Question

A beam of parallel rays from a 32 MHz citizen's band radio transmitter passes between two electrically conducting (hence opaque to radio waves) buildings located 42 m apart. What is the angular width of the beam when it emerges from between the buildings?
°

Answer 1

The spacing between the two building is like a single slit and will cause the radio waves to be diffracted.

here the speed of the waves=3*10^8 m/s

frequency=32 MHz=32*10^6 Hz

then wavelength=speed/frequency=9.375 m

slit width=a=42 m

then formula for the diffraction:

a*sin(theta)=p*lambda

where p=order (1,2,3,... etc)

lambda=wavelength

taking p=1 ,

hence sin(theta)=lambda/a=0.2232

==>theta=12.8978 degrees

as angular width will include both the first order maxima at each side of the central maximum,

angular width of the beam=2*theta=25.7956 degrees