What will be the true actual distance for both lens in terms of actual meters?
closed as not constructive by Mark Whitaker, jrista♦ Oct 11 '12 at 1:32
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. If this question can be reworded to fit the rules in the help center, please edit the question.
I think I understand what you're getting at here. The puzzle is: the superzoom camera has a big "times zoom" number, whereas the SLR lens much smaller. That's explained in the answers to What does 'how much zoom' mean?, and reading that should make this somewhat more clear. The key point is that the "zoom times" number is just a measure of flexibility, not an indicator of magnification.
What you want is a way to tell how much bigger that telephoto lens will make stuff in your pictures — how far away you can be from your subject and make a decent picture. The answers to What does "angle of view equivalent to that of some lens in 35mm format" mean? will tell you some of that, but it's in detached, mathematical terms of angles and so on.
Fortunately, those angles can be translated into the real world. You can work out the math yourself, but there's also handy online (and smartphone) tools to do it for you so you don't have to remember the formulas. There's a nice simple one at http://www.tawbaware.com/maxlyons/calc.htm#fov_calculator. For this purpose, scroll down to the Dimensional Field of View Calculator.
First, put in
200 for the lens focal length, and
1.5 for the "Focal length multiplier" (which is another word for sensor crop factor) — this matches a Nikon, Sony, or Pentax entry-level DSLR with a 200mm lens. Put in
25 ft for the distance to subject, and press compute. You can see in the results that at that distance, this focal length gives you a frame of about 3 feet wide and 2 feet tall.
Next, change the focal length to
180, and change the "focal length multiplier" (remember, crop factor) to 5.6 to match the superzoom. Leave the distance the same, and hit compute again. Now, the imaginary captured rectangle is much, much smaller: 10.7" across and 7.1" high.
Imagine two prints, both the same size, taken from each picture. The superzoom image would be a much narrower angle, but printed to cover the same area as the wider angle, so there'd be more magnification.
In this case, it's about 3× in each direction — if you imagine dividing the field of view of the 200mm DSLR lens in thirds, and then taking just one third and expanding it, there's your direct comparison.
But wait! Before you run off to buy the superzoom based on this alone, consider that here's where Why doesn't it make sense to compare an entry-level DSLR with a super zoom? might kick in. Because that 55-200mm is physically much bigger, and because it has less of zoom range, it may have less compromise in image quality. Plus, the DSLR has a much larger sensor, which will matter particularly if you're not in full sunlight. For these reasons, it may be that a tiny fragment of the DSLR image cropped and expanded is as good or even better than the full image straight from the superzoom.
It may be the case that either gives perfectly adequate results for what you want to do. In that case, there are other factors: size, price (especially price!), and zoom flexibility vs. system flexibility to consider, and that's what you should base your decision on.