An electromagnetic wave
Question:

An electromagnetic wave of frequency $3 \mathrm{GHz}$ enters a dielectric medium of relative electric permittivity $2.25$ from vacuum. The wavelength of this wave in that medium will be ___________$\times 10^{-2} \mathrm{~cm}$.

Solution:

$\lambda$ in vacuum $=\frac{\mathrm{c}}{\mathrm{f}}=\frac{3 \times 10^{8}}{3 \times 10^{9}}=0.1 \mathrm{~m}$

$\therefore \lambda$ in medium $=\frac{0.1}{\mu}$

Where refractive index

$\mu=\sqrt{\mu_{r} \varepsilon_{r}}$

Assuming non-magnetic material $\mu_{\mathrm{r}}=1$

$\therefore \quad \mu=\sqrt{2.25}=1.5$

$\lambda_{\mathrm{m}}=\frac{0.1}{1.5}=\frac{1}{15} \mathrm{~m}=6.67 \mathrm{~cm}$

$=667 \times 10^{-2} \mathrm{~cm}$

Ans. 667

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