How does increased magnification affect brightness? (1 Viewer)

I know this is the case but I would like to know the scientific reason behind it. Where [and how] does the light-loss occur?

And is there a formula for brightness loss / magnification?

Be gentle, as I suspect there is a very obvious answer staring me in the face.......although I enjoy the explanations here more than Wiki!

Brightness is reduced as magnification increases.

It's magnification (not the size of the exit pupil as most people will say) that controls the brightness of an image if, as is most often the case in daylight, your eye's entrance pupil is smaller than the exit pupil of the bin.

Why? The eyepiece spreads the same amount of light over a larger angle with higher magnifications (so there is "less to go around" on each spot of the image).

Brightness changes as the inverse square of the magnification (as you magnify by the same amount in two directions.)

Just like a light source gets four times less bright when you go double the viewing distance to the source.

Isn't the postulate incorrect? If it were correct, we would be able to see even the dimmest stars in the night sky with the naked eye, and the more powerful magnification we use, the fewer we would see.
This is contradictory to the twilight number idea.

I'm still inclined to believe that it is all about the needle's eye (the exit pupil). And a larger frontlens also grasps a wider light cone, hence it will collect more photons. With a larger exit pupil it will appear at least as bright, if not brighter than the lower mag bin or scope.

Kevin Purcell said:

Just like a light source gets four times less bright when you go double the viewing distance to the source.

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Well, the illuminated area of a bright object will actually have the same illumination per square inch, but when distance is doubled, the projected area it resembles on the retina will be a quarter of its former size.
But if you look at a very big grey wall, that covers the entire field with a 6x and 12x bin, both having the same exit pupil size, you will not be able to tell the difference in brightness.

Kevin Purcell said:

Just like a light source gets four times less bright when you go double the viewing distance to the source.

Click to expand...

This can not be applied for terrestrial viewing, where the distance to the sun will remain unchanged even if we back off half a mile. So the incident light will be the same over all of the area that could be over-viewed with binoculars.

And the reflectivity of terrestrial objects, commonly considered being the average of 18% (neutral grey) will not change, independent from whatever distance we observe them.

..now I'm confused...

james holdsworth said:

..now I'm confused...

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Well...I used to be an amateur photographer once upon the time.
I learned that the f/stop numbers are used to calculate the exposure.

A 50mm f/2 lens has a maximum aperture of 25 mm.
A 100mm lens has to have a maximum aperture of 50 mm to be an f/2, and if it has a maximum of 25mm it is a f/4.

So if you divide the focal length with the f number you get the aperture.
And an aperture that changes to half its size will let in a quarter of the light amount.

So, the f number does not refer to a diameter, it is a ratio.
Anytime you calculate an exposure, be it in daylight or flashbulb, the f number is used. And even if you change the lens for another, the same f number is used to get the same exposure.

So, I would say that the f number of a photo lens translates to the size of the exit pupil, and the size of the full aperture translates to the front lens diameter of binoculars. Sadly, I can't really explain why. Where's Henry?

This is my understanding: If we ignore the transmission losses in the optics, each point on the image seen through the binoculars should appear just as bright as the corresponding point on the object if we look at it directly, as long as the exit pupil is larger than the pupil openings. So if we look at a bird, the binocular with the greater magnification will make each feather look just as bright as through a lower-powered binocular but with a larger area, thus making the bird stand out more over the background and giving the impression of a brighter image (again, an exit pupil larger than the pupil opening is assumed).

spacepilot said:

This is my understanding: If we ignore the transmission losses in the optics, each point on the image seen through the binoculars should appear just as bright as the corresponding point on the object if we look at it directly, as long as the exit pupil is larger than the pupil openings. So if we look at a bird, the binocular with the greater magnification will make each feather look just as bright as through a lower-powered binocular but with a larger area, thus making the bird stand out more over the background and giving the impression of a brighter image (again, an exit pupil larger than the pupil opening is assumed).

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...more confused|:S|. Maybe I am misreading you but are you saying that a higher mag. bin will give an image as bright as a lower power one, but with more 'apparent' brightness?

I was expecting just the opposite.

spacepilot has a thing going on there.

I recall lunar photography:
The surface of the Moon is illuminated with the same intensity as the Earth. Thus, the exposure value (EV) should equal that of daylight photography.

But if you take a shot with an ordinary wide angle snapshot camera, the sensor will not be able to detect that little white dot in the sky, but make the exposure just like if it had not been there - with a looooooong exposure time. Thus the Moon will be immensely over-exposed, but the image will still appear more or less black.

If we imagine another shot through a medium telephoto lens, and with the correct EV, the Moon would show some detail, but the image would be dominated by the black sky, hence generally a dark image.
Then again, if we crop the image, or get even closer by using a super telephoto, but still with the same level of exposure, the image would show only the surface of the Moon, no black sky, hence it will seem a lot brighter than the initial snapshot.

Does it all make sense?

james holdsworth said:

...more confused|:S|. Maybe I am misreading you but are you saying that a higher mag. bin will give an image as bright as a lower power one, but with more 'apparent' brightness?

I was expecting just the opposite.

Click to expand...

This is why the twilight number was invented.

looksharp65 said:

spacepilot has a thing going on there.

I recall lunar photography:
The surface of the Moon is illuminated with the same intensity as the Earth. Thus, the exposure value (EV) should equal that of daylight photography.

But if you take a shot with an ordinary wide angle snapshot camera, the sensor will not be able to detect that little white dot in the sky, but make the exposure just like if it had not been there - with a looooooong exposure time. Thus the Moon will be immensely over-exposed, but the image will still appear more or less black.

If we imagine another shot through a medium telephoto lens, and with the correct EV, the Moon would show some detail, but the image would be dominated by the black sky, hence generally a dark image.
Then again, if we crop the image, or get even closer by using a super telephoto, but still with the same level of exposure, the image would show only the surface of the Moon, no black sky, hence it will seem a lot brighter than the initial snapshot.

Does it all make sense?

Click to expand...

This makes perfect sense. If you are only one mile away from the moon when you take a picture, each pixel may be only for 1 square foot of the lunar surface. What this means is that the lens is collecting light reflected off an area of 1 square foot of the surface and focusing it onto the pixel on the sensor. This much light multiplied by the EV you happen to be using, determines the light energy the sensor pixel receives. Take several photos and find the ideal EV, write it down on your notebook. Now move to 2 miles away from the surface, at which time each pixel is corresponding to 2x2 = 4 square feet of the lunar surface. That means the lens is gathering light from 4 times the area of the surface as before, which naturally reflect 4 times the amount of light when compared to 1 square foot. But don't forget that, since we are 2 times farther away, our lens only collect 1/4 of the fraction of the light reflected off the lunar surface as before. So in the end, the 4 and 1/4 cancels out. If we use the same EV that we found before, each pixel will receive the same amount of light energy, and we will arrive at the ideal exposure again. Now move to 4 miles, 8 miles, 16 miles ... away. Each time we can use the same EV to ensure each pixel receive the same amount of light energy, although each time the same pixel is corresponding to more and more "real" area on the lunar surface. We do this over and over again, each time moving further away from the moon, until we are taking photos of the moon on earth. Even now, we should still use the same EV we used when we were 1 mile away from the moon, as that will give the perfect amount of the exposure for the moon face, no matter how big (or small) it may look on our photo

Now, what if we don't move away from the moon, but move closer instead? Each time we move closer to the moon we should also use the same EV to get a good photo, for the reason state above. We move closer and closer until we are actually standing on the surface of the moon, photographing some lunar rock. We should still use the same EV to get a good picture. Should this EV be any different than the EV we should use when we are standing on the surface of the earth, photographing some terrestrial rock? No. Because the earth is the same distance from the Sun as the moon, so each unit area of the moon rock receives the same amount of light power as a unit area of the earth rock, ignoring the effects of the atmosphere.

So if we put everything together, it's obvious that we should use the same EV to photograph the moon as when we photograph a rock on the earth.

It's more or less the same thing with the binoculars. Assuming that our pupil opening stay constant, and that the exit pupil is larger than our pupil, the same cell on the retina will receive the same amount of light energy, regardless of the magnification. The only difference with a greater magnification is that it will project the object of interest onto more cells. As a result, our brain has more chance to pick the object out from the back ground, or can perceive more details on the object.

Of course, when we take into account to the size of the exit pupil of the higher magnification binocular and how it compares to our pupil opening, things gets more complicated, and the perception of each individual may be different. That's why we have debates about the usefulness of higher magnification vs. larger exit pupil at twilight conditions.

spacepilot,

that's a brilliant explanation and you've put into words what I was unable to express.
After having closed my computer last evening, lying in bed I also realized I should have mentioned that the EV would be the same, independent from the photographer's distance to the Moon.

Thank you very much!

/L

Guys,

I think you may be intermixing two meanings for brightness. One refers to the amount of light coming through the binocular (or telescope), i.e., the brightness of the instrument itself. The other refers to the amount of light that enters the eye, i.e., the brightness of the retinal image.

Disregarding all other considerations, binoculars having the same exit pupil size will be equally bright — assuming they have the same transmission capability. This means that at the eyepoint they will present the same amount (quantity) of light.

As we know, the exit pupil, EP, is the ratio A/M, where A = aperture (objective size) and M = magnification.

It should be evident, therefore, that as:

• A is increased the ratio A/M increases as well as brightness.
• M is increased the ratio A/M decreases as well as brightness.

Another way of thinking about this is that one can compensate for the loss of brightness at higher magnification by increasing the aperture (objective) by the same factor. Conversely, at lower power a smaller objective is required to maintain equal brightness.

The effect of the eye's pupil is simply to reduce the 'effective aperture' of the system when it's smaller than the EP. This is equivalent to reducing A in the ratio mentioned above, thereby reducing image brightness. This is consistent with the primary survival purpose of the eye's pupil, incidentally, which is to protect it from over-stimulation. Note that it closes much more quickly than it opens.

Ed

Thanks Ed for clarifying and for bringing the thread back to the original subject.

As usual, I like the way you approach these things.

/L

If you want to go with the "exit pupil theory of binocular brightness," there's another monkey to throw into the "equation" and nobody nitpicks that monkey better than Edz.

Read what he has to say about "illumination of the exit pupil," and how much of the exit pupil is fully illuminated affects brightness and how this differs in various bins.

Be prepared to have your preconceptions blown away! What more can I say? I didn't start the fire!

http://www.cloudynights.com/ubbthreads/showflat.php?Cat=0&Number=3332851&page=5&view=collapsed&sb=9&o=all&vc=1

I think Edz is aware now that his posts about "illumination of the exit pupil" were based on a misconception. What he was actually measuring was vignetting of the FOV. Unfortunately, the old posts are still there to cause confusion

henry link said:

I think Edz is aware now that his posts about "illumination of the exit pupil" were based on a misconception. What he was actually measuring was vignetting of the FOV. Unfortunately, the old posts are still there to cause confusion

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WHAT!? Edz was (dare I say it?).... wrong. Let's put that in the Book of Guinness World Records!

I've lost track of how many times I've seen his "illumination of the exit pupil" hypothesis repeated. I think there may have even been a technical report at some point, which he would refer people to when the subject was broached, but I couldn't find it on CN, and now I know why.

I guess I'm not the only one who tries to explain phenomena with "speculations".

Except Edz is like E.F. Hutton, when he speaks, people listen. When I speak, I get one of these after my after my quote. :-O

There goes another example of your axiom: repetition does not always mean verification.

Thanks for that update.

Surely the simple common sense approach is that for a given objective lens size the greater the magnification the less bright the image. Simple experiment with my old Nickel Supra 15-60x 60 telescope Exit pupil 4-1 mm.
The greater the manification the dimmer the image. This belies the old twilight formula (square root of magnification times size of objective lens).
However does the twilight formula apply when comparing models with the same size exit pupil e.g. 8x40 compared to 10x50?
I also assume their is no advantage with regard to brightness in using binoculars with a larger exit pupil than the maximun size of your dilated pupil?

Here is one nerd's way of looking at this.

You're looking at a distant stop sign, naked eye. Its apparent surface brightness depends on how wide your eye's pupil is opened up.

Now, through a 7x binocular, the stop sign looks 7x bigger. A lot more light than the eye collects is required to illuminated the magnified image as brightly as the real thing. If the binocular objective is exactly 7 times wider than your eye, its exit pupil will match that of the eye. And, the light collected exactly compensates for the magnification, so the surface brightness will appear unchanged from that of the naked eye view.

If the objective of the binocular is even bigger, so that at 7x the exit pupil is bigger than the eye's entrance, some of the light collected by the objective spills onto the iris and is wasted. But the usable, effective, central part of the objective is still 7x wider than the eye, and still functions just like the smaller objective of the first binocular: the eye is filled with light, and the brightness of the image is the same as in that case.

There's another common case where surface brightness will appear reduced. If the first binocular's exit pupil exactly matches the eye's pupil at 7x, consider what happens if its magnification is increased to 10x. All the collected light still enters the eye, since the exit pupil is now smaller than the eye's pupil. But the image, being larger, has the light diluted over a greater area, and the surface brightness will appear reduced. A similar reduction in brightness happens if the magnification was held at 7x, but the size of the objective was reduced.

So, increased magnification leads to diminished brightness in the range where the exit pupil is smaller than the eye's entrance. When the exit pupil is equal or larger than the eye, the surface brightness will be the same as seen by the naked eye. The image cannot be made brighter, since any extra light will only spill outside the eye, ineffectively.

I have assumed that no light is lost in passing through the binocular. No coatings are that good, of course, I just wanted to keep it simple, as far as it goes. And, people don't always see in agreement with simple predictions, there are always complicating factors. But, that's the basic argument.
Ron