A digital camera has a 1.6 multiplier effect due to a smaller field area than a 35mm camera.?

What I need to know is if a 100mm lense used with a 35mm camera then has an effective focal length of 160mm, does it really have more power to magnify than the lense number itself, or does the smaller image area on the digital sensor just have less coverage and it appears larger. I don't understand how the power increases just because the sensor area is smaller. Please explain.

The link to the dpreview site is a pretty good explanation. A better term than "multiplier factor" is crop factor, because that is essentially what is happening. You're not increasing the magnification, you are "cropping" the image/field of view.

It's similar in principle as when you have let's say a 4x6 picture of a person and you crop just their head and make a 4x6 of that. Since the sensor is smaller, it is essentially a "crop" of a full 35mm frame.

So, while this is nice because it makes your telephoto lenses even longer, it also has the converse effect of making it difficult to get a good wide angle shot. This is why there are special lenses that are extremely wide (Canon's 10-22mm EF-S for example) to try and address this "issue".

Here's a good explanation:
http://www.dpreview.com/learn/?/Glossary...

A 100 mm lens on a 35 mm camera gives a magnification X2. 150 mm gives a magnification X3. This is only true for negative size of 24mm X 36mm i.e. normal 35mm negative. Because the sensor is smaller a 100 mm lens will give a magnification of just over X3. This does not say anything about quality of picture if it is enlarged. This is determined by the mega pixel rating. The higher the better. On larger format negatives like 645 80 mm is the standard lens and 160 gives magnification X2