(a) What is the wavelength of 100 MHz radio waves used in an MRI unit? (b) If the frequencies are swept over a $\mathbf{\pm }\mathbf{1}\mathbf{.}\mathbf{00}$range centered on 100MHz , what is the range of wavelengths broadcast?

Wavelength: The distance between identical points (adjacent crests) in adjacent cycles of a waveform signal propagated in space or along a wire is defined as the wavelength. This length is typically specified in wireless systems in metres (m), centimetres (cm), or millimetres (mm).

The radio waves have a frequency of $f=100\mathrm{MHz}\left(\frac{{10}^6\mathrm{Hz}}{1\mathrm{MHz}}\right)=1.00\times {10}^8\mathrm{Hz}$.

The speed of the light is$c=3.00\times {10}^8\mathrm{m}/\mathrm{s}$

(a)

The relationship between the wavelength and frequency can be expressed as,

$$\begin{gathered} f=\frac{c}{\lambda }..................\left(1\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{given}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{will}\mathrm{get}, \\ \mathrm{\lambda }=\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{1.00\times {10}^8\mathrm{Hz}} \\ =3.00\mathrm{m} \\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{wavelength}\mathrm{is}3.00\mathrm{m} \end{gathered}$$………………(1)

(b)

Considering the range, the lower bound of the frequency is,

$$\begin{gathered} f_1=\left(1.00\times {10}^8\right)-\left(\frac{1}{100}\right)\left(1.00\times {10}^8\right) \\ =9.9\times {10}^7\mathrm{Hz} \\ \mathrm{Therefore}\mathrm{using}\mathrm{the}\mathrm{wavelength}’\mathrm{s}\mathrm{higher}\mathrm{limit}\mathrm{using}\mathrm{equation}\left(1\right)\mathrm{is}, \\ {\mathrm{\lambda }}_1=\frac{\mathrm{c}}{{\mathrm{f}}_1} \\ =\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{9.9\times {10}^7\mathrm{Hz}} \\ =3.03\mathrm{m} \\ \mathrm{The}\mathrm{frequency}’\mathrm{s}\mathrm{higher}\mathrm{limit}\mathrm{is}, \\ {\mathrm{f}}_2=\left(1.00\times {10}^8\right)+\left(\frac{1}{100}\right)\left(1.00\times {10}^8\right) \\ =1.01\times {10}^8\mathrm{Hz} \\ \mathrm{Therefore}\mathrm{using}\mathrm{the}\mathrm{wavelength}’\mathrm{s}\mathrm{higher}\mathrm{limit}\mathrm{using}\mathrm{equation}\left(1\right)\mathrm{is}, \\ {\mathrm{\lambda }}_2=\frac{\mathrm{c}}{{\mathrm{f}}_2} \\ =\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{1.01\times {10}^8\mathrm{Hz}} \\ =2.97\mathrm{m} \\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{range}\mathrm{is}2.97\mathrm{to}3.03\mathrm{m}. \end{gathered}$$