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 99.0 MHz, perfectly in phase with each other. The two antennas are arated by a distance d- 12.40 m. An observer, P, is Tocated on the x axis, a distance x 82.0 m 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? Submit Answer Units required. Tries 0/10 Previous Tries 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? As P gets closer A, the path length difference gets larger. What's the smallest path length difference that gives destructive interference? Submit Answer Incorrect. Tries 2/10 Previous Tries ihe derectoine in thle rndio signa. euer es desteactive intenerendey places along the x axis (Including the place you found in the previous problem) Submit Answer Tries 0/10

Explanation / Answer

a)

lambda=v/f

=3*10^8/99*10^6

=3.03 m

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

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

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

=sqrt(12.4^2+82^2)

x2=82.93 m

Putting in ..(i)

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

=(2*3.14/3.03)*(82.93-82)

=1.928 radian (ans)

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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)

=(12.4^2/3.03)-(3.303/4)

= 49.99 m (ans)

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C.

First finf the wavelength = c/f = 3.03 m

Path difference at P = X1 = 82.93-82= 0.93 m

Path difference at the point A = X2= 12.4 m

X1/Lambda = 0.93/3.03 = 0.307

X2/Lambda = 12.4/3.03 = 4.092

For minimas X/Lambda can take the values (0.5, 1.5, 2.5, 3.5 , so on)

so betweem 0.307 and 4.092 we can have three value namely (0.5, 1.5, 2.5,3.5)

So, 4 minimas will be there.

n= 4. (Ans)

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