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प्रश्न
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
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उत्तर
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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संबंधित प्रश्न
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(A) equal width with same intensity
(B) unequal width with varying intensity
(C) equal intensity\
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A Young's double slit experiment is performed with white light.
(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(c) The fringe next to the central will be red.
(d) The fringe next to the central will be violet.
In Young’s double-slit experiment, using monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 x 10-2 m towards the slits, the change in the fringe width is 3 x 10-5 m. If the distance between the two slits is 10-3 m, calculate the wavelength of the light used.
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`beta = (lambda "D")/"d"` where the terms have their usual meaning.
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