Ray tracing in an optical system

Ray tracing in an optical system
•Ray matrix
•OPTICAL CAVITIES
•STABILITY DIAGRAM
•THE UNSTABLE REGION
•EXAMPLE OF RAY TRACING IN A STABLE CAVITY
•REPETITIVE RAY PATHS
•INITIAL CONDITIONS: STABLE CAVITIES
•INITIAL CONDITIONS: UNSTABLE CAVITIES
•ASTIGMATISM*
•CONTINUOUS LENS-LIKE MEDIA
•Propagation of a Ray in an Inhomogeneous Medium
•Ray Matrix for a Continuous Lens
•WAVE TRANSFORMATION BY A LENS
•Applications
Ray matrix
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What we need to describe a ray in an optical system.
1. Where is it with respect to some arbitrarily chosen axis?
2. In what direction is it heading?
We start with inspection of the length of free space.
We assume that all rays are paraxial, tan( )  sin(  )   thus the angle 
equals the slop of the ray, r’.
r2  1 r1  d  r1 '
r2 '  0  r1  1 r1 '
•write this in matrix form as:
•In general, the relation between the output and input parameters
of a general optical system is given by the ABCD matrix of the form:
•Thus ray tracing through a sequence of optical components is reduced to
simple matrix multiplication.
•For instance, if we consider two lengths of free space.
•For optical rays AD — BC = 1 is true provided that the index of refraction at
the exit plane is the same as the entrance plane.
Thin lens of focal length f
•Here we assume that the lens is so thin that there is negligible distance between the
entrance plane 1) and exit plane 2).
•Thus no matter what the slope of the incoming ray, the output position is always
equal to the input position, or
Therefore, A = 1 and B = 0.
•Now consider the circumstances provided by ray α in the diagram.
For this special case the input slope
, yet it is obvious that the output
slope is
In the other case, ray β comes in with a slope of
exits parallel to the axis. Thus, r'2 = 0.
Therefore, D = 1
and obviously
Hence the ray matrix of a thin lens is given by:
Combination of a lens plus free space.
spherical mirror
•The entrance and exit planes are on the same
side of the surface of the mirror. the effect of the
spherical mirror of radius R is to direct the ray
toward the axis just like a thin lens .the focal
length of a spherical mirror is just one-half of the
radius of curvature.
Thus the transmission matrix is given by:
APPLICATIONS OF RAY TRACING: OPTICAL CAVITIES
•If the ray position stays "close" to the optical axis , the system is stable;
• if the ray naturally "walks off' one of the mirror surfaces, it is unstable;
•if the mirrors must be perfectly aligned, it is conditionally stable.
equivalent lens-waveguide
Transmission matrix for a unit cell
Given the information of preceding sections, it is easy to compute the
transmission matrix of the unit cell shown; it is just the product of two
matrices of the form given by:
second-order difference equation for the ray
•Let us find a second-order difference equation for the ray as it passes the various
planes of the succeeding unit cells that correspond to observing a ray as it makes
successive round-trips through the cavity. We do this by eliminating the slope
from these equations:
Does the equation have solutions in which the
magnitude of r is less than some maximum value?
•It is important to realize why we should ask the question.
•If the answer to it is yes, then the position of the ray form the axis
undulates as it propagates along the lens waveguide (or between the two
mirrors).
• If the ray's position is not bounded, it will eventually become so big that
it will "miss" one of the components and thus walk off the mirror.
the position r must be real.
•Now r0 is specified by the initial conditions. It cannot be set equal to
zero for the general case. The exponential term is not zero; hence, the
second factor must be zero. That is a quadratic equation in exp(jθ).
The two solutions are:
STABILITY
Stability diagram
•We can tell by a glance at this
graph whether the system is
stable or not, if the g
parameters is inside the crosshatched region, the cavity is
stable; if outside, it is unstable;
and if on the border, it is
conditionally stable, requiring
perfect alignment.
•One of the surprises that this diagram provides is that the confocal geometry,
consisting of two identical mirrors and separated by the sum of the two focal
lengths, is on the borderline of stability (g1 = g2 = 0).
Nevertheless, it is on that borderline, and most lasers are made to avoid that
situation by increasing or decreasing d slightly.
THE UNSTABLE REGION
resonators operating in the unstable region have become very useful
for high-gain laser systems. In that case, the rays that walk
off the mirrors constitute the output.
•A very crude example.
•After 10 reflections this beam missed one of the mirrors.
•the medium inside the cavity increases the power by a factor of 5.
•the emerging beam that misses a mirror is amplified 10 times .
•If the initial beam contained only 1 mW of power, then the emerging beam has
nearly 10 kW of power.
•Although the example is crude, it illustrates that unstable resonators have their
place.
EXAMPLE OF RAY TRACING IN A STABLE CAVITY
•The incoming ray is perpendicular to the flat mirror ) R1= oo)
and displaced from the axis by a specified amount — r0. Thus
this ray is directed toward the focal point of M2, reflects from
the flat mirror, and heads toward the spherical mirror along a
radius. Once this ray reaches M2, it starts a retrace of its initial
path and forevermore stays along paths indicated. This is an
example of a repetitive ray path.
•The beam stays within a maximum displacement of 2r0 from
the axis, consistent with the ideas of stability developed earlier.
Why the maximum displacement is r0, where as the simple walk through the
cavity indicated that the maximum displacement was 2ro. Why is there a
difference?
•The ABCD matrix for a unit cell relates the position and slope of the ray as it enters
and leaves the reference planes. It tells us nothing about what happens in between.
•If we need to know about the position between the reference planes of the unit cell
(spherical mirror), we must apply the individual ray matrix to obtain that information.
the ray is impinging on the spherical mirror at
After one round trip, p = 1,
2𝑟0 sin −30 = −𝑟0
2𝑟0 sin 120 − 30 = 2𝑟0
At the end of the second round trip, p = 2,
r=2𝑟0 sin 2𝑥120 − 30 = −𝑟0
•Note that this last choice of the unit cell tells us nothing about the trials
and tribulations of the ray at the flat mirror.
REPETITIVE RAY PATHS
The preceding example illustrates a case whereby the beam retraces its path after
a discrete number of round trips. In the particular case shown, the ray went three
complete round trips and then was in the same position, with the same slope, as it
was when it started. We can generalize this result for any cavity.
•If s is increased by m units, corresponding to m round trips, then the ray
returns to its original position after these m round trips, when θ satisfies:
•Note that inequality guarantees that θ is always less than π,
in accordance with the principal value of cos θ = (A + D)/2.
INITIAL CONDITIONS: STABLE CAVITIES
•Let us suppose that we have unfolded a complex cavity and have found the
transmission matrix for a unit cell, and that the ray's position a and the
slope m are known at a given reference plane (call its = 0).
•If a cavity is stable, we can find the angle θ from the transmission matrix of
the unit cell:
•After traversing
one unit cell
(or round trip):
INITIAL CONDITIONS: UNSTABLE CAVITIES
Thus there are two solutions (again), but now they are both real:
The general solution becomes:
After one round trip.
ASTIGMATISM*
•When a material body is placed in the path of a ray and is tilted we must
account for the change in the optical path in two orthogonal directions.
•Astigmatism of a window, (a) Side view, (b) Top view.
•For instance, windows placed at the Brewster angle are very common in gas lasers.
The angles θx and θy paraxial to the optic axis.
•However, if the dielectric material is a Brewster's angle window on a gas laser tube,
then
, and for quartz with n = 1.45845,1' 8 = 55.56°.
•Obviously, the angle θ-θx is not small, and we must account for bending and
resulting displacement of the beam in the xz plane .
•It is left as a problem to show that optical paths traversed by the two
rays through a Brewster's angle window are different.
•The fact that these distances are not equal gives rise to astigmatism.
•Refraction angle - α
•For example :A curved mirror used in ring laser cavities . The mirror focuses parallel
rays in the two planes at different locations, leading to different effective focal lengths
in the xy and xz planes.
•Astigmatism leads to elliptical beams in ring lasers and plays a critical
role in dye-laser cavities.
CONTINUOUS LENS-LIKE MEDIA
• a medium in which the index of refraction is nonuniform in
the transverse direction.
•Optical fiber with an index of refraction
depending on r.
Variation of the index of refraction of a gas due
to local heating by the current density
distributed as in (b).
Propagation of a Ray in an Inhomogeneous Medium
•propagation of a ray in an
inhomogeneous dielectric medium.
Ray Matrix for a Continuous Lens
A simple thin positive lens
•more mass and a longer optical path there than
at the edges.
•Similarly, a positive continuous lens has an
index of refraction that decreases with r.
•For a cylindricallly symmetric positive lens, the first two terms of
Taylor series expansion for n(r) would have the following format:
•the parameter L is simply a scale factor that
indicates how fast n varies with r.
•For instance, if a = 50 μm for the radius of the fiber and n(a) = 1.42 with n0 = 1.47,
then Δn = 0.05. The relationship between Δn and L is given by simple arithmetic:
•
For the numbers just quoted, I = 191.7 μm which is large when compared with the
radius of the fiber. Most fibers have a change in index much less than the value
chosen here, and hence L is even larger.
•Because of the inequality we can use the
first term for n and the derivative of the
second for dn/dr:
•The solution is straightforward. Assume that z= 0 is the input plane to
this optical component where the position r and slope r' are known:
•Then, by differentiation, we obtain the slope at any position z:
•Thus the ray matrix for a length z = d is:
then the medium is uniform
•If the index of refraction increases with r (as in the case of a gas discharge), then the
geometric factor L must take on imaginary values to predict n(r) from the equation:
•If we let L = jL and use the fact that cos iθ = cosh θ ,sin iθ = i sinh θ,
we obtain the ray matrix for a negative lens directly from the T matrix:
The C term in this matrix is the negative
of the focal length of this lens:
•Thus the focal length is positive (i.e., converging) provided that the argument of
the sine function is less than π radians.
•If d is long enough) as in a fiber-optic communication system( the focal length is
alternately positive and negative, corresponding to a converging and diverging
system.
• Thus we could anticipate the "beam" undulating as shown in the Fig as it
propagates down the fiber.
• Going one step further, anticipate a situation where the natural divergence of a
small beam is continuously counteracted by the convergence of the medium.
• Such fibers exist and are used in fiber-optic communication links.
WAVE TRANSFORMATION BY A LENS
•The effect of a lens on a limited extent uniform plane wave.
•What is the shape of the phase front
after passing through the lens?
•The Fermat principle; the light always takes the path that makes the
transit time a minimum.
•The ray at B' has farther to travel; hence, its phase must be ahead of that of A'.
•This extra distance is Δd.
Applications
•In computer graphics, ray tracing is a technique for generating an
image by tracing the path of light through pixels in an image plane and
simulating the effects of its encounters with virtual objects.
•The technique is capable of producing a very high degree of visual
realism, but at a greater computational cost.
•Ray tracing is capable of simulating a wide variety of optical effects, such
as reflection and refraction, scattering, and chromatic aberration.