Index of Refraction in a prism

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Homework Help Overview

The discussion revolves around determining the index of refraction for light incident on a 45-degree prism, particularly in the context of total internal reflection. Participants are exploring the relationship between angles of incidence and refraction in the prism setup.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply Snell's law to find the index of refraction, questioning the angles involved in the prism's geometry. There are discussions about the implications of total internal reflection and the specific angles that should be considered.

Discussion Status

The conversation is ongoing, with participants providing insights into the relevant formulas and questioning the assumptions about the angles. Some guidance has been offered regarding the application of Snell's law, but there is no clear consensus on the specific values for the angles involved.

Contextual Notes

Participants are working under the assumption that the prism is made of glass, with the index of refraction for air taken as 1. There is also a mention of difficulties in editing mathematical expressions, which may affect clarity in communication.

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Homework Statement



The light incident on a 45 degree prism undergoes total internal relfection at point P. What can you conclude about the index of refraction in the prism? (Determine either a minimum or maximum)


This is a right triangle, with the two other angles being 45 degrees and point is half on the hypotenuse.

The Attempt at a Solution



The answer in the back of the book is 1.41. I have no idea how they came up with this. Could someone explain about they found this number.
 
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The index of refraction n2/n1 =[tex]\frac{sin \theta_1}{sin \theta_2}[/tex] where theta 1 and theta 2 are the angles from the line normal to the surface (prism). Note that n1 (air) = 1 (for some reason it's not letting me edit the latex)

theta 1 will be the angle the light ray in air makes with the normal, and theta 2 will be the angle it makes with the normal in the prism. So what will theta 1 and 2 equal?
 
Start with what you do know.

An interesting thing to wonder might be what kind of change in index of refraction at the boundary of the prism could affect such a change in angle? Are there any formulas covered in the chapter that might help you in this regard?

Edit: I see someone has already provided you with the formula even while I was typing my message. You should be well on your way.
 
one will be 45 degrees and the other would be 30 degrees?
 
Why would it be 30? If there is total internal reflection, the light beam travels along the edge of the prism (in the miminum case). So what would the 2nd angle be?
 
From my book, I found that as the angle of incident is increased, the angle of refraction evenutally reaches 90. At 90, it just moves along the surface. So with what your saying, the other angle would be 45 degrees, right?
 
n1 = air = 1
n2 = glass ( assuming that the prism is glass)
Sin theta 1 = 45 degrees
sin theat 2 is = degrees

formula used n2 = n1*sin theta 1 / sin theta 2

n2 = (1)*sin 45/sin theta 2
 
If you're saying n2 = glass, then theta1 should be the refracted angle. Remember, your angles start at your normal line, that is at 90 degrees to the surface of the prism. If the light moves along the surface, what is theta1? Then theta2 is going to be the incident ray's angle, which, as you said will be 45 degrees.
 

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