Question

Consider two antennas separated by 9.00m that radiate in phase at 120MHz, as described in Exercise . A receiver placed 150m from both antennas measures an intensity ${\mathit{I}}_{\mathbf{0}}$.The receiver is moved so that it is closer to one antenna than to the other. (a) What is the phase difference$\mathit{\varphi }$between the two radio waves produced by this path difference? (b) In terms of ${\mathit{I}}_{\mathbf{0}}$, what is the intensity measured by the receiver at its new position?

Short answer

1. The phase difference $\varphi$between the two radio waves produced by this path difference is 4.52rad .
2. The intensity measured by the receiver at its new position is 0.404${\mathit{I}}_{\mathbf{0}}$.

Step-by-step solution

formulas used to solve the question

Intensity is given by

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\left(\frac{\varphi }{2}\right)$ (1)

Phase difference is given by

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{∆}\mathbf{r}\mathbf{*}\mathbf{2}\mathbf{\pi }}{\mathbf{\lambda }}$ (2)

Calculate the phase difference

Given:

d = 9.0m

$\mathit{f}\mathbf{=}\mathbf{120}\mathit{M}\mathit{H}\mathit{z}\mathbf{=}\mathbf{120}\mathbf{*}{\mathbf{10}}^{\mathbf{6}}\mathit{H}\mathit{z}$

$$\begin{gathered} {\mathit{r}}_{\mathbf{1}}\mathbf{=}{\mathit{r}}_{\mathbf{2}}\mathbf{=}\mathbf{150}\mathit{m} \\ \mathbf{∆}{\mathit{r}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathit{m} \end{gathered}$$

The speed of electromagnetic waves is as same as the speed of light. So, the speed of wave emitted from both stations is given by

$\mathit{c}\mathbf{=}\mathit{\lambda }\mathit{f}\mathbf{\Rightarrow }\mathit{\lambda }\mathbf{=}\frac{\mathbf{c}}{\mathbf{f}}$ (3)

In equation (2), plug equation (3),

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{f}\mathbf{∆}\mathbf{r}}{\mathbf{c}}$ (4)

When the receiver is moved closer to one antenna than to the other,

Plug the given in equation (4),

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{*}\mathbf{120}\mathbf{*}{\mathbf{10}}^{\mathbf{6}}\mathbf{*}\mathbf{1}\mathbf{.}\mathbf{80}}{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{52}\mathit{r}\mathit{a}\mathit{d}$

Calculate the intensity

From equation (1), plug the given

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{4}\mathbf{.}\mathbf{52}}{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{404}{\mathit{I}}_{\mathbf{0}}$

Thus, the phase difference$\mathit{\varphi }$ between the two radio waves produced by this path difference is 4.52rad. The intensity measured by the receiver at its new position is $\mathbf{0}\mathbf{.}\mathbf{404}{\mathit{I}}_{\mathbf{0}}$.