When the incidence angle is equal to the angle of refraction

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Homework Help Overview

The discussion revolves around the conditions under which the angle of incidence equals the angle of refraction, particularly in the context of Snell's law and the behavior of light at the interface of different media.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the angles being equal and question whether this scenario indicates the same refractive indices. There is also a consideration of critical angles and the conditions under which light passes straight through the interface.

Discussion Status

The discussion includes various interpretations of the relationship between incidence and refraction angles, with some participants suggesting specific angles and others questioning the assumptions behind these angles. There is acknowledgment of Snell's law and its application to the scenario.

Contextual Notes

Participants reference the critical angle and the conditions for light to pass through without refraction, indicating a need for clarity on the definitions and implications of these angles in different media.

zade70
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Homework Statement


When the incidence angle is equal to the angle of refraction?

Homework Equations


sina/sinb=n2/n1

The Attempt at a Solution


I think as a=b => sina=sinb => n1=n2. We should have the same environment? But is that refraction?
 
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At the interface of two different media, there is a certain angle of incidence where the incident ray just passes straight through. What is this angle?
 
blue_leaf77 said:
At the interface of two different media, there is a certain angle of incidence where the incident ray just passes straight through. What is this angle?
when b= pi/2 rad. It is the limit angle for a. sina=n2/n1
 
zade70 said:
when b= pi/2 rad. It is the limit angle for a. sina=n2/n1
No, that corresponds to critical angle refraction. It very simple, you have Snell's law ##n_1 \sin a = n_2 \sin b##. Try to find a trivial angle ##a=b## which will make the left and right hand side of Snell's law equal irrespective of the refractive indices.
 
blue_leaf77 said:
No, that corresponds to critical angle refraction. It very simple, you have Snell's law ##n_1 \sin a = n_2 \sin b##. Try to find a trivial angle ##a=b## which will make the left and right hand side of Snell's law equal irrespective of the refractive indices.
a=b=0 rad? 0=0
 
Yes.
 
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