Physics Help Forum light question

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 Jan 11th 2012, 09:14 PM #1 Junior Member   Join Date: Jan 2012 Posts: 13 light question hi, So, I was trying to work on this question: Last edited by Koko; Jan 24th 2012 at 01:23 PM.
 Jan 12th 2012, 08:55 AM #2 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Does this help for the length of the shadow? __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
 Jan 12th 2012, 07:55 PM #3 Junior Member   Join Date: Jan 2012 Posts: 13 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.
 Jan 13th 2012, 07:53 AM #4 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 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 ..... __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying? Last edited by Unknown008; Jan 13th 2012 at 09:24 AM. Reason: Typo of m and cm
 Jan 13th 2012, 08:00 AM #5 Junior Member   Join Date: Jan 2012 Posts: 13 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)
 Jan 13th 2012, 09:25 AM #6 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 You didn't make full use the the drawing... the ? that you have to find is half the entire length. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
 Jan 13th 2012, 09:49 AM #7 Junior Member   Join Date: Jan 2012 Posts: 13 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!
 Jan 13th 2012, 10:08 PM #8 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Haha! Okay, all's good __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
 Feb 10th 2012, 09:51 AM #9 Member   Join Date: Nov 2008 Location: PAKISTAN Posts: 79 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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