What is the Refractive Index for Total Internal Reflection in a 45-45-90 Prism?

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Homework Help Overview

The problem involves determining the minimum refractive index required for total internal reflection in a 45-45-90 prism that is immersed in water. Participants are exploring the implications of Snell's law and the conditions for total internal reflection.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Snell's law to find the refractive index of the prism, questioning the assumptions made regarding the medium (water vs. air) and the angles involved in the reflection process.

Discussion Status

There is an ongoing exploration of the calculations related to the refractive index. Some participants have pointed out potential errors in the assumptions about the medium and the application of Snell's law, while others confirm the calculations leading to the refractive index of approximately 1.88.

Contextual Notes

Participants note the confusion regarding the medium in which the prism is immersed, clarifying that it is water rather than air, which affects the critical angle and the refractive index calculations.

Cummings
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The Question:
A 45-45-90 prism is immersed in water. What is the minimum index or fefraction the prism must have if it is to reflect totally a ray incident normally on one of its faces.

Now, we have found this question a bit hard to understand.
We assume that light is hitting one of the sides at 90 degrees (normally) and so will pass into the prism without refraction. Thus, it hits the other side of the prism at an angle of incidence of 45 degrees.

This is where we get stuck, we are assuming that it is totally internaly reflecting and by using snells law nSin(45) = Sin(90) we find that n, the refractive index of the prism is 1.88

This complies with
the fact that to be totally internaly reflected the light must travel from a high refractive index (1.88) to a low refractive index (1 for air)

If we are on the right track, please let us know!
 
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Cummings said:
This is where we get stuck, we are assuming that it is totally internaly reflecting and by using snells law nSin(45) = Sin(90) we find that n, the refractive index of the prism is 1.88
You didn't get n=1.88 using that equation. I assume you meant to write Snell's law as: n_1 sin\theta_1 = n_2 sin\theta_2. And, yes, the critical angle is found by setting \theta_2 = 90.

This complies with
the fact that to be totally internaly reflected the light must travel from a high refractive index (1.88) to a low refractive index (1 for air)
If you assumed going from n=1.88 to air (n=1) I am very curious how you got your answer. (The prism is immersed in water, not air.)
 
Cummings said:
This is where we get stuck, we are assuming that it is totally internaly reflecting and by using snells law nSin(45) = Sin(90) we find that n, the refractive index of the prism is 1.88

If we are on the right track, please let us know!

I think it should be n1*Sin(45) = n2*sin(90), not nSin(45) = Sin(90)

which is

n1*sin(45) = 1.333
n1 = 1.333/sin 45

n1 is approx 1.88
 
Yes, i made a few erros in that post. Had to make it to the next class of uni.

Yes, the prism is in water. relinquished managed to type in what i failed to. He got the same answer to.

I just for some reason replaced water with air when i typed this up.

So, would 1.88 be the minumum index of refraction?
 
Cummings said:
I just for some reason replaced water with air when i typed this up.
I knew you must have, since your answer was correct.

So, would 1.88 be the minumum index of refraction?
Yep.
 

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