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

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Homework Help Overview

The problem involves a beam of light emitted from a point below the surface of an unknown liquid, striking the surface at an angle and undergoing total internal reflection. The objective is to determine the index of refraction of the liquid using Snell's law.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various attempts to apply Snell's law, questioning the assumptions about the critical angle and the validity of calculated values for the index of refraction.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the critical angle and its implications for the refractive index. Some guidance has been provided regarding the minimum possible value of the index of refraction, but no consensus has been reached on the exact critical angle.

Contextual Notes

There is uncertainty regarding whether the calculated angle of 41.2° is indeed the critical angle, which affects the determination of the refractive index. Participants note that the information provided may not be sufficient to ascertain the critical angle definitively.

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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.
 
Last edited:
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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?
 
Last edited:
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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