Right answer is (a) \(θ_n=cos^{-1}([∓1±\frac{4n}{N}])\)
Explanation: The nulls of the N- element array is given by \(θ_n=cos^{-1}(\frac{λ}{2πd} [-β±\frac{2πn}{N}])\)
Since its given broad side array \(β=±kd=±\frac{2πd}{λ}=±\frac{π}{2},\)
\(θ_n=cos^{-1}(\frac{2}{π} [∓\frac{π}{2}±\frac{2πn}{N}])\)
\(θ_n=cos^{-1}([∓1±\frac{4n}{N}])\)