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In YDSE, when whole apparatus is immersed in water (n = 4/3), fringe width:

ABecomes 0
BDoubled
CUnchanged
DDecreased by factor of n (= 4/3): β_water = β_air × 3/4
Answer & Solution
Correct answer: D. Decreased by factor of n (= 4/3): β_water = β_air × 3/4
Wavelength in water = λ₀/n. Fringe width β = λD/d → β_water = β_air/n = (3/4) β_air. Fringes get closer.
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