ECEN 4616/5616Optoelectronic Design

Class website with past lectures, various files, and assignments:http://ecee.colorado.edu/ecen4616/Spring2014/

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Gaussian OpticsGaussian Optics refers to the use of a linear model to represent idealized optical lenses, mirrors, etc. We’ve already shown that the imaging and ray tracing equations are linear in the paraxial limit – how can they be modified to represent ideal finite-sized lenses? This is not an idle mathematical exercise, as most optical designs need to start with an ideal system which solves the desired problem, such as a simple, two-lens zoom camera lens:

Simple zoom lens based on the combination of powers formula:

1 2 1 2 1K K K K K d

The ideal lenses are then replaced with lens combinations designed to be as ideal as possible:

f

K

Recall that the focal length of a (positive) lens is defined as the distance past the lens that an incident plane wave will converge to a point, and that the lens power is defined as the inverse of the focal length: 1K

f

Since the converging wave has a curvature just past the lens, this leads to

the conclusion that the lens adds a curvature of K to an incoming wave:

1c Kf

This observation leads to the imaging equation: o iKl l’

Gaussian Imaging

1 1Kl l

1 1 Kl l

The imaging equation follows from the definition of focal length and power, hence already describes an ideal lens. (We showed that real lenses behave ideally in the paraxial limit, but the imaging equation applies to any lens combination designed to be approximately linear.)

i.e., the problem is to find “angle variables”

The paraxial ray refraction equation, , however, was only shown to hold true in the paraxial approximation where the angles, sines and tangents are all approximately equal:

u u hK

sin(u) tan(u)u

l’

u

l

u’h

K

What definition of u, u’ will allow the linear paraxial refraction equation to also describe finite-sized ideal lenses?

1 1K and u u hK for all hl l

,u u

such that:

Solve both equations for K, and equate:

u u hK 1 1Kl l

u u Kh h

1 1Kl l

Refraction Equation: Imaging Equation:

1 1u uh h l l

h hu ul l

Equating the input variables (unprimed) and the output (primed) variables independently, we see that:

, andh hif u ul l

then the linear ray trace equation and the imaging equation will be true for finite size lenses (h finite).

,,⇒

⇒ ⇒Hence, for Gaussian ray tracing (ideal lenses), we will use the

tangents of the ray angles in the diffraction equation.

Summary of Gaussian Optics (Ideal Lens) Rules

l’

u

l

u’h

K1) Lenses are represented by plane

surfaces. All calculations about a lens are done in the lens’ local coordinates, where the origin is the lens’ intersection with the z-axis.

2) Lens power, K, is the curvature that a lens imposes upon an incident wave. (Hence, in the drawing at right, 1 1K

l l

3) Rays are characterized by their positions and the tangents of the angles they make with the z-axis. (These are generally referred to as the ‘angle variables’)4) The change in the angle variable of a ray is the

negative product of the lens power and distance from the axis: u’ – u = -hK. (Hence rays are bent towards the axis by lenses with positive power and away from the axis by lenses with negative power.)

Sign Rules:1. The (local) origin is at the intersection

of the surface and the z-axis. 2. All distances are measured from the

origin: Right and Up are positive3. All angles are acute. They are

measured from:a) From the z-axis to the ray b) From the surface normal to the

rayc) CCW is positive, CW is negative

4. Indices of refraction are positive for rays going left-to-right; negative for rays going right-to-left. (Normal is left-to-right.)

h1u1 u’1u2 u’2h2

Lens1 Lens2

l1 l’1

l’2l2

d1

Summary of Gaussian Optics (continued)

Image calculation at a lens:

i ii

i i

n n Kl l

Image transfer between lenses:

1i i il l d

Ray refraction at a lens:

Ray transfer between lenses:

i i i i i in u n u h K

1i iu u

1i i i ih h d u

1n 1 2n n 2n

Focal Points of a Lens

f f’

Since lenses can work in either direction, they have two focal points.f is called the primary focal point and f’ the secondary focal point.While f and f’ follow the primed variable rule for positive lenses, they are on opposite sides for negative lenses:

f’ f

Lenses: Real and Virtual Images

f’fReal Object

Real Image

A real image can be projected on a screen.

Projection screen

Lenses: Real and Virtual Images

f’f Real Object

Virtual Image

A Virtual Image appears to come from somewhere where there are no actual rays.

Observer

Lenses: Real and Virtual Images

ff’

Real Object

Virtual Image

Lenses: Real and Virtual Images

ff’

Real Object

Virtual Image

Negative lenses only form virtual images from real objects.

Lenses: Real and Virtual Images

ff’ Virtual Object

A negative lenses can form a real image from a virtual object however.

Real Image

Lenses: Real and Virtual Images

f’

f1

f’2

Real Object

A virtual object for one lens can be a real image from the previous lens in a system.

L1 L2

Real image from Lens 1 and virtual object for Lens 2

Real image from Lens 2 and system.

Optical Devices:The Simple Magnifier: Consider a positive lens with an object less than one

focal length behind it:What is the effective magnification?

S2

Let the object height = h, and assume the user places his/her eye as close to the lens as feasible. Then, the angle that the central ray makes is the angular extent of the image for the user: f’fVirtual

image

S1

Object

ah

1

haS

We need to compare this to the maximum angle that the object can be viewed without using the lens. This is obviously dependent on how close someone can focus their eyes, so is not a universal constant. For the purpose of these kinds of calculations, however, it is usual to use -250 mm as the average close viewing distance achievable for most of the population (the older population may have to use reading glasses!). Hence the maximum angle the object can subtend to the unaided eye is:

250hb

The effective magnification is therefore:1

250aMb S

f’fVirtual image

S1

Object

ah

S2

We will further assume that the distance, S2, of the virtual image is made to also equal -250 mm. This will give us slightly more magnification, since the object can be closer to the lens than for an object at infinity.

The imaging equation then gives us:1 2

1 1 1S f S

21

2

250250

S f fSf S f

The magnification is then:1

250 250 fMS f

Note that, if the object is at the focal distance, then the object is at ∞, and the resulting magnification is: 250M

f

Refract through first element:

1 1 1 1 1 1

1 11 1

1

1 12 2 1 2 1

2

, since 0

, ,

n u n u h Kh Ku un

h Ku u u n nn

Transfer to second element:

2 1 1 1

1 12 1 1

2

h h u dh Kh h dn

Refract at second element:2 2 2 2 2 2

1 1 1 12 1 1 2

2 2

1 1 22 1 1 1 2 1 1

2

n u n u h K

h K h Kn h d Kn n

h K Kn h K h K h df n

Eliminating h1 and using where K is the

equivalent power of the combination, we get:

1Kf

General formula for combination of two surfaces:

1 22 1 2 1

2

K Kn K K K dn

K1K2

f

BFL

u1 =0 u’1=u2

u’2

h1 h2

d1

In the last lecture, we found the combination of power formula by tracing rays through a pair of surfaces:

n1’=n2

n1

n2’

The “Lensmakers” Equation

The ‘thin lens’ is an approximation, as any real lens will have a finite thickness. In the last lecture (1-17) we saw that, by applying the paraxial assumption to a surface separating materials of indices n, n’ that the power of a surface is: K c n n Where c is the curvature (inverse radius) ofthe surface, n the index before and n’ the index after the surface.

c1 c2

d1

A “Thick Lens”:To adapt the surface power combination formula to a thick lens in air, let:

1 2 1 21,n n n n n 1 1n 2 1n

1 2n n n

Substituting for the powers and indices: 1 1 2 21 , 1K c n K c n

We get the formula for the power of a thick lens:

21 2

1 2

11

c c nK n c c d

n