Question

Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150 m from both antennas measures an intensity I0. The receiver is moved so that it is 1.8 m closer to one antenna than to the other. (a) What is the phase difference f between the two radio waves produced by this path difference? (b) In terms of I0, what is the intensity measured by the receiver at its new position?

Answer 1

Answer:

a

  $\Phi =4.524 \ rad$

b

   $I = 0.40637 I_o$

Explanation:

From the question we are told that

  The distance of separation is  $d = 9.0 \ m$

  The frequency is $f = 120 \ MHz = 120 *10^{6} \ Hz$

   The distance of the receiver from the antennas is $D = 150 \ m$

   The intensity measured is  $I = I_o$

    The change in position of the receiver is by  $\Delta D = 1.8 \ m$

Gnerally the phase difference is mathematically represented as

       $\Phi = \frac{ 2 \pi * \Delta D}{\lambda}$

Here $\lambda$  is the wavelength which is mathematically represented as

       $\lambda = \frac{c}{f}$

Here c is the speed of light with value  $c = 3.0 *10^{8} \ m/s$

=>     $\lambda = \frac{3.0 *10^{8}}{ 120 *10^{6} }$

=>     $\lambda = \frac{3.0 *10^{8}}{ 120 *10^{6} }$

=>     $\lambda = 2.5 \ m$

So

 $\Phi = \frac{ 2* 3.142 * 1.8 }{2.5}$

  $\Phi = \frac{ 2* 3.142 * 1.8 }{2.5}$

  $\Phi =4.524 \ rad$

Generally the intensity measured by the receiver is  

     $I = I_o cos^2 [\frac{\Phi}{2} ]$

=>   $I = I_o [cos [\frac{4.524}{2} ]]^2$

=>  $I = I_o [cos [ 2.262]]^2$

=>  $I = 0.40637 I_o$