How limiting is magnitude in astrophotography?

[swag72]

When I am looking at targets and they give me a size as well as a magnitude - How limiting is that magnitude in astrophotography?

If you have two targets of the same size, one with a magnitude of 5 and one with 10 - Will that mean that the 2nd target will just take double the time to image and capture the same data as the first target?

By taking a large number of subs, can you in effect negate the magnitude of a target?

[Macavity]

I would say (on the well known logarithmic scale) a difference of five magnitudes, corresponded to a factor of 100x in brightness. <G> Increase the (conventional) "f-stop", accordingly. But, life is rarely that simple? One deals with point and extended objects - "surface brightness" and stuff. Amidst the many (and erudite!) theoretical discussions re. apertures, f-number, pixel size, nothing seems easily defined...

[ollypenrice]

You can't 'negate' a faint magnitude but more, and longer, subs and faster f ratios will get you deeper. It is very hard to use published magnitudes because of surface brightness being so hard to quantify, I think. Also different cameras have different levels of sensitivty at different wavelengths. There are a lot of 'etceteras' here already! I tend to go and have a look at images on the net and have a look at the sub lengths, total exposure time, f ratio and camera used to see what I'm letting myself in for...

Olly

[swag72]

Thanks both - Just sort of playing around and planning a little, so couldn't help wondering about magnitude.

[JimStan]

Basically the brightness of an extended object is determined by the intrinsic brightness, distance, and surface area of that object.

The way it appears to us is basicaly given a magnitude value by comparing it's brightness to the brightness of a star who's image is defocused to the same area as the object. Thus, if the object was one minute of arc across, and was a bright as a 5th magnitude star, who's image was defocused to a minute of arc in diameter, we would say that the extended object roughly had a magnitude of 5 .

Of course, there are some other factors that enter in, but that is a very basic explanation of how magnitude of extended objects is derived. That would explain, for instance, why the Helix Nebula is almost invisible, even though it has a magnitude of 7.3. It has an apparent size of 16/28 arc minutes making it almost the same size in the sky as the Moon !

Jim S.

[ncjunk]

How the above translates for me is that stars i can usually know down to what magnitude i can image by testing my setup as they arr more or less a point source (hence the magnitude is a good comparison for knowing if i can image it) With Galaxies etc where they are spread out, i fall into the trap trying to image something at magnitude 10 and find it is fainter or more difficult than the 10mag Galaxy i was just imaging.

You have to take into account the size but I also pick random targets and take a 5 min test shot and see how much detail i get to give an indication.

[dph1nm]

Well the original question did say what if the targets were the same size. In which case you would get 10**5/2.5 times as much light from the brighter one i.e. 100x. To image 100x fainter with the same signal to noise, however, requires 100**2 i.e. 10,000 times the exposure. Hence going 5 mags deeper is not easy! At that stage it is much more sensible to buy a larger aperture scope.

NigelM

p.s. I am assuming the exposures are limited by sky noise and not the noise from the objects themselves. This is usually the case for objects at the limit of your exposure (but maybe not for the specific case of 5th and 10th magnitude, which are a bit bright).

[swag72]

Blimey - That's some large figures there!! Thanks for answering the question guys.