What is the Maximum Angle for Total Internal Reflection in a Prism?

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Homework Help Overview

The discussion revolves around determining the maximum angle for total internal reflection in a glass prism with a specific refractive index when immersed in air. The original poster attempts to calculate the critical angle and subsequently the angle alpha, which is necessary for no light to refract out of the prism.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the critical angle and the angle alpha, with some questioning the values used for the refractive index of air. There is also confusion about the definitions of angles in relation to the prism's surfaces.

Discussion Status

Some participants have provided insights regarding the correct refractive index for air and the relationship between the angles involved. There is an ongoing exploration of the geometric relationships and the need for clear diagrams to aid understanding.

Contextual Notes

Participants note discrepancies in the values used for refractive indices and the definitions of angles, indicating a need for clarification on these assumptions. There is also a mention of feedback received regarding the original calculations.

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Homework Statement



Light is incident along the normal to face AB of a glass prism of refractive index 1.47

Find the alpha max, the largest value of the angle alpha such that no light is refracted out of the prism at face AC if the prism is immersed in air.

Homework Equations


n1*sin thetha1 = n2 * sin thetha2


The Attempt at a Solution


since no light is refracted out of the prism, then total internal reflection is involved.
sin critical alpha = (n2/n1)*sin 90
= 1.33/147
alpha = 64.8

but I got it wrong,
please someone help me..
thanks..
 

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What you found is the angle that the incident ray makes with the normal to surface A-C. That's not alpha, but you can use it to find alpha.
 
I just got it..
I just realized that the n for air should be 1.00 not 1.33.
then to get the alpha --> 90 - the angle that the incident ray makes with the normal to surface A-C..

anyway.. thanks.. :)
 
xinlan said:
I just got it..
Good!
I just realized that the n for air should be 1.00 not 1.33.
D'oh! (I didn't even notice that. I must have been sleeping.)
then to get the alpha --> 90 - the angle that the incident ray makes with the normal to surface A-C..
Right.
 
can someone explain a little more abour this problem ... i do no understand why is necessary to use 90 ... i just got one answer but its wrong this is the feedbak ... You have given the angle between the light ray and the normal to the surface AB, not the angle for alpha
 
Last edited:
blue_shark said:
can someone explain a little more abour this problem ... i do no understand why is necessary to use 90 ... i just got one answer but its wrong this is the feedbak ... You have given the angle between the light ray and the normal to the surface AB, not the angle for alpha
What you gave, I presume, is the critical angle, which is not the angle alpha. You've got to use the critical angle to figure out alpha. Draw a clear diagram and look for the triangles.
 

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