Answer :
Answer:
a
[tex]\Phi =4.524 \ rad[/tex]
b
[tex]I = 0.40637 I_o [/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 9.0 \ m[/tex]
The frequency is [tex]f = 120 \ MHz = 120 *10^{6} \ Hz[/tex]
The distance of the receiver from the antennas is [tex]D = 150 \ m[/tex]
The intensity measured is [tex]I = I_o[/tex]
The change in position of the receiver is by [tex]\Delta D = 1.8 \ m[/tex]
Gnerally the phase difference is mathematically represented as
[tex]\Phi = \frac{ 2 \pi * \Delta D}{\lambda}[/tex]
Here [tex]\lambda[/tex] is the wavelength which is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
=> [tex]\lambda = \frac{3.0 *10^{8}}{ 120 *10^{6} }[/tex]
=> [tex]\lambda = \frac{3.0 *10^{8}}{ 120 *10^{6} }[/tex]
=> [tex]\lambda = 2.5 \ m[/tex]
So
[tex]\Phi = \frac{ 2* 3.142 * 1.8 }{2.5}[/tex]
[tex]\Phi = \frac{ 2* 3.142 * 1.8 }{2.5}[/tex]
[tex]\Phi =4.524 \ rad[/tex]
Generally the intensity measured by the receiver is
[tex]I = I_o cos^2 [\frac{\Phi}{2} ][/tex]
=> [tex]I = I_o [cos [\frac{4.524}{2} ]]^2[/tex]
=> [tex]I = I_o [cos [ 2.262]]^2[/tex]
=> [tex]I = 0.40637 I_o [/tex]