4. Spherical Mirror
2.8 Nonparaxial Ray Tracing with Interfaces
2.8.3 Example: Parabolic Interface
As an example, let’s consider a parabolic interface, where the parabola is specified by
z − zc= ay2. (2.84)
We can write this in our standard notation f(r) = 0 for root-finding as
f (y, z) = z − zc− ay2= 0. (2.85)
Here zc is an offset along the optical axis, and a controls the curvature of the ray. In the case where a > 0 (such that the parabola opens to the right), and the refractive index to the right is greater than to the left of the interface, the interface will act to focus incoming parallel rays.
2.8.3.1 Refraction
The diagram below shows the refracted rays for the case a = 1/2, n1 = 1 to the left of the interface, and n2= 2 to the right.
2.8 Nonparaxial Ray Tracing with Interfaces 37
The focusing action of the interface is clear, as well as the severe aberrations for rays far from the optical axis. Note that we have also marked the location of the effective focal point with a green dot. This is the common point on the optical axis through which horizontal rays very close to the optical axis will pass. The effective focal length is given by (Problem2.25)
feff = n2
2a(n2− n1) (2.86)
with respect to the location of the parabola along the optical axis; thus the absolute z location of the focal point is zc+ feff.
A program in the Octave1 mathematical scripting language that traces rays refracting through this surface to generate this disgram is given below. Note that we are using the recipe and the coordinate-independent form of the refraction formula. We only worry about one refraction, and we don’t bother with rays that reverse direction, because this doesn’t happen here. For finding the incident point of rays on the parabola, we define the surfacefunc function, which implements Eq. (2.81). We then use the fzero built-in function to find the root of surfacefunc, and thus the intersection point. Since fzero expects a one-dimensional function, we must pass other parameters to the function via global variables. We also compute the normal vector in gradfunc, by explicitly programming the gradient (∂f/∂y)ˆy + (∂f /∂z)ˆz. In a more general-purpose program, we could just estimate the gradient by computing f(r) at several different points, but the explicit form is pedagogically clearer here. This script is a good basis for ray tracing in other simple systems (e.g., Problem2.27). The script is available at
http://atomoptics.uoregon.edu/~dsteck/teaching/octave/parabolic_surface_refract.m and listed below.
% parabolic_surface_refract.m
%
% Trace incoming parallel rays for parabolic-surface "lens"
% defined by f(y,z) = z-zc - a*y^2 = 0.
%
% Represent rays by a position y at z, and also a direction unit vector
% rhat = [ry; rz]; z is the direction along the optical axis.
%
% Use coordinate-free form of Snell's Law in terms of surface normal vector.
global gY gRhat gZ0 gA gZc
a = 0.5; % parabolic parameter zc = -0.5; % shift of parabola center
dy = 0.06; % increment of initial y position of rays ymax = 6*a; % max y ray to trace
zmin = -6*a; % launch z zmax = 6*a; % max z
n1 = 1; % index to left of parabola n2 = 2; % index to right of parabola
% In the following function we need to pass extra parameters,
% hence the global statement.
% gY -> y at z0
% gRhat -> ray direction (unit) vector, in form [ry; rz]
% gZ0 -> reference coordinate z0
% gA -> copy of A
1Octave is free, open-source, mostly compatible with MATLAB, and runs on most platforms: http://octave.org.
38 Chapter 2. Ray Optics
% gZc -> copy of zc
% Note: global variables are labeled with a 'g' to avoid confusion gA = a;
gZc = zc;
% function for finding intersections with the parabolic surface function fout = surfacefunc(z)
% z is the position along the optical axis global gY gRhat gZ0 gA gZc
% extrapolate ray straight to z
% y = y0 + (z-z0)*ry/rz;
ray_y = gY + (z-gZ0)*gRhat(1)/gRhat(2);
% function to define surface via f(y,z) = (z-zc) - a*y^2 = 0 fout = (z-gZc) - gA*ray_y^2;
end %function
% function for finding the normal vector to the surface
% returns gradient of f = z-zc - a*y^2 function fout_vec = gradfunc(y,z)
global gA
fout_vec = [ ...
-2*gA*y; ... % partial f/partial y
1 ... % partial f/partial z
];
fout_vec = fout_vec/sqrt(dot(fout_vec,fout_vec)); % make unit vector end %function
% plot parabola, using parametric form dt = 0.01;
t=(zmin:dt:zmax)';
plot(a*t.^2+zc, t,'b-'); % plot as blue line
% set axes using z dimensions, and square aspect ratio axis([zmin, zmax, zmin, zmax], "square");
xlabel('z (mm)');
ylabel('y (mm)');
title(sprintf('ray trace of parabolic interface, n = %.2f',n2));
hold on; % subsequent plots will be overlayed
%%% loop over rays
nrays = 2*floor(ymax/dy);
for ray = 0:nrays,
y0 = (ray-nrays/2) *dy;
yarr = [y0]; % array of y positions to plot zarr = [zmin]; % array of z positions
% find first intersection with the parabolic surface
% look for intersection (root) between zmin and 10*zmax rhat = [0; 1]; % direction of initial ray, horizontal = zhat gY = y0; gRhat = rhat; gZ0 = zmin;
2.8 Nonparaxial Ray Tracing with Interfaces 39
z = fzero('surfacefunc', [zmin; 10*zmax]);
y = gY;
% add point to plot
yarr = [yarr; y]; zarr = [zarr; z];
% compute refracted ray direction, update direction vector rhat nhat = gradfunc(y,z); % surface normal vector
nhat3 = [0; nhat]; rhat3 = [0; rhat]; % add x dimension for cross product crossthing = (n1/n2)*cross(nhat3,rhat3); % (n1/n2)*(n x r)
rhat = (n1/n2)*rhat ...
+ ( sign(dot(nhat,rhat))*sqrt(1-dot(crossthing,crossthing)) ...
- (n1/n2)*dot(nhat,rhat) )*nhat;
% extrapolate to the end of the optical axis z0 = z; z = zmax;
yarr = [yarr; y + (z-z0)*rhat(1)/rhat(2)]; zarr = [zarr; z];
% plot ray
plot(zarr, yarr, 'r-'); % plot as red line end %for
feff = n2/(2*a*(n2-n1)); % effective, paraxial focal length
% mark effective focal length, account for offset of surface plot([feff+zc], [0], 'g.');
hold off;
2.8.3.2 Reflection
With relatively minor modifications to the script, we can use it to trace the reflected rays from the parabola.
40 Chapter 2. Ray Optics
We list only the changed section below, which is the last part of the loop over rays where we find the direction of the outgoing ray. We simply change from the refraction to the reflection formula. Also, we must now handle the possibility of a backwards-propagating ray, so we insert a condition based on the sign of the z-component of ˆrr, and propagate the final ray to either zmin or zmax for backward- or forward-propagating rays, respectively.
% compute reflected ray direction, update direction vector rhat nhat = gradfunc(y,z); % surface normal vector
rhat = rhat - 2*dot(nhat,rhat)*nhat;
% extrapolate to the end of the optical axis
% handle backwards rays based on sign of z component of rhat if ( rhat(2) > 0 ), % forward case
z0 = z; z = zmax;
yarr = [yarr; y + (z-z0)*rhat(1)/rhat(2)]; zarr = [zarr; z];
else, % backward case z0 = z; z = zmin;
yarr = [yarr; y + (z-z0)*rhat(1)/rhat(2)]; zarr = [zarr; z];
end %if
The full script to produce the above plot is available at
http://atomoptics.uoregon.edu/~dsteck/teaching/octave/parabolic_surface_reflect.m
2.9 Exercises 41
2.9 Exercises
Problem 2.1
A fish is 1 m beneath the surface of a pool of water. How deep does it appear to be, from the point of view of an observer above the pool? The refractive index of water is n = 1.33.
Problem 2.2
A corner-cube reflector is an arrangement of 3 planar mirrors to form, appropriately enough, the corner of a cube. The reflecting surfaces face the interior of the cube. Commercial corner cubes often use internal reflections to form the mirror surfaces, as in this photograph of the back side of a mounted corner-cube prism.
Show that any ray entering the corner cube is reflected such that the exiting ray is parallel (but opposite) to the incident ray. You may assume that the incident ray’s direction is such that it reflects from all 3 surfaces.
Problem 2.3
Suppose that a ray in air (n = 1) is incident on a planar window of uniform thickness T and refractive index n. If the angle of incidence is θ, show that the transmitted beam is parallel to the incident beam but has a horizontal displacement given by
δ = T sin θ
1 − cos θ n cos θ′
, (2.87)
where n sin θ′ = sin θ.
Problem 2.4
An anamorphic prism pair is used to expand or shrink a beam in one dimension without deflecting its angle, as shown here.
a
a q
42 Chapter 2. Ray Optics
The pair consists of two idential prisms with wedge angle α and refractive index n. The beam enters each prism at normal incidence to the front surface. As a model of the beam, it is useful to consider parallel rays as shown.
(a) Write down an expression that relates the deflection angle θ after the first prism to α and n.
(b) By how much is the beam reduced after the first prism? Write your answer in terms of n and α only.
(c) By how much is the beam reduced after the second prism?
Problem 2.5
A thin lens is submerged in water (n = 1.33). If the focal length is f = 1 m in air, what is the focal length in water? Assume the lens is made of fused silica (n = 1.46).
Problem 2.6
Consider2a reflection off the center P of a parabolic mirror (x = ky2) as shown.
A (a, b) x yx = ky™
P
A¢ (a, -b) P¢
By symmetry of the reflection at the center, we can fix the endpoints of the reflected ray to have coordinates (a, b) and (a, −b), with a > 0.
(a) Consider a point P′ a small distance (with vertical component ε) from P . Show that the optical path length of AP′A′ is an extremum when P′ coincides with P .
(b) Show that AP A′ is a minimum when k < kc and a maximum when k > kc, where kc:= a
2(a2+ b2). (2.88)
(c) The locus of all points B such that ABA′ = AP A′ is clearly an ellipse with foci A and A′. Show that the equation describing this ellipse is given by
(x − a)2 γ2− b2 +y2
γ2 = 1, (2.89)
where 2γ := ABA′. Argue that this result is consistent with the result of part (b).
Problem 2.7
(a) Derive the ABCD matrix for a refractive spherical boundary:
2Adapted from H. A. Buchdahl, Introduction to Hamiltonian Optics (Dover, 1993).
2.9 Exercises 43
R OA
n1 n2
M =
1 0
−(n2− n1) n2R
n1
n2
convex (shown): R > 0 concave: R < 0
The convention is that R > 0 for a convex surface (as shown here) and R < 0 for a concave surface.
Note that in the paraxial approximation, the height of the ray does not change across the boundary.
(b) Derive the ABCD matrix for a refractive planar boundary:
OA n1 n2
M =
1 0
0 n1/n2
(c) Find the determinants of the ABCD matrices in parts (a) and (b).
Problem 2.8
(a) Derive the ray-transfer matrix for free-space propagation, followed by a thin lens, followed by more free-space propagation, as shown in the Figure.
d™
f
d¡
(b) Show that applying the thin-lens law, 1 d1
+ 1 d2
= 1
f, (2.90)
all rays originating from a single point y1 in the input plane reach the output plane at the single point y2, independent of the input angle y1′. Compute the magnification y2/y1.
(c) Show that if d2= f , all parallel incident rays are focused to a single point in the output plane.
Problem 2.9
Let M be the ray matrix for an arbitrary optical system where the input and output refractive indices are n1 and n2, respectively. Prove that
det(M) = n1
n2
(2.91) (as you saw in Problem 6(c)), using the following outline, which exploits the formal equivalence of ray optics to classical Hamiltonian mechanics.
Recall the action principle (Fermat’s principle) for ray optics:
δ Z
n(x, y, z) ds = 0, (2.92)
44 Chapter 2. Ray Optics
where ds2= dx2+ dy2+ dz2 and n(x, y, z) is a refractive-index profile that models the optical system.
Compare to the action principle for classical mechanics. Take the coordinate z to be the “time” variable and the coordinate y to be the position coordinate. Let’s consider the two-dimensional case, so x is an ignorable coordinate, and note that z is also ignorable in the sense of being completely determined by x, y, and s. Then for the optical case, write down the Lagrangian. Show that the conjugate momentum p for the position y is n dy/ds, and then write down the Hamiltonian.
Now consider the following transformation relating the canonical coordinates before and after the optical system,
which of course is valid in the paraxial approximation (where it is also true that s ≈ z). Because y and p are canonical variables and M′ represents “time evolution” of a Hamiltonian system, M′ represents a canonical transformation and in particular M′ is a symplectic matrix, which implies that det(M′) = 1. This is essentially the content of Liouville’s theorem.
Using this result, transform to the standard (noncanonical) variables y and y′, and compute the determinant of M.
A very brief review of variational principles in classical mechanics may help. Recall that the action functional is given by the integral
S[L] :=
Z t2
t1
L(q, ˙q; t) dt, (2.94)
where the Lagrangian L is typically of the form L = T ( ˙q)−V (q) in particle mechanics. The variational principle (Hamilton’s principle) is δS[L] = 0, which for our purposes implies the Euler-Lagrange equation
under the condition that the endpoints of the variation are fixed (δq(t1) = δq(t2) = 0). The Hamiltonian is given by a Legendre transformation of the Lagrangian via
H(q, p; t) := ˙qαpα− L(q, ˙q; t), (2.96) where the conjugate momenta are defined by pα:= ∂L/∂ ˙qα.
Problem 2.10
(a) Suppose that two thin lenses of focal length f1 and f2 are placed in contact. Show that the combination acts as a thin lens with a focal length given by
1
(b) The optical power of a lens is defined as 1/f, where f is the focal length of the lens. Typically the lens power is measured in diopters, defined as 1/f where f is measured in meters (i.e., a lens with a 100 mm focal length has a power of 10 diopters). Based on your answer for part (a), why is the optical power a natural way to characterize a thin lens?
Problem 2.11
Show that the effective focal length feff of two lenses having focal lengths f1 and f2, separated by a distance d, is given by
1
2.9 Exercises 45
Note that this system is no longer a thin lens, so for this to work out you should show that the effect of the two-lens optical system is equivalent to that of a single lens of focal length feff, with free-space propagation of distances d1 and d2before and after the single lens, respectively, where
1
This problem is a simple model for one realization of a “zoom” or variable focus lens, e.g., for still photography. (A “true zoom” lens would also shift the location of the pair to maintain focus as the focal length changes, whereas a simpler variable focus lens merely changes the separation to achieve different focal lengths, possibly requiring a refocusing adjustment.)
Problem 2.12
Because of dispersion, the index of refraction varies slightly with the wavelength of light, and thus the focal length of a thin lens varies slightly with optical wavelength. This effect is called chromatic aberration. A common technique to correct for this aberration is to cement two lenses together of different materials to form an achromatic doublet or achromat.
For this problem, assume a simple linear model of the refractive-index variation:
n(λ) ≈ n(λ0) + dn dλ
λ=λ0
(λ − λ0), (2.100)
where λ0 is some wavelength in the center of the region of interest. The dispersion of an optical glass is often characterized by its refractive indices at three special wavelengths, the Fraunhofer C, D, and F lines, given by λC= 656.3 nm, λD= 587.6 nm, and λF = 486.1 nm, named after Fraunhofer’s catalog of the dark features in the solar spectrum. In terms of the three indices, we can define the Abbé v-constant by
vD:= nD− 1
nF− nC, (2.101)
in terms of which we can write the refractive index as n(λ) ≈ 1 + (nD− 1)
Design a thin, achromatic doublet with the following materials: BK7 borosilicate crown glass (nF = 1.52238, nD= 1.51673, nC= 1.51432) and F2 flint glass (nF = 1.63208, nD= 1.61989, nC= 1.61503).
Use BK7 for the first section in the shape of a biconvex lens, and F2 for the second section in the shape of a plano-concave lens, as shown.
F2 BK7 R¡
R™
R£ = •
Obviously, the radii of curvature at the cemented interface should match. Make the thin-lens approx-imation and choose the two radii of curvature to achieve a lens with f = 100 mm over the visible spectrum.
Problem 2.13
A refracting telescope is an optical system consisting of two thin lenses with a fixed space in between.
Light first enters the objective lens of focal length fo, propagates over a distance L, and goes through the ocular lens (eyepiece lens) of focal length fe. The length of the telescope satisfies L = fo+ fe.
46 Chapter 2. Ray Optics
L
fo fe
objective lens ocular lens
(a) Construct the ABCD matrix for propagation through a telescope (from left to right in this diagram).
(b) Show that a telescope produces angular magnification. That is, incoming rays with angle θ1
from the optical axis exit the system at angle −(fo/fe)θ1, independent of the initial ray position y1. (c) Show that a telescope can also act as a beam reducer (or expander): i.e., show that a bundle of rays of diameter d traveling parallel to the optical axis has diameter (fe/fo)d when exiting the eyepiece.
(d) Sketch a Keplerian telescope, where both focal lengths are positive. Draw in the parallel rays corresponding to part (c). Also sketch a ray that is initially not parallel to the optical axis; use the thin lens law to justify the minus sign in the angular magnification of part (b).
(e) Clearly, a telescope with positive angular magnification (i.e., a telescope that produces an upright image) has exactly one lens with negative focal length so that (−fo/fe) > 0. Such a refracting telescope is known as a Galilean telescope. Which of the two lenses can have negative focal length if the telescope produces a magnified (not reduced) image?
Problem 2.14
A retroreflector is any optic that reflects an incident ray, such that the exiting ray is parallel (but opposite) to the incoming ray. One version of a “cat’s eye” retroreflector uses a thin lens of focal length f and a mirror as shown.
f
Set up the ray matrix for this optical system to prove that it is, indeed, a retroreflector.
Problem 2.15
(a) A ball lens is a sphere of optical material (e.g., glass or sapphire) that is used as a lens, especially for coupling light into or out of an optical fiber. Within the paraxial approximation, derive an expression for the location where a beam of collimated incoming light (aimed at the sphere’s center) will come to a focus, assuming a sphere of diameter D and refractive index n, surrounded by vacuum. (Your answer should be in the form of a distance from the center of the ball lens, thus giving its effective focal length.)
(b) What is the condition for incoming rays parallel to the optical axis to be imaged onto the back surface of the sphere? Argue that under these conditions, the sphere acts as a retroreflector (incoming rays are reflected back along the same direction of incidence). This is the traditional realization of a “cat’s eye” retroreflector, and is the reason, for example, that some painted lines on roads and road signs look so bright at night under headlights (small silica spheres embedded in the paint reflect headlight back to your eyes).
Problem 2.16
Consider a two-mirror resonator, as shown here. One mirror is concave (R < 0) and the other is flat.
2.9 Exercises 47
d R
(a) What is the round-trip ray matrix for this resonator? Derive the matrix for a ray starting just to the right of the curved mirror, traveling to the right.
(b) For what range of d is the cavity stable?
(c) Derive the ray matrix for two round trips for the special case R = −2d. Sketch an example ray that illustrates your answer.
Problem 2.17
Suppose you have two convex, spherical mirrors in a gas laser resonator, with unknown and possibly different radii of curvature R1and R2.
R¡ R™
d
Suppose also that you can vary the length d of the cavity, and that after much labor you find that the laser operates in the ranges d < 50 cm and 100 cm < d < 150 cm. What are the numerical values of R1 and R2? Keep in mind that gas lasers have relatively low gain per pass, and thus proper laser operation requires that the light makes many round trips inside the resonator before leaking out.
Problem 2.18
(a) Consider a cavity consisting of two planar mirrors and identical thin lenses of focal length f, regularly spaced as shown.
f
d/2 d
f f f f f f f
d/2
d d d d d d
For a given set of lenses, what is the range of d for which the cavity is stable?
(b) Write down the eigenvalues of the round-trip ray matrix for this cavity.
Problem 2.19
Consider a symmetric cavity consisting of two planar mirrors separated by 1 m. Suppose that a thin lens of focal length f is placed inside the cavity against one of the mirrors. For what range of f is the cavity stable?
Problem 2.20
Consider the ring cavity shown below, where the optical axis forms an equilateral triangle, with sides of length d, and a single lens of focal length f centered in the bottom leg of the cavity.
48 Chapter 2. Ray Optics
d
f (a) For what range of f is the cavity stable?
(b) Discuss how moving the lens to the left or right influences the stability of the cavity.
Problem 2.21
Suppose an optical resonator is represented by the round-trip matrix M, which satisfies M
1 0
= 3 2
1 0
. (2.103)
(a) What are the two eigenvalues of M?
(b) Is the resonator stable? Explain.
(c) If your answer to (b) was “yes,” are there any rays that diverge after many round trips? If your answer to (b) was “no,” are there any rays that remain stable? In either case, give an example and give a physical interpretation of the rays.
Problem 2.22
In the matrix formalism of optics, we have stuck to modeling only one dimension (y), transverse to the optical axis (z). Suppose we want to model both transverse directions x and y, by an analogous matrix formalism, with a four-vector of the form
x x′ y y′
(2.104)
for the state of the vector at some point along the optical axis. Write down the ray-transfer matrix for
for the state of the vector at some point along the optical axis. Write down the ray-transfer matrix for