Re: [APML] Re:I'm Feeling a Flame Getting Ready ...

[Aplanatic@aol.com]

Roland,

> For stars, a scope with twice the aperture should have 4 times the light 
> intensity. However, if a scope of twice the aperture concentrates the light 
> of a star over twice the area, is not the light per unit area the same as 
> that of the smaller aperture? I understand that visually there is a gain, 
> but 
> can a small aperture with higher inherent resolution compete with a larger 
> instrument? 


There are, of course, various answers depending on what type 
of imaging is being done.  Let's discuss one particular limit, 
taking images of dim extended objects where poor signal-to-noise 
ratio limits the ability to form an image with acceptable information 
content.

The brightness of an extended object is given in 
photons/meter^2/arcsecond^2.  As the aperture increases, 
the number of photons per square arcsecond increases as the 
square of the aperture.  At a constant f/ratio, the increase in aperture 
is offset by an increase in the image scale such that the intensity 
at the detector remains constant.  In other words, at the detector, 
the number of photons per square micron is the same for a 
4" f/8 scope and a 40" f/8.  This fact causes some to come to the 
conclusion that for extended objects, aperture is unimportant.  This 
conclusion is incorrect.

In actuality, the number of photons collected over a unit resolution 
element (say 4 square arcseconds) is proportional to the aperture 
area.  Therefore, even if the 4" and 40" scopes have the same 
resolution (say 2 arcseconds), the 40" scope will have collected 
100 times as many photons per resolution element as the 4" scope.  
This increased signal-to-noise ratio is more information, and this 
information can be used to display deeper images or to aid in 
deconvolution, or some of both.

The typical argument against the above is as follows: Because the 
intensity at the detector is independent of the aperture for extended 
objects, the signal-to-noise ratio for each pixel is the same for the 
4" and 40" case.  Why then does the 40" have better SNR?  The 
answer is:  One can only (fairly) compare SNR at the same 
effective image scale.  For the 40" image, we bin (average) a 10X10 
region to bring it to the same image scale as the 4" image.  In doing 
so, we increase the SNR of the binned image by a factor of ten.

Well made (and well mounted) aperture always wins.

Dave Rowe