I Calcuating retinal irradiance from the sun

[Stoyan Petkov]

Hi all, I want to understand how retinal irradiance (Watts per square centimeters) from the sun is calculated. Some sources calculate 11W/cm^2 like this one: https://books.google.bg/books?id=8T...age&q=calculating retinal irradiance&f=false;
but what I don't understand is how did the calculate the sun's radiance 1000W/cm^2sr? How did they come up with this number?
Quote from Wikipedia: "Dividing the irradiance of 1050 W/m2 by the size of the sun's disk in steradians gives an average radiance of 15.4 MW per square metre per steradian.
How did the calculate that, I mean which solid angle are they using? If I calculate the solid angle regarding a sphere with center of the sun, and area subtended by this solid angle of 1 m^2 on the Earth surface, the solid angle will be so small that dividing 1050W/m2 to it will produce something much more than 15.4MW/m^2*sr. So any help will be appreciated

 

[tech99]

Radiance is at the Sun and Irradiance is at Earth.

 

[Stoyan Petkov]

what is the relation between them, quote again "Dividing the irradiance of 1050 W/m2 by the size of the sun's disk in steradians gives an "average radiance of 15.4 MW per square metre per steradian.", I mean how the steradians are calculated is what I don't understand

 

[tech99]

what is the relation between them, quote again "Dividing the irradiance of 1050 W/m2 by the size of the sun's disk in steradians gives an "average radiance of 15.4 MW per square metre per steradian.", I mean how the steradians are calculated is what I don't understand

Each sq metre of the Earth surface receives 1050 Watts. Next we find the area occupied by a Steradian centred on the Sun at the distance of Earth. From this we find the power intercepted by that area. This is then the power radiated by the Sun into each Steradian. Next we divide that power by the area of the Sun, and this tells us how much power each sq metre of the Sun radiates into one Steradian.

 

[tech99]

Each sq metre of the Earth surface receives 1050 Watts. Next we find the area occupied by a Steradian centred on the Sun at the distance of Earth. From this we find the power intercepted by that area. This is then the power radiated by the Sun into each Steradian. Next we divide that power by the area of the Sun, and this tells us how much power each sq metre of the Sun radiates into one Steradian.

May I mention that at a distance from the Sun D, the surface area of a Steradian is D sq metres.
Why the illumination people do not these days treat it all the same as radio waves, I do not know. In other words, using TX power and antenna gains.

 

[Andy Resnick]

what is the relation between them, quote again "Dividing the irradiance of 1050 W/m2 by the size of the sun's disk in steradians gives an "average radiance of 15.4 MW per square metre per steradian.", I mean how the steradians are calculated is what I don't understand

The solid angle subtended by the sun (or any luminous object) Ω = A/r², where A is the area of the object (here, assumed oriented normal to the viewing axis), and r the distance to the object. The sun and moon both subtend a solid angle of about 7×10⁻⁵ sr.

Hidden away in the solar flux calculations is the angular distribution of emitted radiation: the sun is isotropic, while the moon is (nearly) Lambertian. This plays into the overall radiance because each differential element of the source subtends a solid angle dΩ = dA cos(θ)/r², where θ is the angle between surface normal and viewing axis for that differential surface element.

Also, if you are interested in photometry rather than radiometry, you need to include the spectral weighting distribution (photopic curve/luminosity function) for vision

 

[tech99]

The solid angle subtended by the sun (or any luminous object) Ω = A/r², where A is the area of the object (here, assumed oriented normal to the viewing axis), and r the distance to the object. The sun and moon both subtend a solid angle of about 7×10⁻⁵ sr.

For this calculation I believe we need the angle subtended by Earth at the Sun, not the other way round.

 

[Stoyan Petkov]

For this calculation I believe we need the angle subtended by Earth at the Sun, not the other way round.

I think you're right, but calculating with 7x10^-5 steradians gives approximately 15MW. Otherwise the calculated radiance in Wikipedia is not correct. If we calculate the way you propose the radiance will be multiple times higher

 

[Andy Resnick]

For this calculation I believe we need the angle subtended by Earth at the Sun, not the other way round.

Definitely not. The detector is on Earth.

 

[sophiecentaur]

For this calculation I believe we need the angle subtended by Earth at the Sun, not the other way round.

The area of the Earth is of no consequence. We are only interested in 1m² of the Earth's surface or whatever else area our detector takes up..

 

[Stoyan Petkov]

Well, maybe you can see why I'm confused calculating the sun's radiance. That's why I decided to calculate retinal irradiance in a different way. The irradiance of the sun is 1050W/m². All sources about the power of the sun give this number more or less. The size of the pupil while watching bright light is 2mm, which is 3.14x10^-6 m². If we multiply the irradiance of the sun by the area of the pupil we'll get the power at the pupil in Watts. The result is 3.297x10^-3 Watts. The retinal image of the sun is with diameter 0.159mm, or 0.0159cm which is equal to an area of 0.198x10^-3 cm². The eye focuses the power from 2mm pupil size to 0.159mm retinal size. So the power remains the same, but the irradiance increases, since the area gets smaller. So dividing 3.297x10^-3 Watts to 0.198x10^-3 cm² we get 16.65 W/cm². That's if we assume the transmittance of the ocular media equal to 1. This is about 1.5 times higher than what is calculated here https://books.google.bg/books?id=8T...age&q=calculating retinal irradiance&f=false;, Furthermore my google search for retinal irradiance from the sun gives me results of documents and books that calculate or just state that the retinal irradiance is in this range 10-11 W/cm². So am I right or wrong? One thing is bugging my mind - that the size of the sun is 0.159mm, but even though the eye is focused on the sun this is not the only thing it sees. So maybe the 2mm image size on the pupil is not focused on image size 0.159mm but on rather larger area corresponding to the image of the complete landscape the eye sees, thus resulting to less optical gain. Anyway this is just a guess.

 

[Stoyan Petkov]

Definitely not. The detector is on Earth.

I have to say that the more I search, the more confused I get
I found post in this forum, here it is: https://www.physicsforums.com/threads/example-of-calculation-of-radiance.490538/#post-3248682.
Actually the post redirects here: http://omlc.org/classroom/ece532/class1/radiance_flashlight.html, where is explained how radiance is calculated. The author uses solid angle with centre located at the source, and as far as I read your reply you said that the author's calculations are correct. But in this thread we talk about solid angle with centre located at the observer/detector, i.e. this is the viewing angle. Here are another quotations I have found on the web again with viewing angle:

Quote: "Radiance is a measure of the flux density per unit solid viewing angle, expressed in W/cm2/sr" from http://www.dfisica.ubi.pt/~hgil/fotometria/HandBook/ch07.html.
Another Quote: Radiance is measured by viewing a defined area of the source using an input optic. It is defined as the measure of the flux density per unit solid viewing angle" from http://www.azooptics.com/Article.aspx?ArticleID=459

I have also attached pdf file in which sun's radiance is calculated, again using viewing angle.

Please, can you explain to me which solid angle should be used - centred at the source, or centred at the observer(detector)?

 

[sophiecentaur]

I have to say that the more I search, the more confused I get

I am only using my native cunning in this answer as I have never needed to work this out - but I do know about antennae! You seem to have got hung up on the term 'radiance' and the way people have chosen to define or use it; the second post here points out the important difference between radiance and irradiance. . Perhaps you could ease up on the Research and approach the problem from scratch - at least, attempt to - and you may come to a better understanding. But that link seems pretty clear to me and it has a fairly practical approach. Did you read it through or just scan for some terms? (I do that a lot)

The irradiance of the sun is 1050W/m2.

That refers to the flux density arriving on the Earth's surface and it's all you need to know. It wouldn't matter if the sun was half the diameter and its radiance was 4X as high - same power flux arriving per m² on Earth. The 1kWm^-2 is a measured value and, for the Sun, it's all you need to know. The above link, however, considers a distributed source. (No point-source approximation)

can you explain to me which solid angle should be used - centred at the source, or centred at the observer(detector)

It depends on the situation but, in this (easiest) case:
Ideally, all that gets through the Iris (area) of the Eye will hit the retina. That flux depends on the solid angle, presented by your iris and the radiance of the Sun but you can bypass that step because you know the radiance on Earth and you can scale the 1kW value to the aperture area. (i.e. around 1/π10⁶). The Irradiance on the retina will depend on the size of the image. That will be about 20tan(0.5) ≈ 0.2mm (I reckon). So you can work out the Irradiance at the retina that way.
If you are dealing with a close up, extended and directional source of light then you would need to integrate over the whole source to get the total light landing on the image - i.e. taking the solid angle of the source into account as well as the (above) solid angle of the receiver.

 

[Stoyan Petkov]

sophiecentaur, thank you for the reply.

The goal of my search is to calculate retinal irradiance (W/cm^2) while measuring irradiance of the sun (which is in fact the corneal irradiance) during the course of the day. Unfortunately I didn't find direct correlation between both, here is quote from David Sliney, Ph D, Laser Institute of America, author of articles about retinal damage from extended and other light sources:

This is from his article "Quantifying retinal irradiance levels in light damage experiments "
Note the underlined in "It is not readily related to corneal irradiance", so knowing the solar irradiance is not enough. We have to know the solar Radiance, respectively the solid angle involved.
In general the question here is how is Ls (Radiance of a source), W/(m^2*sr) is calculated? Watts, and square meters are clear to me, but defining the solid angle seems to be a contradiction (even in this forum).

 

[sophiecentaur]

We have to know the solar Radiance, respectively the solid angle involved.

I don't see where you are going here. If you know the irradiance from the Sun then you know the power flux through the cornea. That power is distributed over an area of the retina which is the Sun's image. Isn't that what you want to know?
The sun's brightness over most of its disc does not vary much - just on the outer limb. See http://www.cv.nrao.edu/course/astr534/Brightness.html and many others. So the irradiance on (of?) the retina will be much the same over the whole image and equal to the power flux divided by the area of the image.

I don't understand how that attachment can say that the Source Radiance is affected by the focal length of the eye when the source is the Sun unless they are using some definition of their own. This hyper physics link defines Radiance in terms of Watts/m² steradian, which makes good sense to me because it doesn't involve the receiving equipment at all. You seem to be assuming that those yellow bits of text are appropriate to what you want. (I appreciate that I am not yet considering the transmittance of the optical media - that's something to sort out later, I think). They are dealing with lasers which have narrow beams and so the beam may be smaller than the cornea - not your problem but could be very relevant to them.

The goal of my search is to calculate retinal irradiance (W/cm^2) while measuring irradiance of the sun (which is in fact the corneal irradiance) during the course of the day. Unfortunately I didn't find direct correlation between both

During the day, the iris will change the area of the aperture in order to keep the irradiance on the retina fairly constant (won't it?) so why would you expect there to be a correlation? That comment of mine is, of course, made in the dark and you may have included this your experiment. There's no knowing what sort of feedback could be at work in our 'autoexposure' software. Is this work on humans or animals, btw?

 

[Stoyan Petkov]

I think I have found an answer to what I was looking for. I am not sure about the name of the first document, because I couldn't download it completely. It has something to do with optics geometry, optical design, optical sources etc.

Here is quote:

Edward F. Zalewski
Hughes Danbury Optical Systems
Danbury , Connecticut
CHAPTER 24
RADIOMETRY AND
PHOTOMETRY
24.3 RADIOMETRIC DEFINITIONS AND BASIC CONCEPTS
...
Radiance. Radiance, shown in Fig. 2, is the ratio of the radiant power, at an angle θs to
the normal of the surface element, to the infinitesimal elements of both projected area and
solid angle. Radiance can be defined either at a point on the surface of either a source or a
detector, or at any point on the path of a ray of radiation.
End of quote

And another quote:

I L S C ® 2 0 1 3 C o n f e r e n c e P r o c e e d i n g s
The radiance of the sun, a 1 mW laser
pointer and a phosphor emitter
Karl Schulmeister

For me there are some important outlines here:

1. The vertex of the solid angle involved in calculating the radiance can be at either the surface of the source or detector;
2. The vertex of the solid angle must belong to the surface of interest, i.e. the vertex and the surface area form a pair, so if we decide to use the angle subtended by the detector with vertex at the sun as tech99 suggests we have to consider the irradiance of the projected area of the source(sun), at the source which of course is impractical;
3. Radiance remains constant through optical system with lenses (radiance conservation theorem);
4. The products of the projected areas by the corresponding solid angles are equal due to geometry. (See the yellow highlight above)
This makes it possible to calculate the irradiance at retinal image knowing the solid angle subtended by the pupil.
5. From p.3 and p.4 we derive that Power also remains constant in optical system with lenses (of course if transmittance is equal to 1) Power1/Area1*solid angle1 = Power2/Area2*solid angle 2, since the products of areas and solid angles are equal, then power must be also equal.

If you know the irradiance from the Sun then you know the power flux through the cornea. That power is distributed over an area of the retina which is the Sun's image

.
You are absolutely right. I was on the right path from the very beginning. The difference between my calculation and the one in the book
https://books.google.bg/books?id=8T...epage&q=calculating retinal irradiance&f=true

derives from the fact that I didn't notice sun's radiance is stated to be 1000W/cm^2*sr, which is about 680W/m^2 Irradiance, not 1000W/m^2 which I used in the calculation, because I thought that was the value. The steradian at denominator I thought was a typing mistake (how dumb of me) hence the 1.5 times mistake.

I don't understand how that attachment can say that the Source Radiance is affected by the focal length of the eye when the source is the Sun unless they are using some definition of their own

The radiance of the source, together with the focal length of the eye (which defines the solid angle) both define the retinal Irradiance as you can see from the formula:
Er=pi*Ls*t*de^2/4f^2

Source Radiance is independent of focal length, not sure how you came up with this conclusion.
Basically what this formula does is multiplying the Radiance of the source by the solid angle subtended by the pupil with vertex at the retina ( and of course transmittance is taken into account ). Of course this works fine for small angles as area of the pupil is used for calculating the solid angle.
Both ways - with this formula, and with corneal irradiance yield the same results, I have checked it. The reason for deriving this formula, as explained in "Safety with lasers and other optical sources" is the necessity to have easy to use formula for every source with known Radiance without the concern of viewing angle, viewing distance (i.e. size of retinal image doesn't need to be calculated) and also corneal irradiance measurement is not needed.

They are dealing with lasers which have narrow beams and so the beam may be smaller than the cornea - not your problem but could be very relevant to them.

As far as I can understand Mr. Sliney has been specialized in the field of Non-Ionizing Radiation protection, which means all sources emitting this kind of radiation. This formula has been derived for extended sources like the sun. There is whole chapter devoted to sunlight and its damaging effect on the eyes.
Laser safety is subject to another chapter in "Safety with lasers and other optical sources" which is not relevant for me, but I think the approach is different and this formula is not applicable.

During the day, the iris will change the area of the aperture in order to keep the irradiance on the retina fairly constant (won't it?)

This is basically true, but given the fact that the sun is very bright source I will assume maximum constructed pupil with 2mm size (of course use of drugs and mental problems are not taken into account here) during the exposure as I consider direct gazing at the sun (I'm not sure if I mentioned it). Of course comparative calculations can be made with 1,5mm, 2,5mm, 3mm to reveal how small changes in pupil size affect exposure duration.
There is information that the pupils dilate independently from the brightness as the exposure increases, because of "saturation" I think. This is observed for prolonged exposure, but I think the safe exposure durations which is the goal of my calculations will be far less shorter. Anyway this needs clarification.

 

[sophiecentaur]

The radiance of the source, together with the focal length of the eye (which defines the solid angle) both define the retinal Irradiance as you can see from the formula:

Which do you want, radiance or irradiance? That attachment talks about radiance being dependent on focal length. It may jut be a typo but it cannot be right.

It strikes me that it is a very bad idea to experiment with looking directly at the sun, even with the naked eye. It can very soon lead to blindness.

small changes in pupil size affect exposure duration

Are we talking about eyes or cameras? Edit: sorry. I get it now.

 

[Stoyan Petkov]

Which do you want, radiance or irradiance? That attachment talks about radiance being dependent on focal length. It may jut be a typo but it cannot be right.

The goal is Retinal irradiance, W/cm^2 because you can then calculate the energy dose J/cm^2 for given duration. In the literature all threshold values for retinal lesions are in J/cm^2. However I started this thread because I couldn't understand how Radiance of the source is calculated when Irradiance at a surface is known.
Yes you're right Ls, source radiance is independent from focal length, it doesn't make sense. Maybe you got confused because I have mixed a little bit the attachments, as the text is spread across two consecutive pages. But to be honest I didn't find text segment about radiance of the source depending on focal length. I will attach it again.

It strikes me that it is a very bad idea to experiment with looking directly at the sun, even with the naked eye. It can very soon lead to blindness.

This is correct, but even the scientist who work in the area of eye damage by the sunlight admit that one can gaze at the sun "at very low solar elevations and for short periods of time" or "when the sun is yellow or orange indicating that the hazardous blue light has been scattered, then the sun may be fixated for many minutes without risk".

Of course gazing at the sun during noon time is clearly damaging, no discussion here, although some ophthalmologists claim that damage of the retina appear after about 90 seconds of exposure (a lot can be debated here). You can watch it during sunrise and sunset, but I want to calculate how long. As you can see in the abovementioned quotes there are no exact numbers. "Low solar elevation" - how low, "many minutes" - how many, also what time of year, latitude and longitude of the location, pupil size, etc.

There are safe exposure durations from organizations like International Commission on Non-Ionizing Radiation Protection (ICNIRP), and American Conference of Governmental Industrial Hygienists (ACGIH), but the values are calculated with substantial safety factor, for example maximum gazing time at noon - 0.25 seconds.

 

[sophiecentaur]

Now you have explained the context, I realize that you are well aware of the risks involved. haha! Poor old rats and pigeons.
I still have a problem accepting that the radiance of the Sun is of concern. All that's required is the angle subtended by the sun and the power flux density (modified by atmosphere etc). Those values plus the optics of the eye will give you the irradiance at the solar image on the retina.
AS we have all looked at the Sun at times when we 'shouldn't', I realize that it can't be instant frazzle. I guess that is part of the evolutionary 'design' of our eyes. They absolutely have to be able to deal with accidental overload for short periods.

I couldn't understand how Radiance of the source is calculated when Irradiance at a surface is known.

I think that it would be reasonable, given the irradiance, the aperture and the distances involved, to infer the radiance of the Sun but isn't the important factor of the area of the Sun's disc needed too, when you are dealing not only with irradiance but total power dissipation on areas of the retina? Total flux on the retina over a small area is surely going to affect the severity of any damage.
As you say, it's not a simple problem. It's not the sort of study that yields useful information from the effects of human eye injury because the conditions causing recorded injuries can't often be known - unlike many other situations involving radiation damage of other sorts.

 

[Stoyan Petkov]

AS we have all looked at the Sun at times when we 'shouldn't', I realize that it can't be instant frazzle.

For the sake of correctness the duration is not 0.25 seconds, but 0.6 seconds, my mistake. Not that much of a difference though.

I think that it would be reasonable, given the irradiance, the aperture and the distances involved, to infer the radiance of the Sun but isn't the important factor of the area of the Sun's disc needed too, when you are dealing not only with irradiance but total power dissipation on areas of the retina?

This indeed was something to think about, but I think that the damage will appear in the area of highest irradiance. The blurred image around should have lower irradiance, and thus if we calculate safe levels for the focused image (159um for the sun), the dissipated rays shouldn't be a problem.

 

[sophiecentaur]

I think that the damage will appear in the area of highest irradiance.

Absolutely- but the level of irradiance in the areas immediately surrounding the peak is likely to have an effect. You get similar effects with film and CCD camera sensors, where there is a region around the peak which is affected. It would be reasonable to suspect and a sensor that's subjected to an isolated peak of irradiance could be affected less than a sensor that's in amongst a number of others. Over a longer period, the temperature would be higher (way past the point you're interested in, I would think) but what about charges and chemicals, produced by the overexposure? The effect of the actual image size on a sensor will affect the image so I would expect the retina would react in a similar way. The Sun image is always more or less the same size, of course, when in focus.

 

[Stoyan Petkov]

I will attach a graph representing the relative irradiance on the retina. 1,0 represents 100% of the irradiance calculated by the formula Er=pi*Ls*t*de^2/4f^2. The image of the sun on the retina is 159um, so there will be maximum irradiance at the image. The author notes that the formula breaks down for images smaller than 10 min of arc, or 50um . As you can see the irradiance decreases as we go further from the centre of the image. The important part of the image is where the maximum irradiance is, and the damage will(should?) appear at this area if the threshold values are exceeded. It doesn't make sense to me that the damage will appear at other areas with lower irradiance. (the image is from free preview of book available on google books)