Question

owo identical radio antennas are moved to be only 10m au broadcast frequency is apart 2.) Suppose the two identical radio antennas a poinit mewe ncreased to f-60MHz. At a distance of 700m from thrt an between the antennas and in the direction 0-0, the intensity is lo-0.02W/m2. At this same distance (a) the intensity in the direction 04 the direction near 0-O for which the intensity is (c) the direction in which the intensity is zero.

Answer 1

Because the 700m distance from antennas to the point where intensity is measured in much greater than the distance b/w antennas (d = 10m) the amplitudes of the waves from the two antennas are very nearly equal. C = f lambda lambda = c/f = 3 times 10^2m/s/60 times 10^6 s^-1 = 5.00m The spacing d = 10m b/w the antennas is just twice the wavelength I - I_0 cos^2 (pi d sin/lambda sin theta) = (0.002 w/m^2)(cos^2(pi times 10m times sin theta/5.00m) = (0.020)cos^2 ((2.00 pi rad) times sin theta) (a) when theta = 4.0 degree I = 0.020 cos^2(2.00pi rad)sin 4.0) = 0.016w/m^2 \r\n This is a about theta 2% of the intensity at theta = 0 degree (b) I equals to I_0/2, I = I_0 cos^2(2.00 pi sin theta) I_0/2 = I_0 cos^2(2.00 pi sin theta) plusminus 1/squareroot 2 = cos(2.00 pi sin theta) 2.00 pi sin theta = plusminus pi/4 sin theta = plusminus 1/theta = plusminus 0.125