Fresnel Equations and Snell's Law

[says]

Homework Statement

From the Fresnel equations and Snell’s Law, prove that, when θ = θ_B where tanθ_B = n_t/n_i, (θ_B is the Brewster angle); (a) Reflection coefficient = 0 , and (b) transmission coefficient = n/n’

Homework Equations

reflection coefficient = (n_tcosθ_i - n_icosθ_t) / (n_tcosθ_i + n_icosθ_t)

transmission coefficient = (2n_icosθ_i) / (n_tcosθ_i+n_icosθ_t)

Snell's Law : n_isinθ_i = n_tsinθ_t

Brewster's Angle: θ_i + θ_t = 90°

The Attempt at a Solution

I know that for Brewster's Angle θ_i + θ_t = 90°

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

tan(θ_i - θ_t) / tan(θ_i + θ_t)

For Brewster's Angle tan(θ_i + θ_t) = tan(90°) = 0

So reflection coefficient will equal 0.

Rearranging Snell's Law: n_i= n_tsinθ_t / sinθ_i

Subbing n_i into the reflection coefficient and multiplying both numerator and denominator by sinθ_i / n_t gives:

(sinθ_tcosθ_t - sinθ_icosθ_i) / (sinθ_tcosθ_t + sinθ_icosθ_i)

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

 

[nrqed]

Homework Statement

From the Fresnel equations and Snell’s Law, prove that, when θ = θ_B where tanθ_B = n_t/n_i, (θ_B is the Brewster angle); (a) Reflection coefficient = 0 , and (b) transmission coefficient = n/n’

Homework Equations

reflection coefficient = (n_tcosθ_i - n_icosθ_t) / (n_tcosθ_i + n_icosθ_t)

transmission coefficient = (2n_icosθ_i) / (n_tcosθ_i+n_icosθ_t)

Snell's Law : n_isinθ_i = n_tsinθ_t

Brewster's Angle: θ_i + θ_t = 90°

The Attempt at a Solution

I know that for Brewster's Angle θ_i + θ_t = 90°

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

tan(θ_i - θ_t) / tan(θ_i + θ_t)

For Brewster's Angle tan(θ_i + θ_t) = tan(90°) = 0

So reflection coefficient will equal 0.

Rearranging Snell's Law: n_i= n_tsinθ_t / sinθ_i

Subbing n_i into the reflection coefficient and multiplying both numerator and denominator by sinθ_i / n_t gives:

(sinθ_tcosθ_t - sinθ_icosθ_i) / (sinθ_tcosθ_t + sinθ_icosθ_i)

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??

 

[says]

Yes, for part a) I need to derive the reflection coefficient and part b) the transmission coefficient

 

[nrqed]

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)

 

[says]

something like ... sin(A)cos(A) = 1/2(2sinAcosA) = 1/2(sin2A) = sin2A/2

 

[says]

or 2sinAcosA = sin2A

 

[nrqed]

or 2sinAcosA = sin2A

Right. And then use the identity for sin(A) + sin(B) and so on.

 

[says]

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

 

[nrqed]

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

 

[says]

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

 

[nrqed]

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 :-)

 

[says]

I'm not sure how to get started with the transmission coefficient. There are so many substitutions and trig identities to try...

 

[says]

At brewster's angle n_isinθ_i = n_tcosθ_i

2n_icosθ_i / n_tcosθ_i+n_icosθ_t

= 2n_icosθ_i/n_isinθ_i+n_icosθ_t

= 2cosθ_i/sinθ_i+cosθ_t

I'm kind of stuck here though...

 

[says]

ok, I think I've figured it out using Snell's Law:

n_isinθ_i=n_tsinθ_t

n_isinθ_i=n_tsin(90-θ_B

n_isinθ_i=n_tcosθ_B

OR

n_isinθ_i=n_tcosθ_i

2n_icosθ_i / n_tcosθ_i+n_icosθ_t

subbing in the relation: n_isinθ_i=n_tcosθ_i

2n_icosθ_i / n_isinθ_i+n_icosθ_t

Multiply numerator and denominator by 1/n_i

2cosθ_i / sinθ_i+cosθ_t

subbing in relation: cosθ_t = sinθ_i

cosθ_t = sin(90-θ_t) = sinθ_i

2cosθ_i / sinθ_i+sinθ_i

2cosθ_i / 2sinθ_i

= n_i / n_t