Question

The first null beam width in the E-plane of a rectangular aperture of a×b is given by _______________

(a) 2sin^-1⁡(λ/b)

(b) sin^-1⁡(λ/a)

(c) 2sec^-1⁡(λ/b)

(d) 2cos^-1⁡(λ/a)

Answer 1

Correct answer is (a) 2sin^-1⁡(λ/b)

To explain I would say: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. \(\frac{kb}{2}\) sinθ=nπ

θ= sin^-1\(⁡(\frac{n\lambda}{b})\)

Therefore, the FNBW in E-plane is given by FNBW=2 θ = sin^-1(⁡\(\frac{\lambda}{b}\)).