Physics Help Forum Reflection Problem

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 Nov 20th 2010, 05:56 PM #1 Junior Member   Join Date: Oct 2010 Posts: 28 Reflection Problem "A person lying at poolside looks over the edge of the pool and sees a bottle cap on the bottom, directly below. If the depth of the pool is 3.2 m, how far below the water surface does the bottle cap appear to be?" What formula should you use for this problem? Is it a reflection problem?
 Nov 21st 2010, 01:54 AM #2 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 No, it's a refraction problem. However, I'm not sure about the way such calculations are done though. n = v_(air)/v_(water) Where n is the refractive index of the velocity v_(air) of light in air to the velocity v_(water) of light in water. 1.33 = (3x10^8)/v_(water) Find the velocity of light in water. Then find the time that light takes to cover 3.2 m in water. Use that time, to find the distance that light would cover in air. From that, you should get the apparent depth of water. __________________ 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?
 Nov 21st 2010, 02:52 PM #3 Member   Join Date: Jul 2009 Posts: 72 i would say 3.2m i would the medium make any difference on the distance? the light doesnt get deflected im not sure though edit: anyway if the distance perceived is different you'll need to use the snell law which will give the same result probably as Unknown method Last edited by Haytham; Nov 21st 2010 at 03:32 PM.
Nov 22nd 2010, 05:30 AM   #4
Junior Member

Join Date: Oct 2010
Posts: 28
 Originally Posted by Unknown008 No, it's a refraction problem. However, I'm not sure about the way such calculations are done though. n = v_(air)/v_(water) Where n is the refractive index of the velocity v_(air) of light in air to the velocity v_(water) of light in water. 1.33 = (3x10^8)/v_(water) Find the velocity of light in water. Then find the time that light takes to cover 3.2 m in water. Use that time, to find the distance that light would cover in air. From that, you should get the apparent depth of water.
Okay so you would do 1.33v=3X10^8 which equals 225563909.8 and then you would divide that by 3.2 meters which would give you 70488721.8 so that would be the time it takes?

 Nov 22nd 2010, 06:03 AM #5 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Um.. the speed is right, but recall that: Speed = Distance / Time Hence, we get: 225563909.8 = 3.2 / t t = 3.2 / 225563909.8 = 1.418667 x 10^-8 s In air, light does how much distance in this time? s = ut s = 3x10^8 x 1.418667x10^-8 = 4.256 m So, I would say the apparent depth is 4.3 m to 2 significant figures __________________ 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?
 Nov 22nd 2010, 11:17 PM #6 Physics Team   Join Date: Feb 2009 Posts: 1,425 If i remember correctly, there is a relation Refractive index = real depth / apparent depth. Please google. The proof must be somewhere. If we use 1.33 as r.i, it works out to 2.406. It cannot be 4.3 as then it would appear deeper than the bottom of the pool !!
 Nov 23rd 2010, 05:47 AM #7 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Maybe it's the opposite then =/ Well, what I found is the formula: n = D_r / D_a D_a is the apparent depth and D_r is the real depth We get: 3.2 / 1.33 = D_a D_a = 2.4 m It's strange though, how 'velocity seems to be inversely proportional to distance'... maybe it's just me... __________________ 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?

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