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Two slits at a distance of `1mm` are illuminated by a light of wavelength `6.5xx10^-7m`. The interference fringes are observed on a screen placed at a distance of `1m`. The distance between third dark fringe and fifth bright fringe will be
A. `.65mm`
B. `1.63mm`
C. `3.25mm`
D. `4.87mm`

1 Answer

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Answer:
`beta=(lambdaD)/(d)=(6.5xx10^(-7)xx1)/(10^(-3)`
`beta=.65xx10^(-3)m=.65mm`
The distance between the fifth bright fringe from third dark fringe.
`=5beta-2.5beta rArr 2.5beta=2.5xx.65=1.63mm`

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