C*
on its path (Fig. 699) - the base downwards - it generates the
spectrum *rV*, *violet below, red
on top**. *We now introduce with *C* a
prism *F *(Fig. 700), the

refracting and colour dispersing action of which we define by:
If *white* light arrives from
the direction *r*, in which *red*
light leaves the prism *C* of Fig. 700 -
unrefracted it would create the white spot *W*' at the
location, where the red ray in Fig. 700 meets the wall - it
should creat the spectrum *r'V'* (*violet
above, red below*) and indeed with the same length as *C*,
so that *r'V'=rV. **If there were
not a prism** F**,
*the white light in the direction *would*
give a white spot at *W*', that is, a red spot, a violet
one and all the others would have converged at *W*'.
Hence, the prism* F **has lifted
each of these spots*: The red one to *r' *and
the violet one by *r'V'* higher up, which, however, equals
*rV*, and the other colours by corresponding distances. -
We place the prism, thus defined, into the position in Fig. 698b
between *C* and the wall. Escaping through *C*, the
bundle *rV *(Fig. 699) of rays (generated by refraction of
the white light) meets *F*. This bundle would have
generated without *F* the spectrum *rV* starting at
*W' *on the wall. The spots, which *F* has to
raise, lie therefore from *r* (Fig. 699), respectively *W'
*( Fig. 700), onwards *apart*, that is, *V*
below *r *by the distance *Vr*, which however is
equal to *V'r'*. However, since *V *lifts *more* by *V'r' *than* *it
lifts* r, *but *V **from
the start** *lies lower
by *V'r' *than *r*, it lifts *V *to the same
point to which it raises *r*. Hence the stretched coloured
image of the opening becomes the white image *W'***.**

If the prism performs *equally
perfectly** *for *all* colours what it
does for red and violet, *all coloured**
*images of the opening would be at *W*', whence the
spot *W'* would have no colour and would be perfectly
sharp. However, by the linkage of two prisms made out of the
normally employed kinds of glass, you can only cause two colours
to cover each other. A combination of two given colours takes
place by effectively *condensing*
the spectrum so that they *both**
*coincide exactly. This comparison displays immediately the
pairwise allocation of short and long wave rays*. *All
rays of different lengths have for the achromatization selected *another** *width of
compaction. The phenomenon when the rays of other wave lengths do
not have the same *compaction width*
as the two selected ones is called the *secondary
spectrum*. It arises, because in the *red* section of the spectrum the
dispersion of crown glass predominates, in the blue part that of flint glass. Naturally,
the colours, which have been selected for the superposition,
influence the more or less narrow union of the other colours. The
choice of the first depends on the purpose of the optical
instrument. For the physiological action of light on the retina,
other kinds of rays are decisive than the *actinic*
actions which are important for photography. The optically most
active rays lie between the lines *C* and *F; *they
have maximum brightness between *D* and *E*. If you
cover *C *and *F* by each other, you join
simultaneously the brightest parts of the spectrum between *D*
and *E*. As a result of this good combination of the
brightest part, you call the achromasy of *C* and *F *or
similarly located lines *optical* and employ it in
instruments meant for subjective use. For photographic purposes,
the violet actinicly active part of the spectrum is decisive with
the maximum of action in the vicinity of *G*'. In order to
place there the minimum of the secondary spectrum, you must join
the lines *F* and *Viol*_{Hg}, when you
obtain a *pure actinic** *colour
correction*. *(For astrophysical photography, its
comparatively weakly bright objects demand as perfect as possible
concentration of the actinically active rays.)

Just as you combine a crown
glass *prism* with a flint glass *prism*, you combine a crown glass *lens** *with a flint glass *lens* (Fig. 701). As a rule, they are glued together and the free surface of
the concave lens is given a (computable) curvature corresponding
to the intended action. Just as the combination of *two prisms* enables only covering up of *two* colours, this is also true for the
combination of two *lenses*. In order to join *three *colours, you must, in general, join at least
three lenses. Only a few kinds of Jena glass allow joining of
three colours by two lenses, so that a negligible remainder of
colour is present.

The achromatization of lenses
is decisive for the perfection of microscopes, telescopes and
photographic objectives (first, Chester Moor Hall
1703-1771 1733, then John
Dolled 1706-1761). The path
to glass production came from Fraunhofer who was
co-operating since 1809 with Guinand 1748-1824;
entirely new and until then completely inaccessible methods were
opened up by Ernst Karl
Abbe 1840-1905 and Friedrich Otto Schott 1851-1935 1885 by introduction of new
components into glass. Initially, only silicic acid had been
employed as the basic component of glass flows. Schott introduced also phosphoric and boric acid, phosphate
glass types as replacement of crown glass types, boric acid glass
types instead of flint glasses. Photography - ordinary as well as
micro-photography have been improved by the new lenses:
Photographically effective rays intersect at the same location
behind the lens at which also the physiological effective rays
intersect, whence the photographic image arises at the same
location at which the eye *sees*
the image by means of which
the equipment is *adjusted*. Only the coincidence of the *optical** *and the *actinic** *images has made it possible to
adjust equipment without tests withg photographic objects.

In the sequel, without special mention, it will be assumed that all lenses are achromatic.