dsinθ=nλd \sin \theta=n \lambdadsinθ=nλ
dsinθ=(2n+1)λ2d \sin \theta=(2 n+1) \frac{\lambda}{2}dsinθ=(2n+1)2λ
dsinθ=(n+1)λd \sin \theta=(n+1) \lambdadsinθ=(n+1)λ
dsinθ=(n+1)λ2d \sin \theta=(n+1) \frac{\lambda}{2}dsinθ=(n+1)2λ