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(D) Let the electric field amplitude from each slit at point $P$ be $E_0$. The intensity $I = E_0^2$. Since the slits are equidistant from $S$,the wav

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(D) Let the electric field amplitude from each slit at point $P$ be $E_0$. The intensity $I = E_0^2$. Since the slits are equidistant from $S$,the waves reaching $S_1, S_2, S_3$ are in phase. The resultant electric field at $P$ is $E = E_0 + E_0 e^{i\phi_1} + E_0 e^{i\phi_2}$,where $\phi_1$ and $\phi_2$ are the phase differences relative to the wave from $S_1$. Phase difference $\phi = \frac{2\pi}{\lambda} 	imes \Delta x$. For $S_1$ and $S_2$: $\Delta x_1 = S_1P - S_2P = \lambda / 6$,so $\phi_1 = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3}$. For $S_1$ and $S_3$: $\Delta x_2 = S_1P - S_3P = 2\lambda / 3$,so $\phi_2 = \frac{2\pi}{\lambda} 	imes \frac{2\lambda}{3} = \frac{4\pi}{3}$. The resultant amplitude is $E = E_0 (1 + e^{i\pi/3} + e^{i4\pi/3})$. $E = E_0 (1 + (\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) + (\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}))$. $E = E_0 (1 + (\frac{1}{2} + i \frac{\sqrt{3}}{2}) + (-\frac{1}{2} - i \frac{\sqrt{3}}{2})) = E_0 (1 + 0 + 0) = E_0$. The resultant intensity $I_{res} = |E|^2 = E_0^2 = I$.

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