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Old Jan 11th 2012, 09:14 PM   #1
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light question

hi,

So, I was trying to work on this question:

Last edited by Koko; Jan 24th 2012 at 01:23 PM.
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Old Jan 12th 2012, 08:55 AM   #2
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Does this help for the length of the shadow?

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Old Jan 12th 2012, 07:55 PM   #3
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Hi,
I tried different things and I'm just not sure if I'm doing the right thing; but I found the area for the unknown lengths which you put on the image, and I found it to be .250m^2. Then I noticed that I get the right answer only if I subtract the area of the rectangle. Other ways are similar... and how do you know which part you are supposed to find because it says 1m 'beyond', but you don't know how big the shadow that is produced will be.

Last edited by Koko; Jan 13th 2012 at 06:35 AM.
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Old Jan 13th 2012, 07:53 AM   #4
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You found the area for the unknown lengths!?

A length doesn't have an area!

We know that the distance from the light source to the screen is 1.5 m.

Using similar triangles, we know that:

0.125/0.5 = ?/1.5

? = 37.5 cm

Then the length of the shadow becomes twice that, that is 75 cm.

Now, you do the same with the side as 10 cm.

0.10/0.5 = ?/1.5

.....
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Last edited by Unknown008; Jan 13th 2012 at 09:24 AM. Reason: Typo of m and cm
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Old Jan 13th 2012, 08:00 AM   #5
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Sorry! I don't think I was too clear : S Well, I meant that I found the area using the unknown lengths. The vertical unknown length to be 0.375m and horizontal's known to be 1.0m... I found the area of the whole thing now that I know 0.375m, so 1.5m x 0.375m, and minus the area between the rectangle and the light... so 0.5 x 0.125m... and got 0.250m^2.. but the ans supposed to be 0.225m, I tried subtracting the half area of the rectangle (0.125m x 0.1m) but I only get the right answer if I subtract the whole area of the rectangle (0.25m x 0.1 m)
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Old Jan 13th 2012, 09:25 AM   #6
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You didn't make full use the the drawing... the ? that you have to find is half the entire length.
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Old Jan 13th 2012, 09:49 AM   #7
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Hi!
I just got the answer. So my problem was that I didn't understand the drawing fully. I was confused thinking that I had to use the 1.0m x ? m in order to get the area. And I just read the second last post you put on.. I just didn't have time to read it before running to school but yea now that I look it's what I did to get the answer. Thanks so much for your help!
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Old Jan 13th 2012, 10:08 PM   #8
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Haha! Okay, all's good
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Old Feb 10th 2012, 09:51 AM   #9
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Hi, or you may try trignometery, tan theta= perpendicular /base values are given, for smae angle you can find other length anwaar
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