First slide

Young's double slit Experiment

Question

In the Young's double slit experiment, the spacing between two slits is 0.1 mm. If the screen is kept at a distance of 1.0 m from the slits and the wavelength of light is $5000 Å$ , then the fringe width is

Easy

A

1.0 cm
B

1.5 cm
C

0.5 cm
D

2.0 cm

Solution

$β = \frac{λD}{d} = \frac{5000 \times 10^{- 10} \times 1}{0 .1 \times 10^{- 3}} m = 5 \times 10^{- 3} m = 0 .5 cm$

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