Set 56 Problem number 21

Problem

Radio waves of wavelength 2.84 meters are emitted in phase from two antennae separated by 12.7 meters.   They are observed at a distance much greater than the separation.  A line from the antennae to the point of observation makes an angle of 31 degrees with the perpendicular bisector of the line segment joining the antennae. 

• By what fraction of a wavelength will the wave emitted from the antenna further from the observer lag the wave emitted from the closer antennaf?
• What are the first three angles at which the strength of the signal reaching the observer would be minimized?
• What are the first two angles at which the signal reaching the observer would be maximized?

Solution

At angle 31 degrees, the wave from the further of the sources will travel a distance

• path difference = 12.7 meters * sin( 31 degrees) = 6.540997 meters

further than the wave from the closer.  This path difference is

• number of wavelengths = path diff / wavelength = 6.540997 meters / 12.7 meter = `nWavelengths

wavelengths.  Constructive interference occurs when the number of wavelengths is a whole number; destructive interference occurs when the number of wavelengths is a whole number of wavelengths plus 1/2 wavelength.

An interference minimum will occur at any angle such that number of wavelengths is a whole number of wavelengths plus 1/2 wavelength.  The first will occur when the path difference is 1/2 wavelength, the second when the path difference is 1 + 1/2 wavelength = 3/2 wavelength and the third at path difference 5/2 wavelength.

• The path difference will be 1/2 wavelength at angle `theta1, such that 12.7 meters * sin(`theta1) = 1/2 ( 2.84 meters).  Solving for `theta1 we get

sin(`theta1) = 1/2 ( 2.84 meters) / ( 12.7 meters) so

`theta1 = arcsin( 1/2 ( 2.84 meters) / ( 12.7 meters) ) = 6.423 degrees.

• The path difference will be 3/2 wavelength at angle `theta2, such that 12.7 meters * sin(`theta2) = 3/2 ( 2.84 meters).  Solving for `theta2 we get

sin(`theta2) = 3/2 ( 2.84 meters) / ( 12.7 meters) so

`theta2 = arcsin( 3/2 ( 2.84 meters) / ( 12.7 meters) ) = 19.6 degrees.

• The path difference will be 5/2 wavelength at angle `theta3, such that 12.7 meters * sin(`theta3) = 5/2 ( 2.84 meters).  Solving for `theta3 we get

sin(`theta3) = 5/2 ( 2.84 meters) / ( 12.7 meters) so

`theta3 = arcsin( 5/2 ( 2.84 meters) / ( 12.7 meters) ) = 34 degrees.

Similarly an interference maximum will occur whenever the path difference is 0, 1 or 2 wavelengths.  Thus we see that

• The path difference will be 1 wavelength at angle `theta4, such that 12.7 meters * sin(`theta4) = 1 * ( 2.84 meters).  Solving for `theta4 we get

sin(`theta4) = 1 ( 2.84 meters) / ( 12.7 meters) so

`theta4 = arcsin( 1 ( 2.84 meters) / ( 12.7 meters) ) = 12.92 degrees.

• The path difference will be 2 wavelengths at angle `theta5, such that 12.7 meters * sin(`theta5) = 2 ( 2.84 meters).  Solving for `theta5 we get

sin(`theta5) = 2 ( 2.84 meters) / ( 12.7 meters) so

`theta5 = arcsin( 2 ( 2.84 meters) / ( 12.7 meters) ) = 26.58 degrees.