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

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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.

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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.)

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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

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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 …

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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

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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

= −

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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

= − −

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Summary: Ray-transfer Matrix of Basic Optical Elements (I)

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Summary: Ray-transfer Matrix of Basic Optical Elements (II)

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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:

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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.

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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.