Two antennas located at points A and B are broadcasting radio waves of frequency

Question

Two antennas located at points A and B are broadcasting radio waves of frequency 96.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d=9.30m. An observer, P, is located on the x axis, a distance x=57.0m from antenna A, so that APB forms a right triangle with PB as hypotenuse. What is the phase difference between the waves arriving at P from antennas A and B?

Now observer P walks along the x axis toward antenna A. What is P's distance from A when he first observes fully destructive interference between the two waves?

Explanation / Answer

a)

lambda=C/f

=3*10^8/96*10^6

=3.125 m

phase difference =(2pi/lambda)*path difference

=(2pi/lambda)*(x2-x1)

here x2=sqrt(d^2+x1^2)

=sqrt(9.3^2+57^2)

=sqrt(86.49+

x2=57.75 m

now

phase difference=(2pi/lambda)*(x2-x1)

=(2*3.14/3.125)*(57.75-57)

=1.5072 rad...........

b)

to get destructive interference ,

path difference =lambda/2

x2-x1=lambda/2

and

x2=sqrt(d^2+x1^2)

now

sqrt(d^2+x1^2)-x1=lambda/2

(d^2+x1^2)=(x1+lambda/2)^2

d^2=(x1*lambda)+(lambda^2/4)

====>

x1=(d^2/lambda)-(lambda/4)

=(9.3^2/3.125)-(3.125/4)

=26.89 m.....is answer

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