Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Chapter 10, Problem 73P
To determine

Prove the given reflection and transmission coefficients for oblique incidence.

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Explanation of Solution

Calculation:

Consider the expression of reflection coefficient in the parallel polarization of oblique incidence.

Γ=η2cos(θt)η1cos(θi)η2cos(θt)+η1cos(θi)        (1)

Here,

η1 is the intrinsic impedance of the medium-1,

η2 is the intrinsic impedance for medium-2,

θt is the angle of transmission, and

θi is the angle of incidence.

Consider the expression for intrinsic impedance for medium-1 (nonmagnetic medium).

η1=μoεoεr1

Consider the expression for intrinsic impedance for medium-2 (nonmagnetic medium).

η2=μoεoεr2

Substitute μoεoεr1 for η1 and μoεoεr2 for η2 in Equation (1).

Γ=μoεoεr2cos(θt)μoεoεr1cos(θi)μoεoεr2cos(θt)+μoεoεr1cos(θi)=1εr2cos(θt)1εr1cos(θi)1εr2cos(θt)+1εr1cos(θi)=cos(θt)εr2εr1cos(θi)cos(θt)+εr2εr1cos(θi)=cos(θt)sin(θi)sin(θt)cos(θi)cos(θt)+sin(θi)sin(θt)cos(θi) {εr2εr1=sin(θi)sin(θt)}

Simplify the expression.

Γ=sin(θt)cos(θt)sin(θi)cos(θi)sin(θt)cos(θt)+sin(θi)cos(θi)=2sin(θt)cos(θt)2sin(θi)cos(θi)2sin(θt)cos(θt)+2sin(θi)cos(θi)=sin(2θt)sin(2θi)sin(2θt)+sin(2θi)=sin(θtθi)cos(θt+θi)sin(θt+θi)cos(θtθi)

Simplify the expression.

Γ=[sin(θtθi)cos(θtθi)]{1[sin(θt+θi)cos(θt+θi)]}=tan(θtθi)tan(θt+θi)

Consider the expression of transmission coefficient in the parallel polarization of oblique incidence.

τ=2η2cos(θi)η2cos(θt)+η1cos(θi)

Substitute μoεoεr1 for η1 and μoεoεr2 for η2.

τ=2μoεoεr2cos(θi)μoεoεr2cos(θt)+μoεoεr1cos(θi)=21εr2cos(θi)1εr2cos(θt)+1εr1cos(θi)=2cos(θi)cos(θt)+εr2εr1cos(θi)=2cos(θi)cos(θt)+sin(θi)sin(θt)cos(θi)

Simplify the expression.

τ=2sin(θt)cos(θi)sin(θt)cos(θt)[sin2(θi)+cos2(θi)]+sin(θi)cos(θi)[sin2(θt)+cos2(θt)]=2sin(θt)cos(θi)[sin(θt)cos(θi)+sin(θi)cos(θt)][sin(θi)sin(θt)+cos(θi)cos(θt)]=2cos(θi)sin(θt)sin(θt+θi)cos(θtθi)

Consider the expression of reflection coefficient in the perpendicular polarization of oblique incidence.

Γ=η2cos(θi)η1cos(θt)η2cos(θi)+η1cos(θt)

Substitute μoεoεr1 for η1 and μoεoεr2 for η2.

Γ=μoεoεr2cos(θi)μoεoεr1cos(θt)μoεoεr2cos(θi)+μoεoεr1cos(θt)=1εr2cos(θi)1εr1cos(θt)1εr2cos(θi)+1εr1cos(θt)=cos(θi)εr2εr1cos(θt)cos(θi)+εr2εr1cos(θt)=cos(θi)sin(θi)sin(θt)cos(θt)cos(θi)+sin(θi)sin(θt)cos(θt) {εr2εr1=sin(θi)sin(θt)}

Simplify the expression.

Γ=sin(θt)cos(θi)sin(θi)cos(θt)sin(θt)cos(θi)+sin(θi)cos(θt)=sin(θtθi)sin(θt+θi)

Consider the expression of transmission coefficient in the perpendicular polarization of oblique incidence.

τ=2η2cos(θi)η2cos(θi)+η1cos(θt)

Substitute μoεoεr1 for η1 and μoεoεr2 for η2.

τ=2μoεoεr2cos(θi)μoεoεr2cos(θi)+μoεoεr1cos(θt)=21εr2cos(θi)1εr2cos(θi)+1εr1cos(θt)=2cos(θi)cos(θi)+εr2εr1cos(θt)=2cos(θi)cos(θi)+sin(θi)sin(θt)cos(θt)

Simplify the expression.

τ=2cos(θi)sin(θt)sin(θt)cos(θi)+sin(θi)cos(θt)=2cos(θi)sin(θt)sin(θt+θi)

Conclusion:

Hence, the given reflection and transmission coefficients for oblique incidence are proved.

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Chapter 10 Solutions

Elements Of Electromagnetics

Ch. 10.10 - Prob. 11PECh. 10.10 - Prob. 12PECh. 10.11 - Prob. 13PECh. 10 - Prob. 1RQCh. 10 - Prob. 2RQCh. 10 - Prob. 3RQCh. 10 - Prob. 4RQCh. 10 - Prob. 5RQCh. 10 - Prob. 6RQCh. 10 - Prob. 7RQCh. 10 - Prob. 8RQCh. 10 - Prob. 9RQCh. 10 - Prob. 10RQCh. 10 - Prob. 1PCh. 10 - Prob. 2PCh. 10 - Prob. 3PCh. 10 - Prob. 4PCh. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 7PCh. 10 - Prob. 8PCh. 10 - Prob. 9PCh. 10 - Prob. 10PCh. 10 - Prob. 11PCh. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - Prob. 15PCh. 10 - Prob. 16PCh. 10 - Prob. 17PCh. 10 - Prob. 18PCh. 10 - Prob. 19PCh. 10 - Prob. 20PCh. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - Prob. 25PCh. 10 - Prob. 26PCh. 10 - Prob. 27PCh. 10 - Prob. 28PCh. 10 - Prob. 29PCh. 10 - Prob. 30PCh. 10 - Prob. 31PCh. 10 - Prob. 32PCh. 10 - Prob. 33PCh. 10 - Prob. 34PCh. 10 - Prob. 35PCh. 10 - Prob. 36PCh. 10 - Prob. 37PCh. 10 - Prob. 38PCh. 10 - Prob. 39PCh. 10 - Prob. 40PCh. 10 - Prob. 41PCh. 10 - Prob. 42PCh. 10 - Prob. 43PCh. 10 - Prob. 44PCh. 10 - Prob. 45PCh. 10 - Prob. 46PCh. 10 - Prob. 47PCh. 10 - Prob. 48PCh. 10 - Prob. 49PCh. 10 - Prob. 50PCh. 10 - Prob. 51PCh. 10 - Prob. 52PCh. 10 - Prob. 53PCh. 10 - Prob. 54PCh. 10 - Prob. 55PCh. 10 - Prob. 56PCh. 10 - Prob. 57PCh. 10 - Prob. 58PCh. 10 - Prob. 59PCh. 10 - Prob. 60PCh. 10 - Prob. 61PCh. 10 - Prob. 62PCh. 10 - Prob. 63PCh. 10 - Prob. 64PCh. 10 - Prob. 65PCh. 10 - Prob. 66PCh. 10 - Prob. 67PCh. 10 - Prob. 68PCh. 10 - Prob. 69PCh. 10 - Prob. 70PCh. 10 - Prob. 71PCh. 10 - Prob. 72PCh. 10 - Prob. 73PCh. 10 - Prob. 74PCh. 10 - Prob. 75PCh. 10 - Prob. 76PCh. 10 - Prob. 78PCh. 10 - Prob. 79PCh. 10 - Prob. 80PCh. 10 - Prob. 81PCh. 10 - Prob. 82PCh. 10 - Prob. 83PCh. 10 - Prob. 84P
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