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Young's double slit experiment is performed in water, instead of air, then fringe width
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decreases.
(C)
$\mathrm{X}=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
In water the wavelength $\lambda$ decreases and hence the fringe width decrease.
$\mathrm{X}=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
In water the wavelength $\lambda$ decreases and hence the fringe width decrease.
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