Ophthalmic Optics Note - Dispersion Equations

[Darryl Meister]

This thread describes various "dispersion equations," which are used to calculate the refractive index of a lens material at different wavelengths. The refractive index in the blue end of the spectrum is actually higher than the refractive index in the red end. This leads to an optical effect known as chromatic dispersion, which results in the breaking up of white light into its component colors. Chromatic dispersion is responsible for lateral and axial chromatic aberration in spectacle lenses.

The Abbe value or constringence is a measure of chromatic dispersion in lens materials. The Abbe value (v) is calculated by comparing the mean refractive index of the lens across the visible spectrum to the difference in refractive indices between the blue and red ends of the spectrum:



where n_d refers to the refractive index of the lens material for the helium d line at 587.56 nm, a wavelength produced by exciting helium gas, n_F refers to the hydrogen F line at 486.13 nm, and n_C refers to the hydrogen C line at 656.27 nm.

In some countries, the mercury e line at 546.07 nm is used as the reference wavelength for determining the mean refractive index, instead of the helium d line. (This is also why some manufacturers refer to certain high-index lens materials as "1.67" while others use "1.66.") Because of the effects of chromatic dispersion, the actual power of a lens or prism will depend upon the chosen reference wavelength. This concept is also important when measuring lenses using a device calibrated for either wavelength.

It is possible to estimate the refractive index of a lens material at a given wavelength, if you know the refractive index of the material at other wavelengths, using various dispersion equations. One common dispersion equation is Cauchy's equation, which has the form:



Of course, these dispersion equations require you to know the refractive index of the lens material at multiple wavelengths in order to determine the coefficients of the series (a, b, c...). Unfortunately, we are typically given only the Abbe value and the helium d refractive index at 587.56 nm.

A few months ago, I derived a "general" form of Cauchy's equation for any lens material, based on only the Abbe value (v) and helium d index (n), using the first two terms of the equation:



where the Greek lambda symbol represents the wavelength of interest. Only the first two terms from Cauchy's series can be used, since any additional terms would require more simultaneous equations than we could solve using only the Abbe value and refractive index.

So, using this equation, we can "estimate" that the refractive index of the blue hydrogen line (at 486.13 nm) in polycarbonate is:



While the refractive index of the red hydrogen line (at 656.27 nm) is:



Similarly, we can estimate that the mean refractive index of the lens material for surfacing or power calculations using the mercury e line (at 546.07 nm), instead of the standard helium d line used in the US, with this equation:

[HarryChiling]

abbe = (nyellow-1)/(nblue-nred)

so pluggin in our givens:

30 = 0.589 / (nblue-nred)
(nblue-nred) = 0.589 / 30
(nblue-nred) = 0.01963

Now we can make a fairly accurate assumption that the wavelengths for the red and the blue fall equidistance from the yellow wavelength except on opposit sides of the spectrum so then we can make the assumption that the indices are also equadistant from the yellow index. So lets put that word problem into a formula so that we can reduce the above even further (I used the average but it may be more accurate to use the mean instead, I don't know how much accuracy would be lost but you could always tweak the formula around here to see what the outcome would be):

nyellow = (nblue+nred) / 2

Now if we were to solve the above equation for both red and blue:

nblue = 2*nyellow - nred

and

nred = 2*nyellow - nblue

Now we can use these formulas in the top equation to further reduce to get our indices for red and blue:

2*nyellow - nred - nred) = 0.01963
2*nyellow - 2*nred = 0.01963
nyellow - nred = 0.01963 / 2
1.589 - nred = 0.009817
nred = 1.579

Now to solve for our blue index:

nblue - (2*nyellow - nblue)= 0.01963
nblue - 2*nyellow + nblue= 0.01963
2*nblue - 2*nyellow = 0.01963
nblue - nyellow = 0.01963 / 2
nblue - 1.589 = 0.009816
nblue = 1.599

I did a linear approximation some time ago trying to come up with the same. I am glad you did more research and went further into derivign a more accurate formula. Thanks Darryl.

[Dave Nelson]

Now can we conclude that the residual colour in ar coatings is a factor of the same principle? Since AR coatings are odd multiples of 1/4 wavelengths in thickness, and equal to the square root of the index of the carrier material, and since each band of the spectrum is indeed at a different frequency, it would appear that single layer coatings would be closest to one frequency, but falling away from other frequencies. This is why multi-layer coatings are the norm... they allow more efficiency over the spectrum and its changing frequencies.

[Darryl Meister]

Now can we conclude that the residual colour in ar coatings is a factor of the same principle?

The variation in refractive index as a function of wavelength plays a role, but the primary effect is due to the variation in the wavelength of light and the overall difference in refractive index between the antireflection coating and the lens material. In order for destructive interference to occur, several conditions must be satisfied (or at least nearly so).

To satisfy the path condition, the optical thickness of the coating must be equal to an odd-multiple of one-quarter wavelength of the incident light, as you noted. Since the wavelengths of light vary, however, from 380 to 760 nm, no single coating thickness will destructively eliminate every wavelength of light. The residual wavelengths not destructively eliminated by the coating are therefore partially reflected, which gives the reflex color its hue.

To satisfy the amplitude condition, on the other hand, the reflectance at the boundary of the air-coating interface must be equal to the reflectance at the boundary of the coating-lens interface. Typically, this means that the refractive index of the coating should be equal to the square-root of the refractive index of the actual lens. When this match between the coating and the lens is not achieved, full destructive interference does not occur, which gives the reflex color its intensity.

The refractive index of materials commonly used for antireflection coatings is typically not exactly equal to the square-root of the mean (helium d) refractive index of the lens material. However, because the refractive index of light varies as a function of wavelength, you would still not necessarily satisfy the amplitude condition for every wavelength, even if your coating material did match the lens material at the mean refractive index. Of course, the refractive index of the coating also varies as a function of wavelength, so the differences are probably small.

[Darryl Meister]

ps,

Although I included all of the constants in the expression:



both for the sake of showing how these numbers were derived and for allowing one to substitute the equivalent values from the mercury e system, this expression simplifies in the helium d system to a slightly less intimidating form:

[dbracer]

Daryl,

I was the guy with the big mouth (or too fast keys) from the other thread. In short I was wrong. Your equation review is quite good and gets me to thinking about optics again, which I love as much as I do the medicine aspect of optometry.

Hope you don't care if I join in again and make a few comments and ask a few questions.

Remember I can't be too smart. I'm an optometrist. Bear with me. Harry always does.

Not to mention I used to live just east of you there in Kansas.

Respectfully,
dbracer

[dbracer]

The variation in refractive index as a function of wavelength plays a role, but the primary effect is due to the variation in the wavelength of light and the overall difference in refractive index between the antireflection coating and the lens material. In order for destructive interference to occur, several conditions must be satisfied (or at least nearly so).

To satisfy the path condition, the optical thickness of the coating must be equal to an odd-multiple of one-quarter wavelength of the incident light, as you noted. Since the wavelengths of light vary, however, from 380 to 760 nm, no single coating thickness will destructively eliminate every wavelength of light. The residual wavelengths not destructively eliminated by the coating are therefore partially reflected, which gives the reflex color its hue.

To satisfy the amplitude condition, on the other hand, the reflectance at the boundary of the air-coating interface must be equal to the reflectance at the boundary of the coating-lens interface. Typically, this means that the refractive index of the coating should be equal to the square-root of the refractive index of the actual lens. When this match between the coating and the lens is not achieved, full destructive interference does not occur, which gives the reflex color its intensity.

The refractive index of materials commonly used for antireflection coatings is typically not exactly equal to the square-root of the mean (helium d) refractive index of the lens material. However, because the refractive index of light varies as a function of wavelength, you would still not necessarily satisfy the amplitude condition for every wavelength, even if your coating material did match the lens material at the mean refractive index. Of course, the refractive index of the coating also varies as a function of wavelength, so the differences are probably small.

So using these values and having some knowledge of the A/R fabricators coatings (which seems shrouded in mystery) can we predict a A/R hue.

Respectfully,
dbracer

[Darryl Meister]

I was the guy with the big mouth (or too fast keys) from the other thread. In short I was wrong.

Dbracer, there is no need to apologize; I certainly took no offense to the concern you expressed, and I'm sure that you had the best intentions. It may not be obvious to newcomers at first, but Harry and I have been bantering back and forth over optical topics like this for many years now.

Hope you don't care if I join in again and make a few comments and ask a few questions.

I don't mind at all.

Remember I can't be too smart. I'm an optometrist.

Well, the optometrists I've worked with over the years were all pretty intelligent. ;)

So using these values and having some knowledge of the A/R fabricators coatings (which seems shrouded in mystery) can we predict a A/R hue

Yes, assuming that you couldn't simply obtain an actual lens sample for direct measurement, you could theoretically predict the hue of an antireflection coating with enough knowledge of the coating stack and with a sufficient background in the mathematics involved. Though you would almost certainly need thin-film design software to do the actual calculations, which can become extremely complex for modern, multi-layer coatings. (Although, with experience, a thin-film engineer can probably estimate the effects of certain coating combinations from experience.)

[dbracer]

... Though you would almost certainly need thin-film design software to do the actual calculations, which can become extremely complex for modern, multi-layer coatings. (Although, with experience, a thin-film engineer can probably estimate the effects of certain coating combinations from experience.)

Ya know the thin-film engineering they are doing, I find just fabulous.

When I graduated from SCO in the 70's we didn't even study Abbe numbers. Then when they came out with the hi-index polymers, it seemed nice, but there were so many problems.

Hi-index glass? Give me a break.

And those old single layer A/R's were a complete joke. Some where about the mid-80's the optical industry started cleaning itself up. Frames were actually well engineered. In lenses they established better coatings and worked up polymers with better optics. Everything got so much better. And photo chromics... wow! It took'em four tries over about 10 to 15 years to work that one out in acrylics.

There just shouldn't be much whining about how things are coming along now. Not that we should try to do better, but it's exciting.

I actually find practice now so much more fun than it was 15 - 20 years ago.

Respectfully,
dbracer

[Darryl Meister]

And those old single layer A/R's were a complete joke. Some where about the mid-80's the optical industry started cleaning itself up.

Yes, antireflection coatings for spectacle lenses have certainly come a long way. Some of the most significant advancements over the past decade or two in antireflection coatings have been the move toward ion-assisted deposition to create more durable and adherent coating films and the move toward thermal-cured polysiloxane hard coatings to create a more durable and mechanically compatible interface between the coating film and the actual lens material. The newer, super-slick top coatings have also improved the overall performance of these coatings quite a bit.

[dbracer]

Like I said before, constringence isn't something we studied during training. And, I have always understood Abbe numbers as an indication of aberration.

Only recently have I given it enough attention to recognize, it really isn't. Its purely a measure of chromatic dispersion. I suppose we can separate lateral (mag) from longitudinal (focus), but that is purely and academic exercise of similar implicative properties.

Is this really a good indication of total 3rd and 4th order aberration? Is there a method of combining chromatic, oblique, coma, spherical, transverse and possible reflection problems in a single value to indicate the optical quality of a refractive lens?

Respectfully,
dbracer

[Darryl Meister]

Like I said before, constringence isn't something we studied during training. And, I have always understood Abbe numbers as an indication of aberration... Its purely a measure of chromatic dispersion. I suppose we can separate lateral (mag) from longitudinal (focus)...

You are actually correct on both accounts. As you note, the Abbe value (or constringence) is a measure of dispersion. More specifically, the Abbe value is equal to the reciprocal of the relative dispersion of the lens material, which is a quantity defined using those refractive index values that we were discussing earlier (that is, n_d - 1 and n_F - n_C).

Additionally, chromatic aberration, as defined in optical design applications, is inversely proportional to the Abbe value. Consequently, as the Abbe value decreases, the chromatic aberration increases, so that reducing the Abbe value by 50% results in twice as much chromatic aberration. This applies to both axial (longitudinal) and lateral (transverse) chromatic aberration. So, the Abbe value is a direct measure of the amount of chromatic aberration present.

Is this really a good indication of total 3rd and 4th order aberration? Is there a method of combining chromatic, oblique, coma, spherical, transverse and possible reflection problems in a single value to indicate the optical quality of a refractive lens?

In ophthalmic optics, the effects of chromatic aberration are often calculated independently of the effects of the monochromatic aberrations, mainly because chromatic aberration cannot be significantly influenced for a given lens material. This applies to the chromatic aberration of the eye of the eye as well. Also, whereas monochromatic aberrations can be compared directly on the basis of the distribution of energy in an image point, using a point spread function or similar metric, the contribution of chromatic aberration to image quality, particularly in the presence of monochromatic aberrations, is probably less understood right now.

Nevertheless, the effects of chromatic aberration on the individual monochromatic aberrations are well documented, although I haven't seen many metrics that actually combine the net effects of both aberration types into a single measure of performance. One early attempt to combine chromatic and monochromatic aberrations into a single metric was proposed by John Davis, who developed a measure of optical performance many years ago for single vision lens design that he referred to as a blur index. In the precision optics industry there may be others.

More recently, vision scientists (Applegate, Thibos, and guys like that) have been working on various metrics and models of optical performance based on the wavefront aberrations of the eye, including the low-order and high-order aberrations. They often discuss the possibility of incorporating chromatic aberration terms in their models, but I haven't seen any definitive examples yet. Although you'll often run across a lot of wavefront aberration metrics pertaining to image quality in the literature right now, including the "RMS wave error," "point spread function," "Strehl ratio," and so on, these are typically computed for a single wavelength of light.

[dbracer]

Pretty good answer. Where can I bone-up on the latter aberration "functions," "metrics" / "wave-front analysis" --- at least review their ideas.

Respectfully,
dbracer

[Darryl Meister]

Pretty good answer. Where can I bone-up on the latter aberration "functions," "metrics" / "wave-front analysis" --- at least review their ideas.

If you want to private message me with your e-mail address, I'll send a few articles along. A good book on the subject of wavefront aberrations and vision quality metrics is Wavefront Customized Visual Correction. This book is written primarily from a clinical perspective, so it should be easy to digest for anyone with a basic knowledge of visual optics.

[Falstaff]

Have just finished deriving your formula from the hints you gave. It is ingenious, and will be very useful to us. Thanks for sharing it.

One thing that puzzles me, is that we usually work with a Cauchy dispersion relation in the form:

n = n0 + n1/l + n2/l^2 + n3/l^3 + ...

where l is the wavelength.

It is not obvious to me how the two are equivalent, as n1, n3 etc. are not usually zero.

Best Regards,

Falstaff

[Darryl Meister]

n = n0 + n1/l + n2/l^2 + n3/l^3

I am familiar with a few different dispersion equations, including Cauchy's, Hartmann's, Sellmeier's, etcetera, although I haven't run across the particular form that you've provided. I'd really have to see a numerical example using specific values in your formula. It seems unlikely that you could substitute arbitrary refractive index values into n0, n1, n2, and n3 for specific wavelengths, although I could see this working if n0, n1, n2, and n3 represent coefficients that must be determined.

Here are the details of my own derivation using the first two terms of Cauchy's equation:



where a and b must be determined using the refractive index n_d and Abbe value v of the lens material.

We know that the Abbe value v is calculated from the reciprocal relative dispersion formula:



Further, n_d is defined at 587.56 nm, n_F is defined at 486.13 nm, and n_C is defined at 656.27 nm. Therefore,





Since the mean refractive index is provided by the manufacturer, n_d is known at 587.56 nm, which establishes the first simultaneous equation:



Begin by rearranging to solve for the coefficient a using the mean refractive index (n_d) of the lens material:



Now, rearrange the reciprocal relative dispersion formula:



But, when expressed as the difference between two of Cauchy's equations for n(486.13) and n(656.27), n_F - n_C is equal to:



Which is the second simultaneous equation. So, after canceling the a coefficients and rearranging to solve for the coefficient b:



Finally, after substituting the coefficients a and b back into the first two terms of Cauchy's equation and rearranging, we arrive at:



Again, only the first two terms of Cauchy's equation can be utilized, because we can only solve for two unknowns (that is, the coefficients a and b) when given only two known values of the function (that is, the mean refractive index and Abbe value).

[Falstaff]

Many thanks.

Yes, the n values are coefficients which we try to determine from spectroscopic analysis. I understand that the relation originates from an expansion in terms of photon energies (e):

n(l) = a + be + ce^2 + de^3 + ...

Using e = hc/l, where h is Plank's constant and c is the speed of light gives the previous relation.

Best Regards,

Falstaff

[Darryl Meister]

Yes, the n values are coefficients which we try to determine from spectroscopic analysis. I understand that the relation originates from an expansion in terms of photon energies (e)... n(l) = a + be + ce^2 + de^3 + ...

Yes, the terms have the form of a typical Taylor or Maclaurin series expansion. Sounds like you guys are basically fitting a curve using numerical analysis. If you are determining the coefficients from refractive index measurements at different wavelengths, you would need to know the measurement at a different wavelength for each term that you include in the expansion. Also, you would need to take measurements at sufficiently spaced distances in the far blue and red ends of the spectrum, since it could become considerably less accurate to extrapolate for wavelength measurements beyond the range used to determine the original coefficients.