m₁>m₂
The angular separation of fringes for λ₁ is greater than λ₂
β₂>β₁
From the central maximum, 3rd maximum of λ₂ overlaps with 5th minimum of λ₁
β=Dλ/d
λ₂>λ₁ ⇒ β₂>β₁
Also m₁β₁=m₂β₂ ⇒ m₁>m₂
Also 3(D/d)(600nm)=(2×5+1/2)(D/2d)400nm
Angular width θ=λ/d