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Question
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
Solution
Fringe width is the distance between two consecutive dark or bright fringes,
so we have fringe width `=(lambdaD)/d`.
If the whole apparatus is immersed in water and refractive index of water is n then,
`v/c =1/n` = Where v is velocity of light in water
`=> n =(vlambda)/(vlambda_omega) lambda =` wavelength of light in air
`=> n =(vlambda)/(lambda_omega) lambda_omega =` wavelength of light in wetar
`lambda_omega = lambda/n v ` = frequency of light in air and water
Hence
`beta_omega =(lambda_omegad)/D =(lambda) =(lamdad)/(nD)`
`beta_omega = 1/n beta`
This shows fringe width will be reduced by the factor of the refractive index of water.
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