First slide

Young's double slit Experiment

Question

When Young's double slit experiment is conducted in vacuum, fringe width is found to be $β$ . When separation between the slits is halved, distance between the slits and the screen is doubled and the whole experiment is conducted in water $(μ = \frac{4}{3})$ , the new fringe width will be

Moderate

A

$\frac{β}{3}$
B

$3 β$
C

$\frac{2 β}{3}$
D

$\frac{4 β}{3}$

Solution

$β = \frac{λ D}{d} and β^{1} = \frac{λ^{1} D^{1}}{d^{1}} = \frac{(\frac{λ}{μ}) (2 D)}{(\frac{d}{2})} = \frac{4}{μ} (\frac{λ D}{d})$
$∴ β^{1} = \frac{4}{4 / 3} β = 3 β$

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