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(C) In a $YDSE$ experiment,the intensity at any point on the screen is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference

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(C) In a $YDSE$ experiment,the intensity at any point on the screen is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference between the two waves.At the mid-point of the screen,the path difference due to the geometry is zero.When a slab of thickness $t$ and refractive index $\mu$ is placed in front of one slit,an additional path difference $\Delta x = (\mu - 1)t$ is introduced.The corresponding phase difference is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} (\mu - 1)t$.The intensity at the mid-point is $I = 4I_0 \cos^2(\frac{\pi t}{\lambda}(\mu - 1))$.As $\mu$ increases from $1$,the term $(\mu - 1)$ increases,causing the phase difference $\phi$ to increase.The intensity $I$ follows a $\cos^2$ variation,starting from a maximum value $(4I_0)$ at $\mu = 1$ (where $\phi = 0$) and decreasing as $\mu$ increases,following the shape of a squared cosine function.

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