MHTCET 15 May 2024 Shift 2 Paper

Concept:Introducing a thin transparent plate of refractive index $\mu$ in front of one slit adds an extra optical path $(\mu - 1)t$, which shifts the fringe pattern without changing the fringe width.Explanation:A plate of refractive index $1.6$ and thickness $t$ placed before one slit increases the optical path travelled by light from that slit by $(\mu - 1)t$.This introduces an additional phase difference $\Delta \phi = \frac{2\pi}{\lambda}(\mu - 1)t$ between the two interfering waves.Due to this extra phase delay, the central maximum is no longer at the geometric centre; it shifts laterally toward the slit in whose path the plate is introduced.The fringe width is given by $\beta = \frac{\lambda D}{d}$, which depends only on wavelength $\lambda$, slit separation $d$, and screen distance $D$.Since a uniform phase shift does not change these parameters, the fringe width remains constant.The interference pattern does not disappear, and the number of fringes does not decrease, as long as the plate is thin and transparent.Thus, the entire fringe system shifts toward the side of the plate.Answer:(B) The central maximum will shift towards this side.