Файл: Ersoy O.K. Diffraction, Fourier optics, and imaging (Wiley, 2006)(ISBN 0471238163)(427s) PEo .pdf

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124

 

 

 

GEOMETRICAL OPTICS

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 8.8. The travel of a ray through an optical system between z ¼ z1 and z ¼ z2.

In the paraxial approximation, ðy1; y1Þ and ðy2; y2Þ are related by

y2

¼

C D

y1

ð8:6-1Þ

y2

 

A B

y1

 

where the matrix M with elements A, B, C, D is called the ray-transfer matrix. The ray transfer matrices of elementary optical systems are discussed below.

1.Propagation in a medium of constant index of refraction n: This is as shown in Figure 8.9.

Figure 8.9. Propagation in a medium of constant index of refraction n.

M is given by

M ¼

1

d

ð8:6-2Þ

0

1

2. Reflection from a planar mirror: This is as shown in Figure 8.10.

Figure 8.10. Reflection from a planar mirror.

M is given by

M ¼

1

0

ð8:6-3Þ

0

1

MATRIX REPRESENTATION OF MERIDIONAL RAYS

125

3. Refraction at a planar boundary: This is as shown in Figure 8.11.

Figure 8.11. Refraction at a plane boundary.

M is given by

1

0

5

 

4

n2

 

M ¼ 2 0

n1

3

ð8:6-4Þ

 

4. Reflection from a spherical mirror: This is as shown in Figure 8.12.

Figure 8.12. Reflection from a spherical mirror.

M is given by

 

1

0

 

4 R

5

 

M ¼ 2

2

1 3

ð8:6-5Þ

 

5. Refraction at a spherical boundary: This is as shown in Figure 8.13.

Figure 8.13. Refraction at a spherical boundary.

126

 

 

 

 

 

GEOMETRICAL OPTICS

M is given by

 

 

 

 

 

 

M ¼

 

1

0n1

3

ð8:6-6Þ

2 n1 n2

 

4

 

 

 

5

 

 

n2R

n2

 

6. Transmission through a thin lens: This is as shown in Figure 8.14.

Figure 8.14. Transmission through a thin lens.

M is given by

M ¼

2 1

1

01

3

ð8:6-7Þ

 

 

4

f

 

5

 

7.Cascaded optical components: Suppose N optical components are cascaded, as shown in Figure 8.15.

Figure 8.15. Transmission through N optical components.

The ray-transfer matrix for the total system is given by

M ¼ MN MN 1 M1

ð8:6-8Þ

EXAMPLE 8.9 Show that the ray-transfer matrix for a thin lens is given by Eq. (8.6-7).

Solution: A thin lens with an index of refraction n2 has two spherical surfaces with radii R1 and R2, respectively. Suppose that the index of refraction outside the lens

MATRIX REPRESENTATION OF MERIDIONAL RAYS

127

equals n1. Transmission through the lens involves refraction at the two spherical surfaces. Hence, the total ray-transfer matrix is given by

M ¼ M2M1

¼

2 n2 n1

 

n2

32 n1 n2

 

n1 3

¼

2

1

1

3

 

 

1

0

1

0

 

 

 

1

 

0

 

 

4

 

 

 

54

 

 

 

5

 

4

 

 

5

 

 

n1R2

n1

n2R1

n2

 

 

 

 

 

 

 

f

 

where the focal length f is determined by

1

 

n2 n1

1

1

 

 

¼

 

 

 

 

 

 

 

f

 

n1

R1

R2

EXAMPLE 8.10 Determine the ray-transfer matrix for the following system:

Input

Output

d

Solution: The system is a cascade of free-space propagation at a distance d and transmission through a thin lens of focal length f. Hence, the ray-transfer matrix is given by

 

1

 

0

1

d

 

1

 

d

 

4

f

5

4

f

 

f

5

 

 

 

M ¼ M2M1 ¼ 2

1

1 3 0

1 2

1

1

d

3

 

 

 

 

 

 

EXAMPLE 8.11 (a) Derive the imaging equation for a single thin lens of focal length f, (b) determine the magnification.

Solution: The object is assumed to be at a distance d1 from the lens; the image forms at a distance d2 from the lens as shown in Figure 8.16.

Figure 8.16. Imaging by a single lens.

128

GEOMETRICAL OPTICS

For imaging to occur, the rays starting at a point A must converge to a point B regardless of angle. The total ray-transfer matrix is given by

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

d

 

 

 

 

d d

2

3

 

 

 

 

 

1 1

0

 

 

 

 

 

 

 

 

1

2

 

 

 

d1 þ d2

 

1

M

 

 

1 d2

 

 

 

1 d1

 

 

 

 

f

 

f

 

 

¼

 

0 1

2

 

 

1

3

0 1

 

¼

 

6

 

1

 

1

 

d1

 

7

 

f

 

 

 

 

 

 

 

 

 

f

f

 

 

 

 

 

4

 

 

 

 

5

 

 

 

 

 

 

 

 

6

 

 

 

7

y2 and y1 are related by

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d

 

 

 

 

 

 

 

 

 

 

 

 

d d

2

 

 

 

 

 

 

 

 

 

 

y2 ¼ 1

2

y1 þ d1 þ d2

1

y1

 

 

 

 

 

 

 

 

 

f

f

 

 

 

 

 

 

 

y2 cannot depend on y1. Hence,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d1 þ d2

d1d2

 

¼ 0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f

 

 

 

 

 

 

 

 

 

 

 

 

 

or

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

1

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¼

 

 

þ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f

d1

d2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(b) When imaging occurs, we have

y2 ¼

1

d2

y1

f

Hence, the magnification A is given by

A¼ y2 ¼ 1 d2 : y1 f

EXAMPLE 8.12 Two planes in an optical system are referred to as conjugate planes if the intensity distribution on one plane is the image of the intensity distribution on the other plane (imaging means the two distributions are the same except for magnification or demagnification). (a) What is the ray-transfer matrix between two conjugate planes? (b) How are the angles y1 and y2 related on two conjugate planes?

Solution: (a) In the previous problem, the ray transfer matrix M was calculated in general. Under the imaging condition, M becomes

 

2

m

0

 

3

M ¼

 

1y

m

 

 

 

 

 

f

 

 

4

 

y

5

where my and my equal ð1 d2=f Þ, ð1 d1=f Þ, respectively.

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