Index of Refraction: n1 * sin(theta1) = n2 * sin(theta2)

AI Thread Summary
The discussion revolves around calculating the index of refraction of an unknown liquid based on a light beam's total internal reflection. The angle of incidence is determined to be 41.2 degrees, which raises questions about whether this angle is the critical angle. It is established that if the angle of incidence exceeds the critical angle, total internal reflection occurs, but the exact critical angle cannot be definitively determined from the given data. Consequently, the calculated index of refraction of 1.5 is identified as the minimum possible value, with the actual refractive index potentially being higher. The conversation emphasizes the uncertainty surrounding the critical angle and its implications for determining the refractive index.
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Homework Statement


A beam of light is emitted 8.0 cm below the surface of an unknown liquid. The light beam is angled in a way that it strikes the surface 7.0 cm from the point directly above the light source. The light beam is totally internally reflected. What is the index of refraction of the unknown liquid?

Homework Equations


n1sinΘ1° = n2sinΘ2°

n = c/velocity in medium

The Attempt at a Solution


tanΘ = 7/8
Θ = 41.8°

Attempt 1:
n1sin41.2° = n1sin41.2°
Cancels out...

Attempt 2:
n1sinΘ° = 1sin90°
Two variables!

Attempt 3:
n1sin41.2° = 1sin90°
n1 = 1.518166
n1 = 1.5
Could do this, although I don't even know whether it is correct, since it does not use the actual critical angle.
 
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Attempt 2:
n1sinΘ° = 1sin90°
Two variables!

Here angle is 41.2 degree. That itself is the critical angle. Find n1.
 
rl.bhat said:
Attempt 2:
n1sinΘ° = 1sin90°
Two variables!

Here angle is 41.2 degree. That itself is the critical angle. Find n1.

Can you assume that for sure, though? It could be that the 41.2° is bigger than the actual critical angle, which is why it reflects... =|
 
Even if the angle of incidence slightly more than critical angle, the light will be totally internally reflected. If the angle 41.2 degree is more than critical angle, it is not possible to determine the refractive index of the medium.
 
So that would mean that there's no guarantee whether 41.2° really is the critical angle? And Attempt 3 would only work if 41.2° is the critical angle, right?[/size]
 
Yes. Bur in your attempt 3, how did you get the value of n1= .02?
 
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Ah, sorry. I made a little calculator mistake with brackets... No wonder the value for n1 looked odd...
 
You have calculated that when angle of incidence is 41.2 deg, then the light is totally internally reflected. But it is impossible to know from the given data what will happen if angle of incidence is less than 41.2 deg. This means that we cannot know the exact critical angle. At the most we can say that it is not more than 41.2 deg.
Therefore the answer you got in the third attempt (i.e. 1.5) is the minimum possible value of n. Actual value may be equal to this or more than.
Can you find why 1.5 is the minimum possible value? Why not maximum?
 
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