Overlapping diffraction maxima imply same d sinθ for integer orders m of each wavelength

Steps:

• Use grating equation d sinθ = mλ; for overlap at θ=31°, m₁λ₁ = m₂λ₂ with λ₁=420 nm, λ₂=630 nm.
• Solve m₁(420) = m₂(630); simplify to 2m₁ = 3m₂; smallest integers m₁=3, m₂=2.
• Compute d sinθ = 3(420) = 1260 nm.
• d = 1260 / sin(31°) ≈ 1260 / 0.515 ≈ 2447 nm = 2.4 μm.

Why C is correct:

• Matches d=2.4 μm from grating equation for m=3 (420 nm) and m=2 (630 nm) at θ=31°.

Why the others are wrong:

• A: 1.2 μm gives sinθ ≈1.05 (>1, impossible).
• B: 1.6 μm corresponds to first-order overlap, not at 31°.
• D: 3.7 μm fits higher orders but not smallest integers matching 31°.

Final answer: C