Fresnel Equations and Snell's Law

In summary: Transmission coefficient = ni/ntIn summary, at Brewster's Angle, the reflection coefficient is 0 and the transmission coefficient is ni/nt.
  • #1
says
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12

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.
 
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  • #2
says said:

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
Yes, for part a) I need to derive the reflection coefficient and part b) the transmission coefficient
 
  • #4
says said:
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
something like ... sin(A)cos(A) = 1/2(2sinAcosA) = 1/2(sin2A) = sin2A/2
 
  • #6
or 2sinAcosA = sin2A
 
  • #7
says said:
or 2sinAcosA = sin2A
Right. And then use the identity for sin(A) + sin(B) and so on.
 
  • #8
nrqed said:
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
says said:
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
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
says said:
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 :-)
 
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  • #12
I'm not sure how to get started with the transmission coefficient. There are so many substitutions and trig identities to try...
 
  • #13
At brewster's angle nisinθi = ntcosθi

2nicosθi / ntcosθi+nicosθt

= 2nicosθi/nisinθi+nicosθt

= 2cosθi/sinθi+cosθt

I'm kind of stuck here though...
 
  • #14
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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1. What are the Fresnel Equations?

The Fresnel Equations are a set of equations that describe how light reflects and refracts when it encounters a boundary between two different mediums. These equations were developed by French physicist Augustin-Jean Fresnel in the 19th century.

2. What is Snell's Law?

Snell's Law is a formula that relates the angles of incidence and refraction for a light ray passing through a boundary between two mediums with different refractive indices. It states that the ratio of the sines of the angles is equal to the ratio of the refractive indices of the two mediums.

3. How are the Fresnel Equations and Snell's Law related?

The Fresnel Equations are derived from Snell's Law and provide a more detailed description of how light is reflected and refracted at a boundary between two mediums. Snell's Law is a special case of the Fresnel Equations when the incident light is perpendicular to the boundary.

4. What factors affect the reflection and refraction of light at a boundary?

The reflection and refraction of light at a boundary can be affected by several factors, including the angle of incidence, the refractive indices of the two mediums, and the polarization of the incident light. The surface roughness of the boundary can also play a role in the amount of reflection.

5. How are the Fresnel Equations and Snell's Law used in real-world applications?

The Fresnel Equations and Snell's Law are used in many applications, including optics, photovoltaics, and telecommunications. They are also used in the design and manufacturing of lenses, prisms, and other optical components. These equations are essential for understanding how light behaves at boundaries and can help engineers and scientists optimize the performance of various devices and systems.

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