Question

A microwave antenna radiating at $\mathbf{10}\mathbf{\hspace{0.33em}}\mathit{G}\mathit{H}\mathit{z}$is to be protected from the environment by a plastic shield of dielectric constant$\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{.}$ . What is the minimum thickness of this shielding that will allow perfect transmission (assuming normal incidence)? [Hint: Use Eq. 9.199.]

Short answer

The minimum thickness of the shielding$9.5mm.$

Step-by-step solution

Expression for the minimum thickness of the shielding:

Write the expression for minimum thickness

$\mathit{d}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{k}}$…… (1)

Determine the minimum thickness of the shielding:

Substitute $k=\frac{\omega }{v}$and $\omega =2\pi v$in equation (1).

$d=\frac{\pi }{\left(\frac{\omega }{v}\right)}$

$d=\frac{\pi }{\left(\frac{2\pi v}{v}\right)}$

$d=\frac{v}{2v}$

Substitute $v=\frac{c}{n}$and $n=\sqrt{\frac{\epsilon \mu }{{\epsilon }_0{\mu }_0}}$in the above expression.

$d=\frac{\left(\frac{c}{n}\right)}{2v}$

$d=\frac{c}{2v\sqrt{\frac{\epsilon \mu }{{\epsilon }_0{\mu }_0}}}$…… (2)

Here, $\mu \approx {\mu }_0\text{and}\sqrt{\frac{\epsilon }{{\epsilon }_0}}=\sqrt{{\epsilon }_r}\text{.}$

Substitute $\mu \approx {\mu }_0$and $\sqrt{\frac{\epsilon }{{\epsilon }_0}}=\sqrt{{\epsilon }_r}$in equation (2).

$d=\frac{c}{2\nu \sqrt{{\epsilon }_r}}$

Substitutelocalid="1657471538175" $=3\times {10}^8\hspace{0.33em}m/s,=10GHz$, and ${\epsilon }_r=2.5$in equation (3).

$d=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{2\left(10\mathrm{GHz}\times \frac{{10}^9\mathrm{Hz}}{1\mathrm{GHz}}\right)\sqrt{2.5}}$

localid="1657471850034" $$\begin{gathered} d=9.49\times {10}^{-3}\mathrm{m}\times \left(\frac{{10}^3\mathrm{mm}}{1\mathrm{m}}\right) \\ d=9.5\mathrm{mm} \end{gathered}$$

Therefore, the minimum thickness of the shielding is $9.5mm$.