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Old Nov 20th 2010, 05:56 PM   #1
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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?
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Old Nov 21st 2010, 01:54 AM   #2
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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.
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Old Nov 21st 2010, 02:52 PM   #3
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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.
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Old Nov 22nd 2010, 05:30 AM   #4
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Originally Posted by Unknown008 View Post
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?
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Old Nov 22nd 2010, 06:03 AM   #5
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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
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Old Nov 22nd 2010, 11:17 PM   #6
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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 !!
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Old Nov 23rd 2010, 05:47 AM   #7
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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...
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