Constructive interference quadruples intensity at the center Steps:

• At the center, path lengths from both slits are equal, causing constructive interference.
• Each slit produces wave amplitude $a$, so combined amplitude is $2a$.
• Intensity $I \propto (2a)^2 = 4a^2$, defining the maximum as $I$.
• Covering one slit leaves amplitude $a$, so intensity $\propto a^2 = I/4$.

Why A is correct:

• Coherent waves add amplitudes at constructive maxima, yielding I_\max = 4I_\single per the superposition principle.

Why the others are wrong:

• B: Assumes incoherent intensity addition ($I_1 + I_2 = I$), but sources are coherent.
• C: Matches no standard interference or diffraction result here.
• D: Ignores that interference boosts intensity beyond single-slit value.

Final answer: A