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asa!!

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- Thread starter asa!!
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asa!!

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- #2

Doc Al

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[tex]\frac{1}{o} + \frac{1}{i} = \frac{1}{f}[/tex]

where [itex]o[/itex] is the distance from the object to the lens, [itex]i[/itex] is the distance from the lens to the image, and [itex]f[/itex] is the focal length of the lens. Parallel rays hitting the lens are equivalent to an object an infinite distance away; in that special case, the rays are focused at the "focal point" which is a distance [itex]i = f[/itex] from the lens.

Here's more about lenses: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenseq.html

- #3

El Hombre Invisible

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Basically, the shorter the focal length, the stronger the lens. If you have a source from infinity passing through a converging lens, a stronger lens will focus the parallel rays at a shorter distance than a weaker one. Likewise if the parallel rays pass through a diverging lens, the virtual object will appear closer to the stronger lens than the shorter one. So the focal length is really a measure of how strongly refracted the light will be.asa!! said:

- #4

fizixx

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asa.....

Here is a good link that will help illustrate how the focal length, image distance and object distance are all related.

http://www.hazelwood.k12.mo.us/~grichert/optics/intro.html [Broken]

This is one of those physics concepts that really makes more sense mathematically. Unfortunately many folks get confused and immediately wish to rely on the visual instead of making that leap into the more abstract.

You can work with this link in real time. Meaning, you can add lenses and mirrors and an object and move them all around and see what is happening. Very instructive, but deceptively simple. Meaning, you can see how troublesome the concept is to grasp, but with this tool it quickly and easily does the math-work for you.

A drawback is that it's a bit small and you can't get too complicated with your 'designs', but hopefully you will find it helpful and useful.

Enjoy and keep thinking!

fizixx

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