In a Young's double slit experiment, the maximum intensity of light...

Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \dfrac{2\pi}{\lambda} \Delta x$.

The intensity $I$ at a point on the screen is given by $I = I_0 \cos^2\left(\dfrac{\phi}{2}\right)$.

For (A) $\Delta x = \dfrac{\lambda}{6}$, $\phi = \dfrac{\pi}{3}$. $I = I_0 \cos^2\left(\dfrac{\pi}{6}\right) = I_0 \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_0}{4}$. This matches (IV).

For (B) $\Delta x = \dfrac{\lambda}{4}$, $\phi = \dfrac{\pi}{2}$. $I = I_0 \cos^2\left(\dfrac{\pi}{4}\right) = I_0 \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{2}$. This matches (III).

For (C) $\Delta x = \dfrac{\lambda}{3}$, $\phi = \dfrac{2\pi}{3}$. $I = I_0 \cos^2\left(\dfrac{\pi}{3}\right) = I_0 \left(\dfrac{1}{2}\right)^2 = \dfrac{I_0}{4}$. This matches (II).

For (D) $\Delta x = \dfrac{\lambda}{2}$, $\phi = \pi$. $I = I_0 \cos^2\left(\dfrac{\pi}{2}\right) = 0$. This matches (I).

Therefore, the correct matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).

Answer: (A)-(IV), (B)-(III), (C)-(II), (D)-(I)