Question
<p>Consider two antennas radiating 7.7 MHz radio waves in phase with each other. They are located at points S1 and S2, separated by a distance d = 175 m, Fig. 24-61. What are the farthest two points on the positive y axis where there will be destructive interference? (Destructive interference in this case means that a radio receiver placed at that point would pick up no signal).<br />farthest y ?? m</p>
<p>Second farthest y ?? m</p>
<p> </p>
<p> </p>
<p> </p>
<p>S1 at origin and s2 at some point to the right of S1 on the x-axis</p>
Explanation / Answer
distance d = 175m ------------------------------------------------------------ wavelength = c / f = (3*108 m/s)/( 6.0*106Hz) = 38.96 m Let us take P be the point on the y-axis. At this point destructive interference occures. the path difference is = S2 P - S1 P = ( y2+ d2) - y condition for disturctive inerfernce is = ( m + 1/2) ( m + 1/2) = ( y2+ d2) - y ( m + 1/2) + y= ( y2+ d2) squaring on both sides [( m + 1/2) ]2 + y2 + 2y( m + 1/2) = (y2+ d2) [( m + 1/2) ]2 + 2y ( m + 1/2)= d2 y = [d2-[( m + 1/2) ]2]/ [2( m + 1/2) ] For m=0 y = [d2-[( 1/2) ]2]/ [2( 1/2) ] = [(175 m)2-[( 0+ 1/2)(38.96 m) ]2]/ [2( 0+ 1/2) (38.96 m)] = 776.32 m For m=1 y = [d2-[( 1+ 1/2) ]2]/ [2(1+ 1/2) ] = [(175 m)2-[( 1+ 1/2)(38.96 m) ]2]/ [2( 1+ 1/2) (38.96 m)] = [(175 m)2-[( 0+ 1/2)(38.96 m) ]2]/ [2( 1+ 1/2) (38.96 m)] = 232.800m For m =2 y = [d2-[( 2+ 1/2) ]2]/ [2(2+ 1/2) ] = [(175 m)2-[( 2+ 1/2)(38.96 m) ]2]/ [2( 2+ 1/2) (38.96 m)] = 108.5 m For m =2 y = [d2-[( 2+ 1/2) ]2]/ [2(2+ 1/2) ] = [(175 m)2-[( 2+ 1/2)(38.96 m) ]2]/ [2( 2+ 1/2) (38.96 m)] = 108.5 m = [(175 m)2-[( 2+ 1/2)(38.96 m) ]2]/ [2( 2+ 1/2) (38.96 m)] = 108.5 m