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Hi! I am studying some of my brothers physics questions and I need help finding the answer to this one:

You need to use your cell phone, which broadcasts an 800 MHz signal, but you’re behind two massive, radio-wave-absorbing buildings that have a 20 m gap between them. What is the width of the electromagnetic “beam” that emerges 50 m on the other side of the buildings?

Thanks!

Answer
When a wave passes through a narrow slit the angle at which the first node occurs can be predicted from:
N*lambda=d*sin(theta)
Where: N=1 for the first node, lambda=wavelength of the wave, d=width of the slit and theta=angle at which the signal node emerges.
In this case the wavelength lambda is equal to the speed of the wave divided by the frequency of the signal:
lambda=speed/frequency=3.0x10^8m/s/8.0x10^8/s=0.375m
Therefore the angle theta of the emerging beam that corresponds to the first node will be:
N*lambda=d*sin(theta)
1*0.375=20*sin(theta)
theta=invsin(0.375/20)a=1.07deg
At a distance of L=50 m from the gap the beam will have expanded to a width W of:
tangent(theta)=x/L
Therefore, x, the distance between the center of the beam and the location of the first antinode will be:
x=L*tan(theta)=50m*tan(1.07deg)=0.93m
Finally, the total width X of the beam at a distance L=50m from the gap will be twice this distance x:
X=2*x=2*0.93=1.87m

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James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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