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In a YDSE setup with slit separation $d$ and screen distance $D$, the wavelength of the light source is doubled while the slit separation is halved. The new fringe width compared to the original is:
ASame
BDoubled
CQuadrupled
DHalved
Answer & Solution
Correct answer: C. Quadrupled
$\beta = \lambda D / d$. If $\lambda \to 2\lambda$ and $d \to d/2$, then $\beta_\text{new} = (2\lambda) D / (d/2) = 4 \lambda D / d = 4\beta$. The fringe width becomes four times the original.
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