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Step-by-Step Solution
Step 1: Identify the given quantities
• Frequency of the wave, f = 3\,\text{GHz} = 3 \times 10^{9}\,\text{Hz} .
• Relative electric permittivity, \varepsilon_r = 2.25 .
• Speed of light in vacuum, c = 3 \times 10^{8}\,\text{m/s} .
Step 2: Calculate the wavelength in vacuum
The wavelength \lambda in vacuum is given by:
\lambda = \frac{c}{f} = \frac{3 \times 10^{8}\,\text{m/s}}{3 \times 10^{9}\,\text{Hz}} = 0.1\,\text{m}.
Therefore,
\lambda = 0.1\,\text{m} = 10\,\text{cm}.
Step 3: Determine the refractive index of the medium
For a non-magnetic dielectric, \mu_r = 1 . The refractive index n of the medium is:
n = \sqrt{\varepsilon_r \mu_r} = \sqrt{2.25 \times 1} = 1.5.
Step 4: Calculate the wavelength in the dielectric medium
The wavelength in the medium, \lambda_m , is:
\lambda_m = \frac{\lambda}{n} = \frac{10\,\text{cm}}{1.5} = \frac{10}{1.5}\,\text{cm} = 6.67\,\text{cm}.
Step 5: Express the final result in the requested form
We have \lambda_m = 6.67\,\text{cm} . If we wish to express this as \_\_\_\_ \times 10^{-2}\,\text{cm} , notice that:
6.67\,\text{cm} = 667 \times 10^{-2}\,\text{cm}.
