Calculate the reflection coefficient for light at an air-to-silver interface (μ1=μ2=μ0,ε=ε0,σ=6×107(Ωm)-1)at optical frequencies(ω=4×1015/s).

Short Answer

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The reflection coefficient for light at an air air-to-silver interface is93% .

Step by step solution

01

Expression for the reflection coefficient for light to conducting surface: 

Write the expression for the reflection coefficient for light to conducting surface.

R=|E~0RE~0I|2=|1-β~1+β~|2 …… (1)

Here,β is the complex quantity which is given as:

β~=μ1v1μ2ωk~2β~=μ1v1μ2ω(k2+iK2) …… (2)

02

Determine the value ofk2 and k2:

Write the value of the wave numberk2 .

k2=ωε2μ22[1+(σε2ω)2+1]12

Write the value of the wave numberk2 .

K2=ωε2μ22[1+(σε2ω)21]12

As silver is a good conductor thenσ>>ε2ω, .

The value ofk2 and k2will be,

k2=ωε2μ22(σε2ω)K2=ωε2μ22(σε2ω)

Hence, the value ofk2 and k2becomes,

k2=K2=ωμ2σ2

03

Determine the reflection coefficient for light to an air to silver interface: 

Substitute k2=K2=ωμ2σ2in equation (2).

β~=μ1v1μ2ωσμ2ω2(1+i)β~=μ1v1σ2μ2ω(1+i) …… (3)

Here,

γ=μ1v1σ2μ2ωγ=μ0cσ2μ0ω

Substitute γ=μ1v1σ2μ2ωand γ=μ0cσ2μ0ωin equation (3).

β~=μ0cσ2μ0ω(1+i)β~=cσμ02ω(1+i)

Substitutec=3×108 m/s ,σ=6×107(Ωm)1 , μ0=4π×107 H/mand ω=4×1015 s1in the above expression.

β~=(3×108 m/s)(6×107(Ωm)1)(4π×107 H/m)2(4×1015 s1)(1+i)β~=29(1+i)

Substitute iβ~=29(1+i)n equation (1).

R=|129(1+i)1+29(1+i)|2R=(129)2+(29)2(1+29)2+(29)2R=0.93

Therefore, the reflection coefficient for light at an air air-to-silver interface is93% .

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