Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits \(S_1\) and \(S_2\). In each of these cases \(S_1 P_0=S_2 P_0\), \(S_1 P_1-S_2 P_1=\frac{\lambda}{4}\) and \(S_1 P_2-S_2 P_2=\frac{\lambda}{3}\), where \(\lambda\) is the wavelength of the light used. In the cases \(\mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\), a transparent sheet of refractive index \(\mu\) and thickness \(t\) is pasted on slit \(S_2\). The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point \(P\) on the screen from the two slits is denoted by \(\delta(P)\) and the intensity by \(I(P)\). Match each situation given in Column-I with the statement(s) in Column-II valid for that situation.

(A) \(\rightarrow(\mathrm{p}, \mathrm{s}) \rightarrow\) Intensity at \(P_0\) is maximum. It will continuously decrease from \(P_0\) towards \(P_2\).
\((B) \rightarrow(q) \rightarrow\) Path difference due to slap will be compensated by geometrical path difference. Hence, \(\delta\left(P_1\right)=0\).
\((\mathrm{C}) \rightarrow(\mathrm{t}) \rightarrow \delta\left(P_0\right)=\frac{\lambda}{2}, \delta\left(P_1\right)=\frac{\lambda}{2}-\frac{\lambda}{4}=\frac{\lambda}{4}\) and \(\delta\left(P_2\right)=\frac{\lambda}{2}-\frac{\lambda}{3}=\frac{\lambda}{6}\). When path difference increases from 0 to \(\frac{\lambda}{2}\), intensity will decrease from maximum to zero. Hence, in this case,
\(I\left(P_2\right)>I\left(P_1\right)>I\left(P_0\right)\)
\((\mathrm{D}) \rightarrow(\mathrm{r}, \mathrm{s}, \mathrm{t})\)
\(\delta\left(P_0\right)=\frac{3 \lambda}{4}, \delta\left(P_1\right)=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2}\)
and \(\delta\left(P_2\right)=\frac{3 \lambda}{4}-\frac{\lambda}{3}=\frac{5 \lambda}{12}\)
In this case \(I\left(P_1\right)=0\)