Effect of Magnification on Light Output from a Binocular (1 Viewer)

[Ignatius]

The hyperventilating in this thread, especially by those who have not the wherewithal to refute @grackle314, reminds me a bit of the events following the publication of Copernicus' De revolutionibus orbium coelestium in 1543. One member even went as far as far as PMing me, more or less telling me directly to withdraw my comment. Personally, I consider that to be beyond the pale and not worthy of mature discourse. YMMV.
🍿 ...

Grackle314 has simply taken a known quantity of light, shoved it through some optical instruments at his disposal and used a Newport optical power meter model 840 of known accuracy to measure what came out of the other end. He then proceeded to make a matrix of the parameters and date he collected.
Whether this information is useful is a different matter. His experiment is basically like measuring the transmission of an optical instrument, and for that there is even an ISO norm.

To make this totally abstract: someone gives you a black box with two apertures of different diameters and a quantifiable light source. You then stick the light in one aperture and measure what comes out of the other one. This can be done without knowing anything at all about what happens in the black box. There may be lenses in there, prisms, filters, mirrors or nothing at all. The measurements at the other end will give you data which can be set into relation to, for example, the diameters of the two apertures. The relevance or usefulness of this data can then be discussed. The validity of the data cannot. Neither can the experimental setup.

 

[Tringa45]

reminds me a bit of the events following the publication of Copernicus' De revolutionibus orbium coelestium in 1543.

That is a candidate for the exaggeration of the century!
None of us here has the patience or energy to dissect so many misconceptions.
The OP should look up "Refracting telescope" in Wikipedia or go and read Holger Merlitz' book or "Telescope Optics" from Rutten & van Venrooij.

John

 

[Swedpat]

I think I

Light coming out of a binocular exit pupil is related to the light going into the objective. Minus the transmission losses, the light power collected through the objective is concentrated into the smaller exit pupil. The exit pupil area is approximately equal to the objective area divided by the square of the magnification. For a fixed objective area, the light from a higher magnification will be more intense at the exit pupil, in watts per square millimeter since a higher magnification has a smaller exit pupil area.

As an example, a 42 mm diameter objective will have an exit pupil diameter of 4.2 mm for a 10x magnification and 5.25 mm exit pupil diameter for an 8x . The light power output is spread over the area of the exit pupil, which is reduced compared to the objective area by the square of the magnification. Hence, for a fixed objective the light intensity power per square mm at the exit pupil is higher for the 10x compared to the 8x by the ratio of the reduction of the areas: 100 / 64. Simply put, the light goes into a smaller area for the 10x than the 8x . Keep in mind that the human eye pupil also restricts the light entering the eye, frequently this eye restriction is much smaller than the exit pupil. In typical bright daylight, most human eyes have pupils of 2 mm diameter or less. The light arriving from the binocular exit pupil which is outside the eye pupil will not enter the eye, this light is wasted for observation except that a larger exit pupil makes it easier to find the image. For the fixed objective where exit pupils are greater than the eye pupil, the greater magnification will have greater intensity entering the eye approximately by the square of the magnification ratio. One would expect for a fixed objective area with exit pupil greater than the human eye that a 10x binocular compared to an 8x in daylight will deliver approximately 100/64 = 1.56 more light per square mm to the eye pupil.

Below are the results of an experiment done with binoculars I have at hand. There are two binoculars with 10x and two with 8x, representing four manufacturers and four sizes of objectives.

For a constant input flat sheet white light source at one meter from the objective with the objective looking down a 7.5 cm diameter 1 m long tube and using a Newport optical power meter model 840, the exit pupil power was measured and the exit pupil power per square millimeter calculated.

One may see in the chart below both 10x binoculars have exit pupil output power of 0.57 to 0.58 microwatts per square millimeter and the 8x binoculars are at 0.40 microwatts per square millimeter. The 10x deliver more light per square millimeter at the exit pupil than the 8x . To compare with the 100/64 expected ratio, 0.40 microwatts per square mm x 100/64 = 0.625 microwatts per square mm which is 8% above the observed 10x power per square millimeter. This is a simple experiment and gives average power density over the exit pupil. Refinements such as radial power profile would be expected to alter the results a bit. And the human eye perception of this power is not discussed here other than to note the human eye also has a radial power detection profile.

Binocular	M x Objective	Exit Pupil Diam.	Power out Exit Pupil (microwatts)	Microwatts per square mm
Swaro NL Pure	10x42	4.2	8.10	0.58
Zeiss VP	8x25	3.125	3.10	0.40
Kowa Genesis	10x22	2.2	2.17	0.57
Leica UVD	8x20	2.5	1.97	0.40

I understand how you think. But I still don't think it works like that. It is considered as a well based fact that relative brightness is proportional to exit pupil area.
For example a 10x42 can never provide as bright image as a 8x42 as long as full exit pupil is used.
If you compare 4,2 to 5,25mm exit pupil I don't think a 4,2mm exit pupil will concentrate the light amount compared to 5,25mm.
If you compare looking at a light source the apparant area of the light source is ~56% higher with 10x than 8x. The total light amount reaching the eye is the same but the intensity of the light source is actually lower with 10x than 8x.
This the case with a pretty small and/or weak light source. When looking at a stronger light source, like a street light pretty close, or the moon, the eye pupil will surely constrict to less than 5,25mm, probably less than 4mm.
In this case you will make use of more of the aperture of the 10x42 than with 8x42, resulting in more light reaching the eye. But the light will never be more intense, just higher amount. Because light intensity is the same as relative brightness, and 10x42 can never get higher relative brightness than a 8x42.
As I understand it...

 

[agus_m]

I think I

I understand how you think. But I still don't think it works like that. It is considered as a well based fact that relative brightness is proportional to exit pupil area.
For example a 10x42 can never provide as bright image as a 8x42 as long as full exit pupil is used.
If you compare 4,2 to 5,25mm exit pupil I don't think a 4,2mm exit pupil will concentrate the light amount compared to 5,25mm.
If you compare looking at a light source the apparant area of the light source is ~56% higher with 10x than 8x. The total light amount reaching the eye is the same but the intensity of the light source is actually lower with 10x than 8x.
This the case with a pretty small and/or weak light source. When looking at a stronger light source, like a street light pretty close, or the moon, the eye pupil will surely constrict to less than 5,25mm, probably less than 4mm.
In this case you will make use of more of the aperture of the 10x42 than with 8x42, resulting in more light reaching the eye. But the light will never be more intense, just higher amount. Because light intensity is the same as relative brightness, and 10x42 can never get higher relative brightness than a 8x42.
As I understand it...

I think both are right. I had optics classes a long time ago, so I can't find good analogies to help explain both "points of view". The thing is that the light at the exit pupil is not focused, if you add a focusing element (like an eye), the image formed at the focus plane would have different sizes (different magnifications) and the power density (power/area of image) would be the same (seen as same brightness), at full aperture. But at the exit pupil, the flux density is higher for the higher mag bino.

 

[Brink]

The idea behind a binocular is pretty simple. Start by thinking about how your vision works;

Let's say you are looking at a white sheet of paper from 10 meters away. It looks white and has a certain size (lets call this Object brightness = 1 and magnification =1). As you move away from the paper, the light intensity drops with distance squared. So from 100 meter, there are 100 times less photons per unit area. Your eye doesn't see the paper get dimmer though, it just sees it get smaller. So at 100 meters, it would have Object brightness =1 and magnification = 0.01.

A binocular has the effect of moving the person closer to the source (it lets you see the exact same light photons as if you were standing x times closer to the source). So if you look at the sheet of paper through a 10x binocular from 100m distance, it will have Object brightness = 1 and magnification = 1. The binocular needs to have a large enough objective lens to fill the pupil though, which means it needs to be 100 times larger than the actual pupil or the brightness will start dropping to less than 1. The 8x binocular at 100m would have Brightness = 1 and magnification = 0.64 (8^2/10^2).

Look at it two ways
1) The eye is dilated small. If these are both 42mm binoculars and the pupil is dilated to 3m: for the 10x at 100m: object brightness = 1, magnification = 1. For the 8x at 100m, object brightness = 1 and magnification = 0.64.

2) The eye is dilated large. If both binoculars are 42mm and the pupil is 6mm: for the 10x at 100m: brightness = ~0.5, magnification = 1. For the 8x at 100m, object brightness = ~0.8 and magnification = 0.64.

Modified versions of your (grackle) experiment could be useful for measuring light transmission, actual magnification of a telescope, or field of view (if your light source subtended an area larger than the FOVs). But here your central point is: "One would expect for a fixed objective area with exit pupil greater than the human eye that a 10x binocular compared to an 8x in daylight will deliver approximately 100/64 = 1.56 more light per square mm to the eye pupil." This is true but it would also be true to say: "One would expect for a fixed objective area with exit pupil greater than the human eye that a 10x binocular compared to an 8x in daylight will deliver approximately 100/100 = 1 equal light per square mm to the eye retina." I would argue that the revision is a more valuable statement since we see with our retinas and not our pupils.

 

[Ignatius]

But here your central point is: "One would expect for a fixed objective area with exit pupil greater than the human eye that a 10x binocular compared to an 8x in daylight will deliver approximately 100/64 = 1.56 more light per square mm to the eye pupil." This is true but it would also be true to say: "One would expect for a fixed objective area with exit pupil greater than the human eye that a 10x binocular compared to an 8x in daylight will deliver approximately 100/100 = 1 equal light per square mm to the eye retina." I would argue that the revision is a more valuable statement since we see with our retinas and not our pupils.

I understand that as far as the eye goes we see with the retina, but it is not clear to me how you arrive at the different numerical values for light reaching the pupil and the retina. For me the implication is that at the pupil different amounts of light/mm² arrive from 8x and 10x binoculars of equal aperture, but that at the retina the same amount of light/mm² arrives both from 8x and 10x binoculars of equal aperture.
How can that be?

 

[Holger Merlitz]

What grackle has measured is the conservation of energy: Whatever light enters the instrument through the entrance pupil (objective) has to exit the instrument through the exit pupil (neglecting transmission losses), and since the exit pupil has an area that is smaller by 1/m², the luminous flux per unit area has to be higher by a factor of m².

This is not the brightness we perceive when observing a target: The exit pupil is the real image of the entrance pupil, but what we perceive is the virtual image in which the target area is expanded by a factor of m². The perceived luminance ('brightness per unit area') of the target is reduced by a factor of m² due to it being spread out over a larger area, but this loss is compensated by the m² increase of the luminous flux per unit area through the exit pupil, so that eventually the target appears to be as bright through the binocular as it appears with the bare eye. Only if the exit pupil is smaller than the eye pupil, then the luminous flux, now to be averaged over the eye pupil area, is less than m² times the bare eye's luminous flux that enters the eye from the target, and then the image of the target begins to turn dull. This is called overmagnification. An increase of the objective diameter or a decrease of the magnification would then be needed to match eye pupil and exit pupil diameter, so that the instrument creates an image that is as bright as the one perceived with the bare eye.

Cheers,
Holger

 

[Maljunulo]

I believe that we see with the brain, which receives impulses from the optic nerve (which are the output of the retina) and magically transforms the whole business from both eyes into what we see.

The retina is just a signal generator.

 

[Brink]

I understand that as far as the eye goes we see with the retina, but it is not clear to me how you arrive at the different numerical values for light reaching the pupil and the retina. For me the implication is that at the pupil different amounts of light/mm² arrive from 8x and 10x binoculars of equal aperture, but that at the retina the same amount of light/mm² arrives both from 8x and 10x binoculars of equal aperture.
How can that be?

Good catch- obviously this can’t be the case. His point is not true. The same amount of light per unit area would arrive at the lens as well as at the pupil if the light source was larger than the FOVs (assume equivalent AFOV of the devices).

 

[agus_m]

The same amount of light per unit area would arrive at the lens

Yes, but if the eye's pupil is 4.2 mm, then the eye would catch 100% of that light for the case of the 10x bino. For the 8x bino, there is some light spillage because the exit pupil is now 5.25 mm, and the eye would catch 64% of that light.

 

[Brink]

Yes, but if the eye's pupil is 4.2 mm, then the eye would catch 100% of that light for the case of the 10x bino. For the 8x bino, there is some light spillage because the exit pupil is now 5.25 mm, and the eye would catch 64% of that light.

No, if the amount of light per unit area is the same, the total amount of light is the same for any pupil diameter less than the 10x exit pupil. For any larger pupil size the 8x lens receives more light.

 

[agus_m]

No, if the amount of light per unit area is the same, the total amount of light is the same for any pupil diameter less than the 10x exit pupil. For any larger pupil size the 8x lens receives more light.

The total amount of light gathered by the instrument is determined by the objective size, it is proportional to the area.

 

[Swedpat]

We have to differ between total light amount and relative brightness. If the eye pupil during a certain observation is max 4,2mm, more light can enter the eye through 10x42 than 8x42. But the brightness/light intensity will remain the same.

 

[Brink]

Neither of these statements is correct.

The same amount of light enters the eye, given a uniform field and equivalent instrument AFOV. This is due to the wider angle of acceptance of the 8x instrument; i.e. it 'gathers' light from an area wider than the the 10x instrument (the area of the FOV is 64% smaller just like the exit pupil is 64% smaller in the case of a 10x42 compared to 8x42).

 

[Swedpat]

Neither of these statements is correct.

The same amount of light enters the eye, given a uniform field and equivalent instrument AFOV. This is due to the wider angle of acceptance of the 8x instrument; i.e. it 'gathers' light from an area wider than the the 10x instrument (the area of the FOV is 64% smaller just like the exit pupil is 64% smaller in the case of a 10x42 compared to 8x42).

Now this starts to be deep...🥴
"Angle of acceptance", not sure what you mean.
And I am still not sure how to calculate when different AFOVs are included.
Some guy explained it somewhere on some thread several years ago, not sure I understood it back then.
I still think the exit pupil area is what we can use to know the relative brightness, it's an established value for optics.

 

[Swedpat]

The example I earlier thought about:
If we have two optical instruments with same magnification and aperture. Which means same exit pupil.
One with 50deg AFOV and the other with 70,7deg. In the latter case twice of the area is covered. If you then watching an even illuminated surface the question is: will twice of covered area mean twice the light amount is reaching your eye, or will the true relative brightness halved with twice field area, so the total light amount remains the same?
The latter should be correct, otherwise it means a narrow AFOV does not make use of the full aperture.
I am still confused because I have never read about a claimed relation between AFOV and brightness.
Wide FOV is one of the most important among many users, but I have never read someone wants a narrow FOV because it will be brighter. Some should do it if it increases the apparant brightness of the image.
I think Holger Merlitz can explain this.

 

[agus_m]

Neither of these statements is correct.

The same amount of light enters the eye, given a uniform field and equivalent instrument AFOV. This is due to the wider angle of acceptance of the 8x instrument; i.e. it 'gathers' light from an area wider than the the 10x instrument (the area of the FOV is 64% smaller just like the exit pupil is 64% smaller in the case of a 10x42 compared to 8x42).

I was talking about an extended object that fits within the FOV. The image you see of it through the 10x bino is obviously bigger than the 8x bino's (unless it is a point source like a star).

If you want the same brightness across the magnified image of the object, you need more total light for the 10x mag bino vs for the 8x.

In daylight, the eye's pupil is smaller than either bino's exit pupil. But because the 10x bino's exit pupil is smaller than the 8x, a higher percentage of the light gathered by the 42mm objective actually reaches the eye's retina.

 

[Brink]

The example I earlier thought about:
If we have two optical instruments with same magnification and aperture. Which means same exit pupil.
One with 50deg AFOV and the other with 70,7deg. In the latter case twice of the area is covered. If you then watching an even illuminated surface the question is: will twice of covered area mean twice the light amount is reaching your eye, or will the true relative brightness halved with twice field area, so the total light amount remains the same?
The latter should be correct, otherwise it means a narrow AFOV does not make use of the full aperture.
I am still confused because I have never read about a claimed relation between AFOV and brightness.
Wide FOV is one of the most important among many users, but I have never read someone wants a narrow FOV because it will be brighter. Some should do it if it increases the apparant brightness of the image.
I think Holger Merlitz can explain this.

The answer is that twice the light will reach your eye. It will still have the same intensity, the image will just be bigger.

By saying equivalent AFOV between 8x and 10x is the same as saying that the 10x true FOV is 64% the size of the 8x FOV

 

[Brink]

I was talking about an extended object that fits within the FOV. The image you see of it through the 10x bino is obviously bigger than the 8x bino's (unless it is a point source like a star).

If you want the same brightness across the magnified image of the object, you need more total light for the 10x mag bino vs for the 8x.

In daylight, the eye's pupil is smaller than either bino's exit pupil. But because the 10x bino's exit pupil is smaller than the 8x, a higher percentage of the light gathered by the 42mm objective actually reaches the eye's retina.

I am not sure the point you are trying to make. For an extended object that fits within the FOV, the brightness of the object will be the same through a 10x or 8x binocular in your daylight viewing situation.

Is the point that you and OP are trying to make that a larger image is composed of more light than an equally bright smaller image? If so, you have a very roundabout way of stating the obvious.

 

[agus_m]

If so, you have a very roundabout way of stating the obvious.

Sorry professor, English is not my native language. Your point is not very clear either