How do I read/use histograms?

Peteroid

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I recently decided to get back into photography. Starting in high school (the late 60's), I was into film cameras; specifically a Minolta 101 SLR.

I've always wanted to get into digital photography (especially since I'm a computer programmer and computer game designer). But up until recent times, the resulting imagery was either inferior to film or the camera was way too expensive. And having used an SLR, I can't really see myself using anything else, both for the 'what you see is what you get' in terms of framing and the ability to change lenses.

Soooo, last week I finally got a Canon Rebet XT. This satisfies all my above concerns. It produces great imagery, is SLR, and wasn't too expensive (in fact cheaper than what I paid for my last film camera about 10 years ago!). I'm very happy with it. Especially since there is no need for buying film, or any cost for development, or and delay! I also bought a Canon printer to make hard copies (iP6600D). Digital photography has made things better and cheaper!

But, it has also made things a bit more complex. For example, I can now view a histogram of a shot. I can do this both within the camera and during 'development' ala programs like Photoshop.

So, my question is this. How do I 'read' such a histogram? What can I derive from it, and what should it 'look like' for 'good shots'? It looks like it is telling me the relative amounts of each frequency of light, but I'm not sure this is the case. Should I be trying to make the histogram look more 'centralized', or 'evenly distributed', or 'shifted', or a 'bell curve', or what? I'm sure it depends on the situation and the desired effect. Also it probably differs whether I'm taking a color or black and white shot...

Realizing the answer is probably too big to be a response, is there on-line places I can go to help me understand how to use histograms?

Thanx in advance! :)
 
Geeee, what took you so long to respond? I must have posted my opener more than 3 minutes ago! LOL :lol:

Seriously, thanx! I'll go check it out now!

[==P==]
 

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