First slide

Young's double slit Experiment

Question

Two slits separated by a distance of 1 mm are illuminated with light of wavelength 6 .0 x 10 -7 m. The interference fringes are observed on a screen placed 1 m from the slits. The distance between third dark fringe and the fifth bright fringe is equal to

Easy

A

0.60 mm
B

1.50 mm
C

3.00 mm
D

4.50mm

Solution

The fringe width p is given by
$β = \frac{λD}{d} (d = distance between the slits)$
$x = \frac{5 λ D}{2 d} = \frac{5 \times (6 \cdot 0 \times 10^{- 7}) \times 1)}{2 \times (1 \times 10^{- 3})} = 1.5 \times 10^{- 3} m$
The fifth bright hinge is separated from third dark hinge by

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