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In Young's experiment for interference o...

In Young's experiment for interference of light the slits 0.2 cm apart are illuminated by yellow light `(lamda=5896A^(@))` What would be the fringe width on a screen placed 1 m from the plane of slits? What will be the fringe width if the system is immersed in water (refractive index `=4//3`)

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The correct Answer is:
0.3 mm, 0.225 mm
`d=0.2cm,lamda=5896Å,D=1m`
Fringe width `beta=(lamdaD)/(d)=(5896xx10^(-10)xx1)/(0.2xx10^(-2))=0.3` mm
If system is immersed in water `(mu=1.33)`, then the fringe width becomes
`beta'=(beta)/(mu)=(0.3)/(1.3)mm=0.225mm`
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