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Question about physics of light / light travel

 
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May20-08, 08:06 PM   #1
 

Question about physics of light / light travel


I have a question which relates to the physics of light. I hope I can explain it well enough for a proper answer!

Let's say there is a 15 feet tall light pole 50 feet from my door step. The light pole has a 300W light bulb. Given these info, is there anyway to calculate the amount of light at my door step? Can I say that the light in front of my doorstep, for example, is equal to a 20W light bulb?

Can anyone please provide a general formula for me to plug in the numbers. Thanks in advance.
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May22-08, 02:41 AM   #2
 
the distance of the bulb from your doorstep is (225 + 2500)^0.5 feet. the power falls off with the square of the distance. so it would work out to 300/(225 + 2500) watts.
May22-08, 09:18 AM   #3
 
Thanks very much for the reply. I hate to ask, but can u please explain more.
I am trying very hard to remember my math and physics from 20 years ago but my brain is useless!

How is the distance between the light and door = (225 + 2500)^0.5 .. that would equal to like 7425625!!!! Shouldn't it be a^2 + b^2 = c^2 or 50^2+15^2= 52^2.

So C is equal to about 52 feet which is the distance from lightbulb to the bottom of the door step.

but anyway, are you saying the light infront of the dorr step is equal to about .11 watts? that is 300/(225 + 2500)= .11 watts.









I need to calculate this to prove something to my insurance company!!!
May22-08, 10:34 AM   #4
 

Question about physics of light / light travel


Quote by switchex View Post

How is the distance between the light and door = (225 + 2500)^0.5 .. that would equal to like 7425625!!!! Shouldn't it be a^2 + b^2 = c^2 or 50^2+15^2= 52^2.


(225 + 2500)^0.5 = 52
May22-08, 10:39 AM   #5
 
The brightness of a bulb at distance D from the source is equal to:

[tex]B=\frac{L}{4\pi{D^2}}[/tex]

The distance is indeed ~52 foot,

B = 5.8x10^-4 W/square foot.

It may bode you well to convert to metres from feet first though!
May22-08, 05:00 PM   #6
 
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Quote by switchex View Post
Let's say there is a 15 feet tall light pole 50 feet from my door step. The light pole has a 300W light bulb. Given these info, is there anyway to calculate the amount of light at my door step? Can I say that the light in front of my doorstep, for example, is equal to a 20W light bulb?
The vital information which is missing and nobody seems have asked for is that the light from the 300W bulb at your doorstep is equal to that from a 20W bulb placed where? On the top of the door step? If so, what is the distance from the 20W bulb to the door step?

In both cases, you also need the angle at which the light falls from either bulb on the flat doorstep. The first is given, and the light falls slanted on the step. The second one will be most probably almost directly on the top of the doorstep. (Of course, this I think may not be important, because you are probably estimating the light in the vicinity of the doorstep, not exactly on it.

Another complication is that the percentage of energy converted to light in a 300W bulb is different from that in a 20W bulb. But this can be ignored -- both will be around 4-5%.
May22-08, 10:08 PM   #7
 
Quote by Shooting Star View Post
The vital information which is missing and nobody seems have asked for is that the light from the 300W bulb at your doorstep is equal to that from a 20W bulb placed where? On the top of the door step? If so, what is the distance from the 20W bulb to the door step?
i think what was meant in the OP was how he can compare a 300W bulb placed 52ft away with a 20W bulb placed 'x' feet away. so if the power received at doorstep by a 300W bulb is 0.11W then for an equivalent power to be received by a 20W bulb at the doorstep it would have to be placed (20/0.11)^0.5 feet away. works out to roughly 13.5 feet.

now that i look at the numbers, they somehow dont seem right..
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