Is the Critical Angle Equal to the Brewster Angle?

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SUMMARY

The discussion centers on the relationship between the critical angle and Brewster angle in optics, specifically when the incident refractive index is air (n=1). The user attempts to derive the refractive index (n) using the equations arcsin(1/n) = arctan(n) and encounters algebraic difficulties. Ultimately, the user concludes that the refractive index is approximately 1.272, confirming the relationship between these angles through trigonometric identities and algebraic manipulation.

PREREQUISITES
  • Understanding of Snell's Law in optics
  • Familiarity with trigonometric identities, particularly sin, cos, and tan
  • Basic algebra skills for manipulating equations
  • Knowledge of critical and Brewster angles in optics
NEXT STEPS
  • Study the derivation of Snell's Law and its applications in optics
  • Learn about the mathematical relationships between critical angle and Brewster angle
  • Explore the implications of refractive indices in different materials
  • Investigate advanced trigonometric identities and their applications in physics
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Students studying optics, physics enthusiasts, and anyone interested in understanding the mathematical relationships between angles of incidence and refraction.

fredrick08
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Homework Statement


when the critical angle=brewster angle, what is the refractive index, if incident n is air=1.

arcsin(1/n)=arctan(n)=>n=1.272...?

how is this? i tried differentiating both side and solving but i just get into a big mess, can anyone help me if i am missing something?
 
Last edited:
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ive tried pythagoras, and trig idendenties, but all i come up with is x=x??
 
If you want to use identities you can take the sin on both sides and then use \cos^2x+\sin^2x=1,\;1+\tan^2x=\sec^2x to simplify the right hand side.
 
ok thanks i think i got it now... just confused about the algebra
 
all solved, thanks very much
 
I'm still at a loss as to how n = 1.272.

so getting to:
1/n = sin(arctan(n))

then squaring and adding cos^2 gets:

1/n^2 - cos^2(arctan(n)) = 1

but any way I rearrange the above formula to get 1/cos^2 for sec^2 ends up resulting in n=1?
 

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