A longer *focal length* lens will project a magnified view of a smaller area of the scene, whereas a short lens will project a view of a larger area of the scene (wide angle).

With a bit of maths, it's possible to determine how much of the scene you will capture, and draw a diagram to help yourself out.

I will compare a 28mm (wide angle), 50mm (normal) and 135mm (telephoto) lens for 35mm film.

Taking a table of viewing angle according to focal length,

FL Long Side Short Side
28mm 65° 45°
50mm 40° 27°
135mm 15° 10°

we can determine the effective amount of the image from
the rule:

w = d×tan(0.5×*theta*)

Where w is the width of the image (or height on the vertical side), d is the distance from camera to object and *theta* is the angle taken from the previous table.
*(We can ignore all constants, since we're only interested in ratios)*.

This gives *(for d=100)*:

FL Width Height
28mm 65 46
50mm 40 27
135mm 15 10

Now draw concentric rectangles of these dimensions (probably using mm as a scale), and the comparison should become obvious if you imagine a scene drawn within. Ideally, you should hold the diagram at a distance *d* from your eye (100mm in this case).

+-------------------------------+
| |
| |
| +-----------------+ |
| | | |
| | +-----+ | |
_______|______|_____|__135|_____|______|__(Scene horizon)
| | +-----+ | |
| | 50| |
| +-----------------+ |
| |
| 28|
+-------------------------------+

Obviously, this is only a rough guide, but may be useful if you're trying to select a lens to complement your others.

Also worthy of note, is that the focal length of a lens is infinitessimally larger than the theoretical minimum distance that an object can be placed to the lens and still focused sharply. At the focal length, the image rays will be parallel. Just inside the focal length, the image rays will converge at a huge distance. Therefore, for practical reasons, the minimum distance of focus is usually significantly larger than the focal length of the lens.