Ray Transfer Matrix
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Transcript of Ray Transfer Matrix
EE485 Introduction to PhotonicsRay-Transfer Matrix in Paraxial Optics1. Definition and application2. Matrices for various optical components3. Matrices for cascaded optical components
F. L. Pedrotti and L. S. Pedrotti, “Introduction to Optics,” 2nd ed., Chapter 4, Prentice Hall
2Lih Y. Lin
What is the Ray-Transfer Matrixy
1 1
1 1
A BC D
x
Input(y1, θ1)
y1
θ1
Output(y2, θ2)
θ2
y22 2
2 2
A BC D …
Optical System
A 2 x 2 matrix that relates the position (y) and angle (θ) of a light ray with respect to the optical axis at the output of an optical component to its input.Valid for paraxial rays only.The rays are assumed to travel only within a single plane.The ray-transfer matrix of a cascade of optical components is a product of the ray-transfer matrices of the individual components.
3Lih Y. Lin
The Matrix Method ― A step-by-step approach
To find the output at Point 7, given the input at Point 0 and the optical system.
Step 8: Refraction matrix by a concave surface at P4.
Step 9: Translation matrix from P4 to P5.Step 10: Refraction matrix by a convex
surface at P5.Step 11: Translation matrix from P5 to P6.Step 12: Reflection matrix by a concave
mirror at P6.Step 13: Translation matrix from P6 to P7.
Step 1: Translation matrix from P0 to P1.Step 2: Refraction matrix by a convex
surface at P1.Step 3: Translation matrix from P1 to P2.Step 4: Refraction matrix by a concave
surface at P2.Step 5: Translation matrix from P2 to P3.Step 6: Refraction matrix by a convex
surface at P3.Step 7: Translation matrix from P3 to P4. (Looks complicated, but with matrices it’s
not that bad.)
4Lih Y. Lin
Translation Matrix and Reflection MatrixTranslation in a homogeneous medium
01
01
10 1
yy L = αα
Reflection by a planar mirror
2 1
2 1
1 00 1
y y = θ θ
Reflection by a spherical mirror1 0'2' 1
y y
R
= α α Sign convention:• Positive angles for rays pointing upward,
negative angles for rays pointing downward.• R > 0 for convex mirrors, R < 0 for concave
mirrors.
A
B O
5Lih Y. Lin
Rear-view Mirror as an Example Again
Exercise: If the rear-view mirror has a focal length of 10 meter, determine the position and the nature of the image of a car that is 1.5-meter tall and 20 meters away.
1
1
1 0 1.52 01
20
y = θ
3
3
201 0 1.540
2 1.5120 40
y × = θ −
Find the intersection of the extended reflection rays.
For ray 1: For ray 3:
(I know all these from ray tracing and the imaging equations. This doesn’t seem to make things easier. Why should I learn this?)
Be patient …
6Lih Y. Lin
Refraction MatrixPlanar boundary
1
2
1 0'
0'y y
nn
= α α
Spherical boundary
1 0'
1 1'' '
y yn n
R n n
= −α α
7Lih Y. Lin
Matrices for Thick-lens and Thin-lensThick lensTranslation
matrix
2 1
1 0 1 01
'0 1
' 'L L L
L L
tM n n n n n n
n R n n R n
= − −
Refraction matrix (Second spherical
boundary)
Refraction matrix (First spherical
boundary)
Thin lens
2 1
1 0 1 01 00 1L L L
L L
M n n n n n nnR n n R n
= − −
Lensmaker’s formula
1 01 1
Mf
= −
8Lih Y. Lin
The Thick-lens Example Againn2 = 1.33
Object
n1 = 1
R = 5 cm
Step 1:Define (y1, θ1) and (y2, θ2).
Step 2:Write down the ray-transfer matrix for the thick lens.
Step 3:Apply the matrix and obtain (y1’, θ1’) and (y2’, θ2’).
Step 4:Find out the intersection of the two output rays.
(y2, θ2)
h
(y2’, θ2’)
(y1, θ1) (y1’, θ1’)
2 1 2 1 2 1
1 1 2 2
1 0 1 010 1t
M n n n n n nn R n n R n
= − −
9Lih Y. Lin
Summary: Ray-transfer Matrix of Basic Optical Elements (I)
10Lih Y. Lin
Summary: Ray-transfer Matrix of Basic Optical Elements (II)
11Lih Y. Lin
Matrices of Cascaded Optical Components
0
0
y θ
M101
01
yy = θθ
1M
M22 1
22 1
y y = θ θ M
M3 MNN
N
y θ
02 1
0 N
N
y y A BC D
= = = θ θ
NM M M M M
nfn0
Works particularly well with computer programming (MathCAD example).
Det o
f
nAD BCn
= − =MA useful tip:
12Lih Y. Lin
The Special Cases of Zero
A = 0, yf = B•α0, independent of y0→ Output plane is the 2nd focal plane
D = 0, αf = C•y0, independent of α0→ Input plane is the 1st focal plane
C = 0, αf = D•α0, independent of y0→ Telescope. D: Angular magnification
B = 0, yf = A•y0, independent of α0→ Imaging. A: Linear magnification.
13Lih Y. Lin
Example: Using a Special Case of zero (B = 0)
Find the location and magnification of the output image.21 0 1 161 1 16 12 31 1.50 1
0 1 0 1 1 24(1.50) 1.50
12 3
x xx
M
− − = =− − − 20 16 0, 24 (cm)3xB x= → − = = Magnification 1 1
12xm A= = − = −
(No more nightmare about having to solve the geometric problem of two-ray intersection.)
Exercise: Using the special case of B = 0, find out the location and magnification of the output image in the previous MatCAD example.