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Angle of Refraction

by cybernerd
Tags: angle, critical, light, optics, refraction
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cybernerd
#1
Jul22-08, 11:49 PM
P: 27
1. The problem statement, all variables and given/known data

Calculate the angle of refraction for a substance that has a critical angle of 50.5 degrees when that substance is in the air.

2. Relevant equations

Apparently the equation I need is:

ni *• sine() = nr • sine ()

My problem is, I have no idea how to use this formula or if it's even the right one - my teacher told me to write it down with no explanation.
Please help me, I just need to know what the equation actually tells me - where I am supposed to input data?


3. The attempt at a solution
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mgb_phys
#2
Jul23-08, 08:36 AM
Sci Advisor
HW Helper
P: 8,953
I think you first need to understand refraction ( see http://en.wikipedia.org/wiki/Refraction) This explains where that equation, whihc you have slighlty wrong, comes from.

Do you have part of the question missing, it doesn't seem to make sense?
cybernerd
#3
Jul23-08, 01:21 PM
P: 27
Alright, I got the correct equation now....it makes a bit more sense with that.

No, that's the question exactly as it was written. My teacher told me to use the index of refraction for air (so, 1.00), the angle of incidence/critical angle (50.5). SInce the refracted angle in 90 degrees from the normal, I was told to use 90.

So if I manipulate the equation (where @ = unknown angle) -

ni (sin@i) = nr (sin@r)
into
sin@i = nr (sin@r) / ni

so

sin50.5 = (1.00)(sin90) / ni
equals
sin50.5 = 1/ ni
or
ni = 1-sin50.5

which gives me an answer of 1.295967, or 1.3

Am I doing this right? Or I am completely on the wrong track?

cybernerd
#4
Jul23-08, 01:23 PM
P: 27
Angle of Refraction

Typo correction on the last line of the equation:

ni = 1-sin50.5

SHOULD READ

ni = 1 / sin 50.5
mgb_phys
#5
Jul23-08, 02:50 PM
Sci Advisor
HW Helper
P: 8,953
That looks right (asuming the arithmatic is correct)
Your original equation didn't have any angles in it - thats what I meant by it being wrong.


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