Q.In Young's double slit experiment, a thin uniform sheet of glass is kept in front of the two slits, parallel to the screen having the slits. The resulting interference pattern will satisfy (A) The interference pattern will remain unchanged
(B) The fringe width will decrease
(C) The fringe width will increase
(D) The fringes will shift.

[!TLDR] A sheet placed in front of BOTH slits equally adds the identical extra optical path to both arms, so the relative path difference (and hence the whole pattern) is comple …

As derived in Section 7.8.5, inserting a transparent plate in front of just ONE slit adds extra optical path $(n-1)t$ to only that one arm, which shifts the whole fringe pattern sideways. But this question specifies the glass sheet is kept 'in front of the two slits' -- i.e. covering BOTH slits equally, parallel to the screen. Since both light paths, from $S_1$ and from $S_2$, now pass through the identical extra thickness of the identical glass, EACH arm gains the exact same extra optical path $(n-1)t$. The relative path difference between the two arms at any screen point -- which is what actually determines the fringe pattern -- depends only on the DIFFERENCE between the two arms' optical paths, and since both arms gain the identical extra amount, …

Recognise that the fringe pattern depends only on the RELATIVE optical path difference between the two arms, not on the absolute optical path length of either arm alone -- so any change applied EQUALLY to both slits (rather than to just one) leaves …

Reflexively applying the single-covered-slit result (pattern shifts, Section 7.8.5 / Example 7.6) to this genuinely different setup, without noticing the questio …