Question 9.37: A microwave antenna radiating at 10 GHz is to be protected f...

A microwave antenna radiating at 10 GHz is to be protected from the environment by a plastic shield of dielectric constant 2.5. What is the minimum thickness of this shielding that will allow perfect transmission (assuming normal incidence)? [Hint: Use Eq. 9.199.]

T1=14n1n3[(n1+n3)2+(n12n22)(n32n22)n22sin2(n2ωdc)]T^{-1}=\frac{1}{4 n_{1} n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}} \sin ^{2}\left(\frac{n_{2} \omega d}{c}\right)\right]                          (9.199)

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T=1sinkd=0kd=0,π,2πT=1 \Rightarrow \sin k d=0 \Rightarrow k d=0, \pi, 2 \pi \ldots The minimum (nonzero) thickness is  d=π/k. But k=ω/v=2πν/v=2πνn/c, and n=ϵμ/ϵ0μ0d=\pi / k . \text { But } k=\omega / v=2 \pi \nu / v=2 \pi \nu n / c, \text { and } n=\sqrt{\epsilon \mu / \epsilon_{0} \mu_{0}} (Eq. 9.69), where (presumably) μμ0. So n=ϵ/ϵ0=ϵr\mu \approx \mu_{0} . \text { So } n=\sqrt{\epsilon / \epsilon_{0}}=\sqrt{\epsilon_{r}} , and hence d=πc2πνϵr=c2νϵr=3×1082(10×109)2.5=9.49×103m, or 9.5mm.d=\frac{\pi c}{2 \pi \nu \sqrt{\epsilon_{r}}}=\frac{c}{2 \nu \sqrt{\epsilon_{r}}}=\frac{3 \times 10^{8}}{2\left(10 \times 10^{9}\right) \sqrt{2.5}}=9.49 \times 10^{-3} m , \text { or } 9.5 mm .

 

nϵμϵ0μ0n \equiv \sqrt{\frac{\epsilon \mu}{\epsilon_{0} \mu_{0}}}                              (9.69)

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