ECEN 4616/5616Optoelectronic Design

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

(The first assignment will be posted here on 1/22)

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‘Apochromatic’ correction

31 2

2

2

1 3

1,

nV

n n

Given three wavelengths, we defined the ‘V-number’ at the central wavelength as:

And, we showed (lecture #10) that the power change of a lens between the smallest and largest wavelength was:

2

1 3

2

KK K

V

From this, it followed that the power shift between λ1 and λ3 can be cancelled, if:

1 2

1 2

0K KV V

This gives us an achromatic lens combination, in which the power shift at λ1 and λ3 cancels between the two lenses. For essentially all available glasses, however, this leaves a significant focal distance variation for other wavelengths, known as the “Secondary Color”.

Traditionally, the wavelengths used are:

1

1

2

3

0.4860.5880.656

m Fm dm C

Where the K’s and V’s are evaluated at λ2.

λ1

λ3

Why Secondary Color is Bad

1. Secondary color limits the lens’ Numerical Aperture: θ NA=sin(θ)In lecture #11, we showed that the depth of focus

(dof) of a lens was: dofNAb

where b = the “acceptable blur diameter”,

and the minimum dof of a lens is the Secondary Color (SC) focal shift.Hence the maximum NA of lens (that is corrected for Chromatic focal shift)

is . For a diffraction-limited system, b = the Airy Pattern

diameter , so for high NA systems (NA → 1), the Chromatic Focal Shift must be on the order of the wavelength. This is an extremely stringent criteria to meet.

maxbNASC

2.44 / # 1.22FNA

The Microscope(Microscopes are the most critical application)

1. Microscope Resolution:θObjective lens

systemNA=sin(θ)Aerial Image

Eyepiece:(Just a specially

designed magnifier) 2sin 2

dNA

From Edward Abbe’s experimental and theoretical work (Lecture 6), we know that the maximum (cutoff) resolution of the objective

lens system is:

If we use this as the ‘acceptable blur’, then the depth of focus becomes:

2dofNA 2(NA)d

With a high-power objective with NA=0.75 and λ=0.55µm, we get a depth of focus ≈ 1µm.

Very few glasses in an achromat will give that small of a Chromatic Focal Shift.

‘Fast’ Camera Lenses

The Camera Equation gives us the irradiance of the resolved image, E’, given an object with radiance of an object: 2 2 

4 #

wEmF

R2

wRm sr

Since the exposure, (j/m2), is the irradiance times the time of the exposure, a higher irradiance of the image allows a shorter exposure time for the same energy deposited on the detector.Hence, low F/# lenses (high NA lenses) are called ‘fast’ lenses.

Df

θThe relationship between F# and NA:

# , sin

1tan2 2 #

1if tan sin , then 2 #

fF NAD

Df F

NAF

The Telescope3. Telescope Resolution:

NA=sin(θ)Aerial Image

θObjective Lens

A telescope works just like a microscope, except the objective lens is a long focal length lens and the Aerial Image is from an object at ∞. (The eyepiece design is identical.)

The limiting angular resolution for the telescope is , where D is the diameter of

the objective. While not requiring the very high NA’s of microscope objectives, there is a need for a large diameter objective, both to increase angular resolution, and increase the irradiance of the image to allow the observation of faint objects. If the NA is too low, the telescope will become inconveniently long.

Also, a 1% chromatic focal shift will result in a relative large separation between colors (given a long focal length), which may make it difficult for a high-power eyepiece to include in it’s DOF.

D

‘Apochromatic’ correction(Continued)

Partial Dispersion: As well as giving us the usual Abbe number (V-number), Zemax also lists (for most glasses) a number known as ‘Partial Dispersion’:

This value is labeled ‘dPgF’, and is defined as: g F

F C

n ndPgF

n n

Where g=0.436 µm, a line in the emission spectrum of mercury.

The general notation is:

1 2

1 2F C

n nP

n n

Note that Zemax’s choice for the secondary wavelength, g, is not universal.

‘Apochromatic’ correction(Continued)

Let’s re-label 1d

FC dF C

nV Vn n

Then,

1dFg

F g

nVn n

Then, since: d dF C F g

FC Fg

K KK K K KV V

Some algebra will show that: d dgF

FC Fg

K KPV V

For a doublet to have the same focal power at the F, C, and g wavelengths, requires:This is called an Apochromat.

1 2 1 2

1 2 1 2

0 and 0FC FC Fg Fg

K K K KV V V V

So that the apochromatic condition can be written:

1 2 1 21 1

1 2 1 2

0 and 0gF gFFC FC FC Fg

K K K KP PV V V V

It is essentially impossible to find two glasses that can meet all these requirements. However, the apochromatic (and achromatic) requirements don’t need to use only two elements. Zemax lists 4 apochromats in its ‘Samples’ directory, and all are 3 elements.

Finding ApochromatsWhile there are detailed analyses of the achromat/apochromat construction methods pre-computer (see section 7.1 of the text, or the reference website www.telescope-optics.net for examples), we will take the easy route and let Zemax do the searching.

Let’s start with the achromat that Zemax came up with last time:

This was designed over the F,d,C lines: And had a chromatic

focal shift of 37 µm:

Finding Apochromats

Zemax’s optimized achromat was near, but not quite, diffraction limited:

Our first step will be to add the ‘g’ wavelength to the Wavelength dialog:

Finding Apochromats

Extending the wavelength range greatly extended the chromatic focal shift range, from 31 µm → 178 µm

The MTF accordingly suffered.

F g

Finding ApochromatsAs an achromat, we had put a line (#2) in the MFE to force the F and C lines to have the same focal length. Now we add lines 3-5 encouraging the C,g, F,g, and d,g lines to have the same focal distance:

It won’t be possible for the optimizer to match all the focal lengths but, since the matching is dependent on details of the dispersion curves that we don’t know, we don’t want to prejudice the optimizer beforehand. It will minimize what it can and leave what it can’t change. Hence the number of ‘AXCL’ operands.

Running the ‘Hammer’ optimizer for a few minutes produces a dramatic improvement:

Finding ApochromatsIt’s still not (quite) diffraction limited:

But the chromatic focal shift has been reduced by a factor of 40.

The doublet looks reasonable But is using some fairly rare and expensive glass types.

Finding Apochromats

Since commercial apochromats (and all of Zemax’s examples) are triplets and/or air spaced, let’s add a third element to the LDE and see what the optimizer makes of it:

Lines 5 & 6 add a slab of BK7 to the back of the lens, and add a Glass Substitution Solve. Naturally, the MTF drops and the Chromatic Focal Shift goes back up to almost 200 µm.

Finding Apochromats

Using the same merit function (unchanged) we run the ‘Hammer’ optimizer once more – the effects are dramatic:

The lens is now diffraction limited(at least for on-axis objects)

The chromatic focal shift is now an amazing 0.2 µm.

The triplet looks a bit strangeAnd is still made of expensive glass:

Achromats and Apochromatsconclusion

We’ve found some interesting information about designing color corrected lenses:

1. The performance is critically dependent on picking the right glasses2. There is not enough information in the glass catalogs to make this job easy or

straight-forward.3. There is a great deal of art and theory on this subject – both in our textbook

and on the web (e.g., www.telescope-optics.com)4. In the end, only trying all glass combinations will show which combinations

work best.

5. And, fortunately, Zemax has automated the glass substitution operation so that it is mostly painless.

From our small sample, it seems that exotic, rare, and expensive glasses work best. For a practical design, it would be best to limit glass selection to what is known to be available (by communicating with your selected suppliers). This can be done by creating new glass catalogs (see Ch. 23 of the Zemax Manual), then limiting the allowable substitutions to one (or more) of those catalogs.

DiffractionThe electromagnetic wave equation for free space is: 2

20 0 02t

EE

In terms of the electric field. There is a corresponding equation for the magnetic field.

One of the simplest solutions to this equation is the Plane Wave:

( , ) cos

where: is the phase at 0, 0

, , is the position vector

, , is the wave vector in radians/meter

Note that ,etc. are direction cosines of the propagation direction

and

x y z

x

E t A t

tx y z

k k k

k

r k r

rrk

k

2

Hence: , cos( )x y zE t A k x k y k z t

k

r

Complex Notation

Euler’s formula: cos( ) sin( )ixe x i x

Richard Feynman: “…one of the most remarkable, almost astounding, formulas in all of mathematics.” (Feynman Lectures on Physics, V 1)

Proof by Taylor expansion: Taylor expansion about x = 0:

2(0) (0)( ) (0) (0)

2! !

nnf ff x f f x x x

n

Expansion of eix : 2 3 5

2 4 6 3 5 7

12! 3! 4! 5!

12! 4! 6! 3! 5! 7!

cos( ) sin( )

ix ix ix ix ixe ix

x x x x x xi x

x i x

Re exp x y zE A i k x k y k z t

, cos( )x y zE t A k x k y k z tr

Complex Notation

Using Euler’s formula, we can write the plane wave solution,

as:

Where Re(x) means ‘the real part of x’, and will henceforth be assumed.

Note that only linear operations and the operation will

give valid answers with complex notation. The real part must be extracted for non-linear operations. Also note that taking the complex

conjugate of E , produces a wave traveling in the opposite direction.Note that by letting the amplitude, A, also be expressed in complex notation, we can let A subsume the phase:

2A E E

i i

, exp( )ix y zE t A k x k y k z tr

The Aperture FunctionNote: If we know the frequency, or free space wavelength of the plane wave, then we only need to know, say kx and ky to determine the entire wave function, since:

2 2 2 2x y zk k k kk

So that: 2 2 2z x yk k k k

Let us ask: What is the field, from a plane wave, on the Z=0 plane at t=0?

exp x yE A i k x k yThe answer is:

Where the phase of the wave is implicitly contained in the complex amplitude: iA A e

Where are the direction cosines of the wave

The Aperture FunctionOne last modification is to define variables with more geometric meaning. Let:

coscos ,

2 2y yx xkku v

,x y

With these substitutions, the E-field on the Z=0 plane at t=0 becomes:

exp 2E A i ux vy

Which is the equation for a spatial frequency on the x,y,z=0 plane, where u,v have dimensions of cycles/length.

Specifically, the period of the pattern is 1/u along the x axis and 1/v along the y axis.

Spatial Frequencies

Here are two views of a spatial frequency with 3 cycles along the x-direction and 2 cycles along the y-direction of the finite array:

exp 2E A i ux vy The field on the z=0 plane due to a plane wave is just such a spatial frequency:

Where the direction cosines of the plane wave are: ,u v

, , exp 2

, , exp 2

A u v E x y i ux vy dx dy

E x y A u v i ux vy du dv

The Plane Wave SpectrumConsider the Fourier Transform Pair (from Fourier theory):

This tells us that an arbitrary function on a plane, E(x,y), can be converted into an equivalent collection of spatial frequencies on that plane, A(u,v), where A is a complex number giving both the magnitude and phase of each spatial frequency defined by u,v.

We conclude, therefore, that we can take any monochromatic (we have to know the wavelength) wave field on a plane and decompose it into a set of spatial frequencies, each of which defines a plane wave passing through the plane in a different direction. (The tilt of the plane wave w.r.t. the plane of interest is what determines the period of the spatial frequencies.)

We term the set of plane waves (which would add up to the original field on the plane of interest) as the Plane Wave Spectrum (PWS).

Evanescent Waves

The integrals in the Fourier Transform go from - to , but direction cosines have a valid range of –1 to 1.

What is the meaning of values of ? , 1u v

These values represent spatial frequencies in the field distribution, E(x,y), that have spatial periods less than , and hence cannot be formed by (and cannot create) propagating waves. These field components are known as evanescent waves, and their amplitude decays exponentially in the +z direction.

In practice, high spatial frequency disturbances are created by obstructions such as the edge of a physical aperture. While there may be very fine detail in the field close to such an obstruction, those details decay exponentially with distance and don’t contribute to the PWS at distances of more than several dozen wavelengths.Typically, we will zero out such components before initiating the propagation algorighm.

Propagating Waves in the Computer

We will use the ability of the fourier Transform to calculate the Plane Wave Spectrum of a field distribution as a simple means of simulating the propagation and resulting diffraction of waves. Say we have a monochromatic wave field in an aperture (for example, the exit pupil of an optical system, as reported by Zemax in the Wavefront Map analysis window) and we want to find what it will be at some other plane further down the Z-axis;

1. We Fourier Transform the wave field to get the PWS. (Using an FFT program such as is available in Matlab, Mathworks, or many other computer languages.)

2. Since the only thing about a plane wave that changes with space (as long as time is held constant) is its phase, we modify the phase of each component of the PWS by a formula determined by the propagation direction of that component and the distance to the next plane.

3. We use the inverse Fourier Transform (iFFT) to add the (phase adjusted) plane waves up on the distant plane to get the total wave field at that position.