This point takes a while to explain so stay with me through this section because it's really really really important.

As shown on the Minimum Magnification page, the maximum surface brightness for extended sources, like planets and nebulae, is achieved with the minimum magnification for the telescope. Let's figure out just how bright that is.

The total gain in light that the telescope collects, over that
of the eye alone, we call it G_{L}, is shown on the
Magnitude Gain page to be

Where D_{O} is the diameter of the objective and
D_{eye} is the diameter of the eye pupil. Also shown on
the Minimum Magnification
page is the equation for magnification in the form

Where D_{ep} is the diameter of the exit pupil.
And it is also noted on that page that the surface brightness of
an object is reduced by the magnification squared, because the
light is distributed over a larger area... there are a lot of
things discussed on the minimum
magnification page, aren't there?...

So the surface brightness of an object that you see with your eye,
which I'm going to call "SB_{eye}" since I'm not exercising
much imagination, is multiplied by G_{L} and then divided
by M², resulting in the surface brightness you see in the scope,
which I'm going to call "SB_{scope}" for the same reason as
above.

When the exit pupil diameter, D_{ep}, equals the diameter
of the pupil of the eye, D_{eye}, like at minimum
magnification, we have the following very interesting result for
the net result in SB_{scope} at minimum magnification:

What the heck is that saying? What -- that the very brightest image I can get with the scope is exactly the same brightness that I see with the naked eye??????

Yup.

Well, remember we're talking about surface brightness, meaning
the brightness per unit area, or the brightness density.
At the minimum magnification, which is the brightest image you
can get, the surface brightness is exactly the same as what you
see with your eye. Of course, the image is 30, 40, 50, even 60+
times bigger at that brightness, so it might be as much as 4000
times the __total__ brightness, depending on the diameter of
the objective. But the surface brightness never exceeds what you
can see with your eye alone.

This means something else that's very important -- the images in
all telescopes operating at minimum magnification have, at best*,
the **same surface brightness**. Then we can use the brightness
at minimum magnification as a reference, and since it is the
maximum brightness, we can -- and we will -- consider it to be
100% brightness.

* Note - some telescopes have better optics, meaning less light loss, than others. So poorer quality scopes will not reach this "standard" brightness even at minimum magnification.

Here comes the even bigger more important point.

As is seen in the example on the minimum magnification page, sometimes it is not only undesirable to operate the scope at minimum magnification, it's actually impossible. For telescopes with an f-ratio greater than 8, you will have a challenging time indeed trying to find the eyepiece to get you to minimum magnification, and it's not so easy to find one for f-ratios above 6.

This means that **scopes are almost never operating at 100%
brightness**.

So... at what brightness is my scope operating? As mentioned above in passing, the surface brightness of the scope depends on the magnification. Let's consider an example.

If I'm looking at a extended source, like Jupiter, at 30 power,
I can change the eyepiece out for one with half the focal length
and double the power to 60x. Jupiter will then be twice the diameter
in my image, and since area is a function of the diameter squared
(area =
^{π}/_{4}×D²),
then when I double the diameter of Jupiter I multiply the area by
2² = 4. The telescope is still collecting the same light, which
is now spread out over 4 times the area, so the surface brightness
will drop by a factor of 4.

Low Magnification | High Magnification |

Since M_{min} is the magnification that gives you 100%
brightness, and any increased magnification M will reduce your
brightness as M², then you can find the surface brightness
**SB = (M _{min}/M)²**. Simple. Keep reading.

As shown on the Minimum
Magnification page, the maximum eyepiece focal length to get
your minimum magnification is simply f_{e-max} =
7×f_{R}, where f_{R} is the f-ratio of the scope.
Since magnification M = f_{O}/f_{e} then we can
substitute into the surface brightness equation to compute the
surface brightness from the eyepiece that we are using.

Also simple. But there is a still better, simpler way to find the surface brightness, using the exit pupil...

You wouldn't think that the exit pupil, the diameter of the light beam leaving the eyepiece, would provide the magic key to the performance of the scope, but check this out:

Substituting the calculation of M_{min} from D_{O}
we get

Also from the Minimum
Magnification page we have that the exit pupil D_{ep}
= D_{O}/M, so we can replace D_{O}/M in the
surface brightness equation with D_{ep} and we get

Given this equation, it’s actually pretty easy to find the surface
brightness from the exit pupil since 7² = 49, or about 50 = 100/2.
Then SB = D_{ep}²/7² = 2×D_{ep}²/100, so

is the surface brightness in percent. Seriously. Take the exit pupil, square it, double it, and you've got it.

Well, heck -- you can just knock that one out in your head... for
example with an exit pupil of 5mm, then 5² = 25, double it and you
get 50, so surface brightness with a 5mm exit pupil is 50% of
maximum. If D_{ep} is 3mm then 3² = 9, double it gives
you 18 so the surface brightness is 18% of maximum and so forth.
Just another good reason to learn those two simple exit pupil
equations: D_{ep} = D_{O}/M and D_{ep} =
f_{e}/f_{R}.

So we have a one-to-one correspondence between surface brightness
and the exit pupil. If I define a specific exit pupil, I have
also defined exactly where I am on the brightness scale, **totally
independent of the scope**.

That means the exit pupil gives me the same information that the surface brightness does, without calculating squares of anything. The exit pupil gives me a simple way of defining and transferring settings from one scope to another, as follows:

- Once the proper settings are found to capture an image
on a scope, determine the exit pupil that was used as
D
_{ep}= f_{e}/f_{R}or D_{ep}= D_{O}/M. - Calculate the magnification and/or eyepiece focal length
for the second scope using M = D
_{O}/D_{ep}and f_{e}= D_{ep}×f_{R}.

Check out "The Power of the Exit Pupil" example below to see how easily you can solve problems this way.

You may have noticed that your eyesight is sharper in daylight than it is in the dark. This is not just some sort of illusion, it is a real effect based on the physics of your eye.

This is because the lens of the eye is part of your body, and it's adjustable -- it's not a computer-generated, rigid lens with mathematically perfect surfaces all the way from the center to the edge. Its shape is good (spherical) at the center, and starts to deviate from "perfection" toward the edges.

The non-spherical shape, or "spherical aberration" combines with an effect called "chromatic aberration", which means different colors passing through the lens focus at slightly different points. Like spherical aberration, chromatic aberration also is more pronounced toward the edges of the lens.

The larger the exit pupil, the more of the lens is used to focus the light, and the more the deviation from "perfect" causes errors in focusing the light. So at a 7mm exit pupil, you get the highest surface brightness, but not the sharpest focus in the eye.

Meanwhile the smaller the exit pupil, the more the diffraction limit of the pupil impedes the resolving power of the eye. So there is an optimum point where the combined effects of spherical aberration, chromatic aberration, and diffraction effects reach a minimum.

It turns out that an **exit pupil of about 2-3mm**
(2.4mm to be precise) is the optimum point for maximizing the
resolving power of the eye. This is about where theory would put it,
and it has been confirmed by observational studies.

Let's see how the one-to-one correspondence between the exit pupil
and brightness scale, and the dependence of the optimum resolution on
exit pupil, can be used to define a performance scale for the scope
based on the diameter of the exit pupil, D_{ep}. Because
the exit pupil is a scope-independent reflection of surface brightness
and optimum resolution, this is a __universal performance scale__
for all scopes.

For this interesting little exercise, we will use the fact that
both magnification, M , and the focal length of the eyepiece, f_{e},
can also be determined by the exit pupil as follows:

Then we have the following four key reference points on our performance scale:

__Maximum magnification__: D_{ep} = D_{O}/M_{max}
= D_{O}/D_{O} = 1 mm. This might be a bit of a
surprise: at max magnification, where the scope's resolution
matches your eye's resolution, the exit pupil is exactly 1 mm. Also
it gets smaller as you go to higher power, so if you can see the
Airy disk the exit pupil is
less than or equal to 1mm. Here we have a magnification of D_{O},
an eyepiece focal length of f_{R}, and a surface brightness
of SB(%) = 2×D_{ep}² = 2×1² = 2%.

__Optimum magnification__: we will assume the optimum
is at 2mm, as this actually is closest to the true optimum (2.4mm),
is an easy point to remember and simple for mental calculations. Bear
in mind, though, that anywhere on the range between 2 and 3mm pretty much
hits it. Then we have M = D_{O}/D_{ep} = D_{O}/2,
f_{e} = D_{ep}×f_{R} = 2×f_{R},
and a surface brightness of SB(%) = 2×D_{ep}² = 2×2²;
= 8%. Note that this optimum is at __half maximum magnification__.

__Maximum Brightness__: We found brightest image, which also has
the widest field of view, by setting the magnification to give us an
exit pupil of 7 mm -- the minimum magnification we will want to use.
At this point we have a magnification of M = D_{O}/D_{ep}
= D_{O}/7, eyepiece focal length f_{e} = 7×f_{R},
and a surface brightness of 100%.

__Wide Field__: by backing down from a 7mm exit pupil to 5mm, we
arrive at a point that is __half maximum brightness__, resulting in an
eyepiece focal length f_{e} = 5×f_{R} that sometimes
is more available than the maximum eyepiece focal length. This point, with
50% max brightness and 70% of widest possible field of view
(so we can still call it "wide field"), represents a good compromise
in brightness vs. magnification for deep-sky observation.

To these four critical points we will add one additional point, an "Extra-high"
resolution point, introduced because max magnification brings the scope's
resolving power to where your eye can *just* see it. We can push the
scope a little beyond its performance range at the high power end, to make
the detail a bit easier to see.

As you go past the limit of the scope's performance you can see the effect,
in the form of progressively greater rounding or blurring of edges. We
therefore will stop at 50% higher magnification, or 1.5×M_{max}
= 1.5×D_{O}, before the blurring (not to
mention the image darkening) becomes objectionable.

Setting an extra-high magnification M = 1.5×M_{max} leads to a
D_{ep} = D_{O}/M = D_{O}/(1.5×D_{O}) =
^{2}/_{3}mm.

Then the dynamic range for the exit pupil is fundamentally **from 1 to 7
mm**, with three additional reference points noted for a total of 5 points
on the scale. These are points to consider when selecting a set of eyepieces
for use with a given scope:

Maximum Brightness | D_{ep} = 7mm |

Half-Maximum Brightness | D_{ep} = 5mm |

Half-Maximum Magnification | D_{ep} = 2mm |

Maximum Magnification | D_{ep} = 1mm |

Extra-High Magnification | D_{ep} = ^{2}/_{3}mm |

This is what the resulting universal performance range for telescopes looks like:

Some comments on this range:

- When I'm looking for faint objects, particularly when I'm using a filter, I want lots of brightness and will work at the bright end of the range (large exit pupil).
- If I have a dark sky I can afford to go with the brightest eyepiece, but if I have a light-polluted sky, or a difficult target, or both, I might back off a bit to a somewhat smaller exit pupil... 4-5mm rather than 7mm. By increasing the magnification, I spread out the background sky glow, and that can improve the contrast with the faint target. Neat trick to remember.
- If I want to split a double, I will work at the dark, high-power
end of the range (small exit pupil). Remember that stars don't
start to get darker until I exceed the maximum magnification
of the scope, or the maximum allowed by the atmospheric seeing
(usually about 150-200X). So increased magnification
*really*improves star contrast against background sky glow. - This performance range assumes that the f-ratio will permit it.
In practice, eyepieces over 40mm are hard to find and above 56mm
virtually don't exist. Therefore, for all practical purposes,
scopes with an f-ratio over 6 are limited on the bright end of
their performance range to an exit pupil of about
40÷f
_{R}. - If I’m a photographer, I can compensate for low brightness with long exposures, so a low f-ratio is less important than optical perfection. Then I may go with a longer focal length in the objective, which is easier to get relatively free of artifacts (chromatic & spherical aberrations).

Since **SB = 2×D _{ep}²**, then if we want the
same surface brightness we need to keep D

We have a way to find D_{ep} using the scope diameter
D_{O} and magnification M:

Then to keep D_{ep} the same we need to keep the ratio of
scope diameter to magnification the same.

That means:

- If we go from a
**smaller**scope to a**larger**scope we can get the same surface brightness at**higher**magnification. For example, if we go from a 4" scope to an 8" scope, we've doubled the scope diameter (D_{O}) so we can get the same surface brightness at double the magnification. - If we go from a
**larger**scope to a**smaller**scope, we need to**reduce**the magnification to get the same surface brightness. We change the magnification by the same factor that we changed the scope diameter. So for example if we're going from an 8" scope to a 4" scope, we changed by a factor of 2 so the magnification will have to be reduced by a factor of 2 to keep the same surface brightness. - A larger scope at the same magnification will give a higher
surface brightness. Since SB goes as D
_{ep}², surface brightness will increase with the scope diameter squared. That means when we go from that 4" scope to the 8" scope, twice the diameter, then at the__same__magnification we will see 2² = 4 times the surface brightness.

We still have **SB = 2×D _{ep}²**, and if we want
the same surface brightness we're still trying to keep D

In this case we will work off the other exit pupil formula, which uses the eyepiece focal length and the f-ratio.

Now we are looking at the effect that a different f-ratio will have,
and we can see that to keep D_{ep} the same we need to keep
the ratio of the eyepiece focal length to the f-ratio the same.

That means:

- If we go from a scope with a
**larger**f-ratio to one with a**smaller**f-ratio, we can get the same surface brightness with a**shorter**focal length eyepiece. For example, if we are using an f/10 scope with a 20mm eyepiece and switch to an f/5 scope, we've cut the f-ratio in half so we would use a 10mm eyepiece to get the same surface brightness in the image. - If we go from a scope with a
**smaller**f-ratio to one with a**larger**f-ratio, we need to**increase**the focal length of the eyepiece (moving back from the image plane and reducing magnification) to get the same surface brightness. What is interesting about these equations is that we only need to know the f-ratio of the scope to determine the eyepiece that will hit our target surface brightness. - A scope with a smaller f-ratio using the same eyepiece will give
a higher surface brightness. Since SB goes as D
_{ep}², surface brightness will__increase__as the reduction in the f-ratio squared. That means when we go from that f/10 scope to the f/5 scope, half the f-ratio, then with the__same__eyepiece we will see 2² = 4 times the surface brightness.

This is a 90mm f/13.9 scope.

- then the minimum magnification is D
_{O}÷7 = 90÷7 = 12.8 ≈ 13 - the eyepiece required to get that magnification is
7×f
_{R}= 7×13.9 = 97mm – which I won’t find, and for that matter I don’t want to because magnification of 13x is about what my binoculars are doing. So I (or anyone else) would never be working at that level with this scope. - let's say we will go with a routine value for a long-focus eyepiece,
like 40mm, and I'll work the problem as I normally would:
- Find the exit pupil for this eyepiece:

D_{ep}= f_{e}÷f_{R}= 40÷13.8 = 2.9mm (call it 3). - Find the magnification with this eyepiece:

M = D_{O}÷D_{ep}= 90÷3 = 30 (decent) - Find the surface brightness with this eyepiece:

SB = 2×D_{ep}² = 2×3² = 18%

Note that since this is about the longest focal length eyepiece I can get, this is about the brightest I can get the image in this scope.

- Find the exit pupil for this eyepiece:
- for fun, let's see where we are with the 26mm eyepiece that came with
the scope:
- Find the exit pupil for this eyepiece:

D_{ep}= f_{e}÷f_{R}= 26÷13.8 = 1.9mm - Find the magnification with this eyepiece:

M = D_{O}÷D_{ep}= 90÷1.9 = 47.8 - Find the surface brightness with this eyepiece:

SB = 2×D_{ep}² = 2×1.9² = 7.2%

Pretty dim.

- Find the exit pupil for this eyepiece:

I just dug up the biggest aperture scope I can find looking through the catalogs on my desk. This is a 406.4mm diameter f/4.5 scope.

- then the minimum magnification is 406.4÷7 = 58 -- which is not a bad magnification value at all.
- the eyepiece required to get that magnification is
7×4.5 = 31.5mm – where a 32mm eyepiece is a common size,
albeit toward the big end of the spectrum. So it is perfectly
possible to operate this scope at
**100% brightness**. - the scope comes with a 26mm eyepiece so let's analyze that as we did above:
- Find the exit pupil for this eyepiece:

D_{ep}= f_{e}÷f_{R}= 26÷4.5 = 5.8mm (much bigger than the ETX) - Find the magnification with this eyepiece:

M = D_{O}÷D_{ep}= 406.4÷5.8 = 70 - Find the surface brightness with this eyepiece:

SB = 2×D_{ep}² = 2×5.8² = 67.3%

- Find the exit pupil for this eyepiece:
- which means that out of the box this scope is operating at a surface
brightness that is about 10 times the brightness of my ETX
**at higher magnification**.

So that very short f-ratio makes this a very bright, sometimes called a "fast", scope -- coupled with the large diameter this scope is a killer for deep-sky exploration.

Both of these scopes come with the same focal length eyepiece -- 26mm -- so how do they compare based on that?

With the same eyepiece, the surface brightness of the two scopes
goes with the ratio of f-ratio's squared, so we get
[f_{R1}÷f_{R2}]² = [13.9÷4.5]² = 9.54. This is
telling us that the Lightbridge is 9.54 times brighter than the
ETX with the same eyepiece, consistent with what we found above,
but computed more directly.

So presuming we can find the appropriate eyepieces to get the two scopes operating at the same magnification, or close enough, how would the two image brightnesses compare?

At the same magnification, the surface brightness of the two
scopes goes with the ratio of the scope diameters squared, so we
get [D_{O2}÷D_{O1}]² = [406.4÷90]² = 20.4. The
LightBridge would show the image at the same magnification at
20 times the brightness of the ETX 90. Yikes.

Before we continue beating up on my poor little
ETX, I would hasten to point out that it cost about a quarter of
what the LightBridge would set me back, and the ETX is highly
portable. But for image quality, it simply has no chance against
the __much__ bigger scope.

I'm going to show you how great the exit pupil is by way of, what turns out to be, a real example.

We will suppose I'm out observing with a friend who has a 15" f/4.5 Obsession telescope, and that I'm using an Orion 8" f/6 Dobsonian. My friend is able to show me the Horse Head Nebula with his Obsession telescope using a Televue 22mm eyepiece. What should I use in the Orion scope to also be able to see this nebula?

So the exit pupil my friend was using on his Obsession scope is
D_{ep} = f_{e}/f_{R} = 22/4.5 = 4.9 ≈
5mm.

Then the eyepiece focal length for the Orion is f_{e} =
D_{ep}×f_{R} = 5×6 = 30mm.

Easy-peasy.

Of course, at the same brightness, the 8" scope will be operating at lower magnification than the 15" -- to be precise it will be set to 8/15, or about half, the magnification.

By the way, this example is based on real experience, although for simplicity I left out the fact that a nebular filter is also used to see the Horse Head. It was Barbara Wilson of the Houston Astronomical Society who discovered that an exit pupil of 5mm, along with using a nebular filter, is the secret to seeing the Horsehead Nebula: check out her Magic Horsehead Eyepiece page for details.

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*Your questions and comments regarding this page are welcome.
You can e-mail Randy Culp for inquiries,
suggestions, new ideas or just to chat.
Updated 16 November 2012*