Correct Answer - (a) maximum in a direction perpendicular to the array, (b) net signal to `EW` direction is zero.
The antennas radiate summetrically in all direction and in phase with each other hence radiated wave from the antennas can suffers constructive or destructice interference
(a) The radiated signal will be a maximum in a direction perpendicular to the array since in this direction the waves are all in phase with each other.
(b) If `f` is the frequency and `C` is the velocity of radio waves, then their wavelength `lambda` si given by
`lambda = (c)/(f) = (3 xx 10^(8))/(1.5 xx 10^(6))` or `lambda = 200 m`. Note that `c =` speed of light since the waves from part of `EM` spectrum
Now the spacing between the antenns in `EW` direction is `50 m = lambda//4`. Hence, in the `EW` direction, antennas `1` and `3` are are out of phase with each other by `lambda//2` and so cancel each other by destructive interference.
similarly `2` and `4` cancel each other.
Thus the net signal in `EW` direction is zero.