In Young's experiment, if the entire setup is immersed in a medium of refractive index $n$, the fringe width:
AStays the same
BIncreases by factor $n$
CDecreases by factor $n$ (since $\lambda$ shortens)
DBecomes zero
Answer & Solution
Correct answer: C. Decreases by factor $n$ (since $\lambda$ shortens)
Wavelength in medium $\lambda' = \lambda/n$, so fringe width $w' = \lambda' D/d = w/n$. Hence immersion in water (n=1.33) shrinks fringes by ~25%.
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