Question

Consider the special case of an interferometer, which can be described as a 2-element phased array antenna, consisting of two isotropic, in-phase, radiating elements separated by a distance d. Assume d is much greater than λ, the wavelength of the transmitted EM wave. Show that the first null off boresight in the far-field antenna pattern occurs at angle θ≈λ /2d radians.(See “The Double Slit” or “Young’s Experiment” for ideas).

Answer 1

For null or minimum intensity points, path difference of two waves upto point of interference,

delta x = (2n-1)$\lambda$ /2

where n is +ve integer, representing n th null point. Hence for first null point,

delta x = $\lambda$/2

S MALL a d