(2n−1)λ4(2 n-1) \dfrac{\lambda}{4}(2n−1)4λ
(2n−1)λ2(2 n-1) \dfrac{\lambda}{2}(2n−1)2λ
nλn \lambda nλ
(2n+1)λ2(2 n+1) \dfrac{\lambda}{2}(2n+1)2λ