Ad Astra Observatory

Hi Jimbo,

Ahh, I am sorry, of course that doubleing the f/number is two stops
difference!
It must be this flue that is consuming my brain cells… 😉

Under one of your M 42 photos you say:
” The reflector has two distinct advantages over the ED80; first, it’s about
2 f/stops faster (f/5 vs f/7.5) … ”

Since I agree with all you wrote me on f/stops and all the rest, my
conclusion is that it must have been a lapsus, it is only one stop.

Thank you very much for your patient reply.

Clear skies, Dag

—–Original Message—–
From: Jimbo S. Harris [mailto:jimbo@jimbo.net]
Sent: Friday, February 15, 2008 6:02 PM

To: dag@dag-orsic.com
Subject: Re: astrophotography

Hi Dag,

At 05:28 PM 2/15/2008 +0100, you wrote:
>I was studying your interesting comparison photos of M 42 with different
>instruments

Thank you for checking out my work.

>when I noticed a mistake.
>
>It is about a concept of f/stop. I must correct you, I hope you don’t mind.
>
>If you have an instrument with an objective f/10, one stop more or less
>means f/20 or f/5 (double or half of the original value). Only in this
>case you can double or half your exposure time. One stop difference from
>f/20 is f/40 and so on, always double.

I hate to disagree, but the f/stop (on a moveable aperture camera) is the

*radius* of the iris compared to the radius of the objective. The amount of
light that gets through the iris depends on the *area* of the opening.
Halving the radius of a circle decreases its area to 1/4 the original size
(note one half squared = one quarter).

“Standard” f/stops on camera lenses are:
…f/2.8 f/4 f/5.6 f/8 f/11 f/16…
And the reason for this is that each represents a halving or doubling of

the amount of light reaching the film (or sensor).

The amount of light reaching the sensor is relative to the *square* of the
f/ratio.
Note:
2.8^2 = 7.84 ~8
4^2 = 16
5.6^2 = 31.36 ~32
8^2 = 64

11^2 = 121 ~128
16^2 = 256

Note that the squares of these numbers are doubling for each f/stop. This
is not a coincidence.

“One stop” faster than f/10 (^2 = 100) is f/7 (7.07^2 = 50). One stop
slower than f/10 is f/14.

f/5 is 4x faster than f/10

f/20 is 4x slower than f/10

http://en.wikipedia.org/wiki/F-stop

The same rules apply when talking about fixed-aperture telescopes. To do a
similar comparison, you’d have to compare either objectives of the same
focal length, but differing aperture (8” f/5 vs 5” f/8, for instance), or

objectives of the same aperture, but different focal lengths (8” f/5 vs 8”
f/10).

The first comparison is easier to grasp; the field of view will be the same
for both, but it’s clear that, in the same amount of time, the 5” scope
will allow a lot less light to reach the film plane than the 8” scope will.

In fact, the 5” will require and exposure 64/25 = 2.56x as long to acquire
the same number of photons as the 8”.

In the case of equal aperture, you have to keep in mind that the 8” f/10
has twice the magnification of the 8” f/5. This means that the light that
falls on a single pixel in the 8” f/5 will be spread over 4 pixels in the

8” f/10. This means that 1/4 as many photons fall on each pixel of the f/10
scope, and thus it will take 4x as long (that’s 100 / 25) to acquire the
same number of photons as in the f/5 scope. The f/10’s image will be 4x
bigger (by area), but also 4x dimmer, for the same amount of exposure time.

Thank you for emailing. I hope that my reply has been informative. I think
that this maybe should have been a post to CN, rather than an email.

Best,
Jimbo

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