asinθ=(2n+1)λ2a \sin \theta=(2 n+1) \dfrac{\lambda}{2}asinθ=(2n+1)2λ
a2sinθ=(2n+1)λ4\dfrac{a}{2} \sin \theta=(2 n+1) \dfrac{\lambda}{4}2asinθ=(2n+1)4λ
asinθ=2nλ2a \sin \theta=\dfrac{2 n \lambda}{2}asinθ=22nλ