First slide

Young's double slit Experiment

Question

In a certain double slit experiment arrangement interference fringes of width 1.0 mm each are observed when light of wavelength $5000 Å$ is used. Keeping the set up unaltered, if the source is replaced by another source of wavelength $6000 Å$ , the fringe width will be

Easy

A

0.5 mm
B

1.0 mm
C

1.2 mm
D

1.5 mm

Solution

$\frac{β_{1}}{β_{2}} = \frac{λ_{1}}{λ_{2}} or \frac{1 .0}{β_{2}} = \frac{5000}{6000} or β_{2} = \frac{6000}{5000} = 1 .2 mm .$

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