) = I_0 \left[ 1 + \cos (\Delta \phi) \right] $

where $I_0$ is the intensity due to each slit, and $\Delta \phi$ is the phase difference.

Equivalently, one can write:

$ I(\Delta \phi) = 4I_1 \cos^2\!\left(\frac{\Delta \phi}{2}\right) $

where $I_1$ = intensity from each slit.

Phase difference:

$ \Delta \phi = \frac{2\pi}{\lambda} \cdot (\text{path difference}) $

At path difference = $\lambda$:

$ \Delta \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi $

So intensity at this point is maximum:

$ I_{\max} = 4I_1 $

They have denoted this maximum intensity value as $I$.

So, $I = 4I_1$.

At path difference = $\lambda/6$:

$ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} $

Now intensity is:

$ I' = 4I_1 \cos^2\!\left(\frac{\Delta \phi}{2}\right) = 4I_1 \cos^2\!\left(\frac{\pi}{6}\right) $

$ = 4I_1 \left(\frac{\sqrt{3}}{2}\right)^2 = 4I_1 \cdot \frac{3}{4} = 3I_1. $

Since $I = 4I_1$,

$ I' = \frac{3}{4}I $

The intensity at path difference $\lambda/6$ is:

$ \boxed{\tfrac{3I}{4}} $

Correct Option: B

Wave Optics

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