Introduction
| Slide 1 | |
In order to understand the way light behaves, it is best to
divide its properties into two different realms. For example, physical optics
helps to explain how light has different colors, how lasers produce such
energy, and how to deal with wavelengths invisible to the human eye, among
other subjects. Geometric optics, on the other hand, describes how light
travels through different materials, how lenses and prisms work, and the optics
of the many instruments we use on a daily basis. This tutorial discusses some
of the basic concepts of geometric optics by defining the various terms used in
optics. More advanced aspects of geometric optics and how they apply to our
practice of ophthalmology are discussed in the Advanced Geometrics tutorial.
| Slide 2 | |
Geometric optics treats light as tiny particles (photons) that
travel in straight lines. These lines, or rays, may be in any direction -
horizontal, vertical, or oblique. Light rays in a vacuum will continue to
travel in a straight line until encountering a surface or interface, where they
may change direction. For simplicity, it is assumed they travel in air the same
as they do in a vacuum.
What happens when light rays pass through an interface? A
good analogy is riding a bicycle on a hard surface and then
encountering a different surface such as sand or mud. Two things happen. First, the bicycle slows down
and, second, the wheel tends to turn and the bike changes direction. Similarly,
when light travels through air and then strikes a transparent surface, like a
piece of glass, the ray of light slows down and changes direction. The ability
of the substance to slow and bend rays of light is described by its refractive
index. The refractive index of a medium is
The refractive index of any medium will be greater than 1.0.
The greater the medium slows light, the higher is its refractive index.
| Slide 3 | |
I have described how light rays are slowed when entering a
medium denser than air, and how refractive index is related to this. Now let's
review how the direction is also changed. The change of direction is related to
a line that is perpendicular to the surface through which the light is passing,
the "surface normal" (Slide 1). Light rays will strike
the surface at an angle (relative to the surface normal) known as the angle of
incidence and, when a ray passes through the interface, it will have a new angle,
the angle of refraction. If the ray is traveling from a medium with a lower
refractive index into one of higher refractive index, it will be bent
(refracted) toward the surface normal (Slide 2).
Conversely, if traveling from a higher refractive index medium to a lower one,
the ray of light will be refracted away from the surface normal (Slide 3). A special circumstance arises when a ray
traveling from a medium of higher refractive index to one of lower index and is
at such a high angle, relative to the surface normal, that it is reflected from
the surface, rather than passed through and refracted (Slide 4). The angle at which reflection first occurs,
rather than refraction, is call the "critical angle," and this event is termed
"total internal reflection."
The refractive index of the medium where the light starts is
called n and that of the second medium is n'. If the angle in the first medium
is i and the angle in the second is i', an equation can be created to describe
the refraction of light at an interface between two media of different
refractive indices. This is Snell's law.
This discussion has been based on refraction at plane (flat)
surfaces. Snell's law also governs refraction at a curved surface, such as an
ophthalmic lens. Using that equation and some trigonometric calculations, the
power of a curved surface can be determined to be described by the equation
| Slide 4 | |
Where r represents the radius of curvature of the refracting
surface in millimeters. For the full derivation of this equation, consult your
favorite text on optics.
To determine where a lens would image a given object, we need
to know more than the refractive power of the lens. We also need to know how
the light rays are converging or diverging before and after they pass through
the lens. The degree of convergence or divergence of light rays can be
designated the dioptric power of the light. The amount of convergence or
divergence of light emitted by a source (or reflected from an object) is
dependent on how far the lens is from the source, and the dioptric power is the
inverse of the distance (in meters) from the source or object. If an object is
1 m from a lens, light from that object will have 1/1 m = 1.0 D (diopter) power
at the lens. Likewise, an object 2 m from the lens will dispense light which
will have 1/2 m = 0.5 D at the lens. If the object is 0.5 m from the lens, the
light will have 1/0.5 m = 2 D at the lens. The closer to the refracting
surface, the greater is the dioptric vergence power; the farther from the
surface of the lens, the less is the vergence power. Considering light after
refraction, its power is determined by the distance to the image the lens creates.
| Slide 5 | |
Nearly all rays of light emanating from sources in nature are
divergent, and by convention divergent light is given a negative power. Rays of
light, which are acted upon by a lens or multiple lenses, mirror, or other
surface, may be either divergent or convergent. Convergent rays are designated
as having positive power. To simplify matters, light rays are described as if
they are moving from left to right, although it is obvious they can travel in
any direction.
The terms focal points and focal planes must be defined to
fully understand the relationship among objects, images, and lenses. They all
flow from the simple equation:
in which U is the vergence power (in diopters) of the rays of
light from an object or source when they strike the lens and P is the dioptric
power of the lens, that is, how much it changes the direction of the light
rays, and V is the vergence power (in diopters) of the rays after they exit the
lens. Obviously, given any two of these variables we may find the third. For
instance, if an object is located 1 m from a +3.00 D lens, the image will be
0.5 m or 50 cm from the lens. Object vergence U = -1.00 (negative because the
light rays are diverging when they strike the lens).
Since V = +2.00 D and vergence is the inverse of the distance
from the lens, then the distance to the image is 1/V = 0.5 m (Slide 5).
Focal Points
| Slide 6 | |
When an object is placed 33 cm in front of a +3.00 D convex
lens, the object rays have a vergence of -1/33 cm = -3.00 D (minus because the
rays are divergent). According to the equation U + P = V, the vergence of the
image (V) is zero, -3.00 + 3.00 = 0. This means the light rays emanating from
the lens are parallel rays. This particular point, the point in space where
light must originate to result in image vergence of zero, is termed the primary
focal point of the lens (Slide 6).
| Slide 7 | |
The secondary focal point of a lens is defined as the point in
space where the lens will focus an object at infinity (thus its rays are
parallel when they reach the lens, and have zero vergence). Using a +3.00 D
lens, the U + P = V equation would be 0 + 3.00 = +3.00. Therefore, the outgoing
rays of light will have a vergence of +3.00 D and will be focused at a point
1/+3.00, or 33 cm, to the right of the lens (Slide 7).
| Slide 8 | |
So far I have been working with plus lenses, but now let us
turn our attention to minus lenses and their primary focal points. Let's start
with the secondary focal point, since it is a bit easier to explain. Remember,
the secondary focal point is where parallel rays of light refracted by our lens
will be brought to a focus. Since the incoming rays of light have a vergence of
zero, if we use an example lens of -2.00 D power, then the equation will be U + P = V, 0 + (-2.00) = -2.00. The rays of light emanating from the -2.00 D lens
will be divergent, since minus power indicates divergent light. They then would
be focused (or seem to be coming from) a point 50 cm (1/-2.00 = 1/2 m = 50 cm) to the left of our -2.00 D lens (Slide 8).
| Slide 9 | |
You've seen how to derive the secondary focal point of a minus
lens, but the primary focal point is slightly more difficult to conceptualize.
However, if we use the formula, U + P = V, it becomes easier. By the definition
of the primary focal point, the vergence of light rays after being acted upon
by a -2.00 D lens will be zero. Therefore, V = 0. If P = -2.00 D, then U + (-2.00) = 0, and U = +2.00. The primary focal point of this lens is then to be
found 1/+2.00 = 50 cm, to the right of the lens (Slide 9). The light must be converging toward that point, 50 cm from the lens,
for it to be refracted by the lens to become parallel.
Focal Planes
| Slide 10 | |
| Slide 11 | |
As I have been discussing lenses, all the objects and images
have been along the axis of the lens. What happens when two object points are
situated off-axis? Let us consider an object at infinity, but off-axis. You
know that the vergence of light rays from this object will be zero, or parallel
rays of light. They will be refracted by the +3.00 D lens and will ultimately
be focused 33 cm to the right of our lens (U + P = V, 0 + 3.00 = +3.00) in a
vertical plane that is perpendicular to the axis. This is the secondary focal
plane of the +3.00 D lens, the plane at which light from infinity will be
focused (Slide 10).
The primary focal plane of a +3.00 D lens is determined in
essentially the same manner as the secondary focal plane. If we take the rays
of light from an off-axis object positioned 33 cm from the lens, these rays
will be acted upon by the lens and will exit the lens with a vergence of zero
(U + P = V, -3.00 + 3.00 = 0). This off-axis object must be situated on a plane
perpendicular to lens axis at the primary focal point, the primary focal plane
(Slide 11).
Lens Systems
| Slide 12 | |
So far you have learned about primary and secondary points and
planes of a single lens. If you were to put two or more such lenses together
and evaluate light passing through all of them, you would need to figure the
vergence at the first lens and where it would image the object. Then that image
would become the object for the second lens, and you would need to determine
the image that lens would create, and so on down the line of lenses, until you
had found the image created by the last lens in the system. With this
information, you could calculate a primary and secondary focal point
representing the entire system (Slide 12).
| Slide 13 | |
As you recall from the definition of the primary focal point,
if you were to draw a ray from our object (O) which passes through
F1, the primary focal point of the system, it must emerge from the
last lens in the row parallel to the axis. Now if you trace a line back from
the image (I), a line which is parallel to the axis, this line will intersect
the ray from O which went through F1, since rays parallel to the
axis after passing through the lenses must have "originated" at the primary
focal point. A plane perpendicular to the axis at this intersection point is
therefore the primary principal plane (L1) of the lens system.
Similarly, we can draw a line parallel to the axis from our object, and extend
it through the entire lens system, until it intersects a ray that passes
through the secondary focal point of the last lens and also the image. That ray
must have been parallel to the axis before the last lens refracted it, since
such parallel rays pass through the secondary focal point by its definition.
The plane perpendicular to the axis where these two lines meet is thus the
secondary principal plane (L2) of the lens system. The advantage of
these principal planes is that they define the properties of the entire lens
system as if it were a single lens, and all the different lenses can be
ignored.
Nodal Point
With a thin lens, the nodal point is
where a ray from an object passes through the lens undeviated. With a thick
lens, this ray will strike the first nodal point, be refracted, pass through
the second nodal point, and finally emerge from the lens at the same angle as
it entered the first surface of the lens (Slide 13).
These points and planes - the primary and secondary focal points and planes,
primary and secondary principal planes, and the nodal points - of a lens system
are termed the cardinal points of the lens or system.