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Fresnel Equations and Snell's Law

  • #1
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Homework Statement


From the Fresnel equations and Snell’s Law, prove that, when θ = θB where tanθB = nt/ni, (θB is the Brewster angle); (a) Reflection coefficient = 0 , and (b) transmission coefficient = n/n’

Homework Equations


reflection coefficient = (ntcosθi - nicosθt) / (ntcosθi + nicosθt)

transmission coefficient = (2nicosθi) / (ntcosθi+nicosθt)

Snell's Law : nisinθi = ntsinθt

Brewster's Angle: θi + θt = 90°

The Attempt at a Solution



I know that for Brewster's Angle θi + θt = 90°

Part a) is basically asking to derive the current version of the reflection coefficient, into it's more known tangent version:

tan(θi - θt) / tan(θi + θt)

For Brewster's Angle tan(θi + θt) = tan(90°) = 0

So reflection coefficient will equal 0.

Rearranging Snell's Law: ni= ntsinθt / sinθi

Subbing ni into the reflection coefficient and multiplying both numerator and denominator by sinθi / nt gives:

(sinθtcosθt - sinθicosθi) / (sinθtcosθt + sinθicosθi)

I don't know how to get from there to tan though. Any help would be much appreciated.
 

Answers and Replies

  • #2
nrqed
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Homework Statement


From the Fresnel equations and Snell’s Law, prove that, when θ = θB where tanθB = nt/ni, (θB is the Brewster angle); (a) Reflection coefficient = 0 , and (b) transmission coefficient = n/n’

Homework Equations


reflection coefficient = (ntcosθi - nicosθt) / (ntcosθi + nicosθt)

transmission coefficient = (2nicosθi) / (ntcosθi+nicosθt)

Snell's Law : nisinθi = ntsinθt

Brewster's Angle: θi + θt = 90°

The Attempt at a Solution



I know that for Brewster's Angle θi + θt = 90°

Part a) is basically asking to derive the current version of the reflection coefficient, into it's more known tangent version:

tan(θi - θt) / tan(θi + θt)

For Brewster's Angle tan(θi + θt) = tan(90°) = 0

So reflection coefficient will equal 0.

Rearranging Snell's Law: ni= ntsinθt / sinθi

Subbing ni into the reflection coefficient and multiplying both numerator and denominator by sinθi / nt gives:

(sinθtcosθt - sinθicosθi) / (sinθtcosθt + sinθicosθi)

I don't know how to get from there to tan though. Any help would be much appreciated.
Wait... for part b) they are asking to find the transmission coefficient, right? Are you calculating the reflection coefficient??
 
  • #3
594
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Yes, for part a) I need to derive the reflection coefficient and part b) the transmission coefficient
 
  • #4
nrqed
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Yes, for part a) I need to derive the reflection coefficient and part b) the transmission coefficient
Ah, I thought you had started part b). Ok.

Note that it is not true that tan(90) = 0.


You have to use trig identities for the product of sin(A) cos(A)
 
  • #5
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something like ... sin(A)cos(A) = 1/2(2sinAcosA) = 1/2(sin2A) = sin2A/2
 
  • #6
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or 2sinAcosA = sin2A
 
  • #7
nrqed
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or 2sinAcosA = sin2A
Right. And then use the identity for sin(A) + sin(B) and so on.
 
  • #8
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Right. And then use the identity for sin(A) + sin(B) and so on.
I don't follow why I need to use sin(A) + sin(B) identity
 
  • #9
nrqed
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I don't follow why I need to use sin(A) + sin(B) identity
Write the next few steps. You will encounter this type of terms
 
  • #10
594
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ok, so I think I have it. I've replaced most of the terms with A and B for easier writing / reading at this point.

2sinAcosA = sin2A

(2sinAcosA - 2sinBcosB) / (2sinAcosA + 2sinBcosB)

= sin2A - sin2B / sin2A + sin2B

= 2sin(A-B)cos(A+B) / 2sin(A+B)cos(A-B)

At Brewster's Angle, A+B = 90°, therefore

= 2sin(A-B)cos(90) / 2sin(90)cos(A-B)

cos(90°) = 0, therefore

Reflection coefficient = 0 at Brewster's Angle
 
  • #11
nrqed
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ok, so I think I have it. I've replaced most of the terms with A and B for easier writing / reading at this point.

2sinAcosA = sin2A

(2sinAcosA - 2sinBcosB) / (2sinAcosA + 2sinBcosB)

= sin2A - sin2B / sin2A + sin2B

= 2sin(A-B)cos(A+B) / 2sin(A+B)cos(A-B)

At Brewster's Angle, A+B = 90°, therefore

= 2sin(A-B)cos(90) / 2sin(90)cos(A-B)

cos(90°) = 0, therefore

Reflection coefficient = 0 at Brewster's Angle
good job :-)
 
  • #12
594
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I'm not sure how to get started with the transmission coefficient. There are so many substitutions and trig identities to try...
 
  • #13
594
12
At brewster's angle nisinθi = ntcosθi

2nicosθi / ntcosθi+nicosθt

= 2nicosθi/nisinθi+nicosθt

= 2cosθi/sinθi+cosθt

I'm kind of stuck here though...
 
  • #14
594
12
ok, I think I've figured it out using Snell's Law:

nisinθi=ntsinθt

nisinθi=ntsin(90-θB

nisinθi=ntcosθB

OR

nisinθi=ntcosθi

2nicosθi / ntcosθi+nicosθt

subbing in the relation: nisinθi=ntcosθi

2nicosθi / nisinθi+nicosθt

Multiply numerator and denominator by 1/ni

2cosθi / sinθi+cosθt

subbing in relation: cosθt = sinθi

cosθt = sin(90-θt) = sinθi

2cosθi / sinθi+sinθi

2cosθi / 2sinθi

= ni / nt
 

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