What Happens When the Sine of the Refracted Angle Exceeds One?

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Homework Help Overview

The discussion revolves around the behavior of light as it transitions between two substances with different refractive indices, specifically addressing the scenario where the sine of the refracted angle exceeds one. The original poster describes a situation encountered during a test involving an incident angle and refractive indices, leading to confusion about the implications of a sine value greater than one.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Snell's Law to determine the refracted angle but encounters a sine value greater than one, prompting questions about the validity of their approach and the implications of this result.

Discussion Status

Some participants provide insights into the nature of total internal reflection and the critical angle, suggesting that if the sine of the refracted angle exceeds one, total internal reflection occurs instead of refraction. There is an exploration of the original poster's reasoning and attempts to clarify the situation without reaching a definitive conclusion.

Contextual Notes

The original poster expresses concern about waiting for test results to understand the problem, indicating a sense of urgency in seeking clarification. There are also mentions of logistical challenges in accessing the instructor for further questions.

steppenwolf
ok so this was in my end of unit test today and i have been stressing as to what i should have done, so as it's not really homework you could always just tell me the answer

pretty basic stuff, a ray of light is incident from a substance upon the surface of another substance with n=1. i can't remember the exact refractive index of the first material but it was about 1.4 or something, hmm. anyway, i worked out the incident abgle to be 47 degrees and to find the refractive angle used
sin incident angle x n1 = sin refractive angle x n2
seeing as n2 = 1 then you get
sin refracted angle = sin incident angle x n1
which ended up being about 1.09 or something, ok, a sine>1 ?

well i just pretended it was 1 and so worked out the refracted angle to be total, ie 90 degrees. was this the right thing to do? was this some measurment discrepancy i was supposed to disregard? help, i don't want to wait a month for results to understand this problem!
 
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Wait a month? Why not just go to your teacher and ask?


I cringed at "i just pretended it was 1" but basically, that is what happens. If the angle of incidence and index of refraction
are such that n*sin(theta)> 1 then the light is completely reflected- there is no angle of refraction.
 
thanks, that's what i thought. i would ask my teacher but i don't know where he lives, in his lab maybe but i doubt it, i will go stake out the science department until he surfaces...
 
Originally posted by steppenwolf
i will go stake out the science department until he surfaces...

You could look at it as a part of the test... apart from learning physics, they want you to get familiar with the institute... :wink:
 
Going to class might be a good way of finding your instructor!

Failing that, checking the syllabus he gave you at the beginning of the year, the schedule posted on his office door or even asking the department secretary just might be ways of determining when he is in.
 
i don't have classes for a few weeks actually!
but thanks anyway, i am now stumped by a certain geometry problem, is it worth a new thread? hmm, alright then!
*goes off to post new thread on particularily nasty geometry shenanigans*
 
Originally posted by HallsofIvy
Wait a month? Why not just go to your teacher and ask?


I cringed at "i just pretended it was 1" but basically, that is what happens. If the angle of incidence and index of refraction
are such that n*sin(theta)> 1 then the light is completely reflected- there is no angle of refraction.

Yep..When doing these problems, when lights going form more dense to less dense, you have to check for the critical angle...and like Halls said, if your angle of incidence is greater than that critical angle, you get reflection...and when it reflects its just incidence=reflection
 

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