Question

A microwave antenna radiating at 10GHz 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.]

Short answer

The minimum thickness of the shielding is 9.5 mm.

Step-by-step solution

Expression for the minimum thicknesof the shielding:

Write the expression for minimum thickness.

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

Determine the minimum thickness of the shielding

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

$$\begin{gathered} \mathrm{d}=\frac{\mathrm{\pi }}{\left({\displaystyle \frac{\mathrm{\omega }}{\mathrm{\nu }}}\right)} \\ \mathrm{d}=\frac{\mathrm{\pi }}{\left({\displaystyle \frac{2\mathrm{\pi \nu }}{\mathrm{v}}}\right)} \\ \mathrm{d}=\frac{\mathrm{v}}{2\mathrm{\nu }} \end{gathered}$$

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

$$\begin{gathered} \mathrm{d}=\frac{\left({\displaystyle \frac{\mathrm{c}}{\mathrm{n}}}\right)}{2\mathrm{\nu }} \\ \mathrm{d}=\frac{\mathrm{c}}{2\mathrm{\nu }\sqrt{{\displaystyle \frac{\mathrm{\epsilon \mu }}{{\mathrm{\epsilon }}_0{\mathrm{\mu }}_0}}}} \end{gathered}$$ ....... (2)

Here, $\mathrm{\mu }\approx {\mathrm{\mu }}_0$and $\sqrt{\frac{\mathrm{\epsilon }}{{\mathrm{\epsilon }}_0}}=\sqrt{{\mathrm{\epsilon }}_{\mathrm{T}}}$.

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

$\mathrm{d}=\frac{\mathrm{c}}{2\mathrm{\nu }\sqrt{{\mathrm{\epsilon }}_{\mathrm{T}}}}$........(3)

Substitute $\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$,$\mathrm{\nu }=10\mathrm{GHz}$and${\mathrm{\epsilon }}_{\mathrm{T}}=2.5$ in equation (3).

$$\begin{gathered} \mathrm{d}=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{2\times \left(10\mathrm{GHZ}\times {\displaystyle \frac{{10}^9\mathrm{Hz}}{1\mathrm{GHz}}}\right)\sqrt{2.5}} \\ \mathrm{d}=9.49\times {10}^{-3}\mathrm{m}\times \left(\frac{{10}^3\mathrm{mm}}{1\mathrm{m}}\right) \\ \mathrm{d}=9.5\mathrm{mm} \end{gathered}$$

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