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Question
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
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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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