In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is

At the original central maximum, the initial phase difference is $0$, so the intensity is maximum. Introducing a glass plate in one path adds optical path difference $$\Delta = (\mu - 1)t$$ Here $\mu = 1.5$, so $$\Delta = 0.5t$$ For the intensity at that same point to remain unchanged, the phase difference must again correspond to a bright fringe: $$\Delta = m\lambda$$ For minimum nonzero thickness, take $m=1$: $$0.5t = \lambda$$ $$t = 2\lambda$$ This matches option $A$.