Refraction Problem

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Homework Statement


http://img98.imageshack.us/img98/371/physicsproblemrh8.jpg [Broken]

Homework Equations




The Attempt at a Solution


Basically, I tried to use triangles to find the angles. But I don't know how as it seems I'm only given one side of the triangles.
 
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  • #2
LowlyPion
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Homework Statement



Homework Equations



The Attempt at a Solution


Basically, I tried to use triangles to find the angles. But I don't know how as it seems I'm only given one side of the triangles.
I think they want you to consider the effect of Snell's Law here.

[tex] \frac{n_1}{n_2} = \frac {Sin \theta_1}{Sin \theta_2} [/tex]
 
  • #3
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I think they want you to consider the effect of Snell's Law here.

[tex] \frac{n_1}{n_2} = \frac {Sin \theta_1}{Sin \theta_2} [/tex]
You mean
[tex] \frac{n_1}{n_2} = \frac {Sin \theta_2}{Sin \theta_1} [/tex]

So lets say n1 = n2 when the glass is empty because the angle of incident is equal to the angle of refraction. When the glass is full, n2 = 1.22 and everything else is unknown. So how would I go about finding just theta r, which is theta_1 in this case?
 
  • #4
LowlyPion
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You mean
[tex] \frac{n_1}{n_2} = \frac {Sin \theta_2}{Sin \theta_1} [/tex]

So lets say n1 = n2 when the glass is empty because the angle of incident is equal to the angle of refraction. When the glass is full, n2 = 1.22 and everything else is unknown. So how would I go about finding just theta r, which is theta_1 in this case?
Yes. Sorry for the hurried typo.

With respect to the problem you know the angle of sight over the top is h/(h2 + d2)1/2

And the angle it makes in the liquid is h/(h2 + d2/4)1/2

And you know they are related by the 1.22 then don't you?

You know D so solve for h. Having h in hand then you can figure the angle directly.
 
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  • #5
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Yes. Sorry for the hurried typo.

With respect to the problem you know the angle of sight over the top is h/(h2 + d2)1/2

And the angle it makes in the liquid is h/(h2 + d2/4)1/2

And you know they are related by the 1.22 then don't you?

You know D so solve for h. Having h in hand then you can figure the angle directly.
Thanks a lot lowlypion.

I had a little trouble figuring out why your values didn't work, but in the end I found that the values were actually:

d/(d2 + h2)1/2 for the top angle

and

(d/2)/(d2/4 + h2)1/2 for the liquid angle.

All in all though, you greatly helped me. Thank you so much!
 
  • #6
LowlyPion
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Thanks a lot lowlypion.

I had a little trouble figuring out why your values didn't work, but in the end I found that the values were actually:

d/(d2 + h2)1/2 for the top angle

and

(d/2)/(d2/4 + h2)1/2 for the liquid angle.

All in all though, you greatly helped me. Thank you so much!
My apologies. I relied on the Hyperphysics link that shows the angle with the surface rather than with the normal.

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/refr.html#c3

snell2.gif
 

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