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Question
A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?
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Solution
Given:-
Refractive index of the paper, μ = 1.45
The thickness of the plate,
\[t = 0 . 02 mm = 0 . 02 \times {10}^{- 3} m\]
Wavelength of the light,
\[\lambda = 620 nm = 620 \times {10}^{- 9} m\]
We know that when we paste a transparent paper in front of one of the slits, then the optical path changes by \[\left( \mu - 1 \right)t.\]
And optical path should be changed by λ for the shift of one fringe.
∴ Number of fringes crossing through the centre is
\[n = \frac{\left( \mu - 1 \right)t}{\lambda}\]
\[ = \frac{\left( 1 . 45 - 1 \right) \times 0 . 02 \times {10}^{- 3}}{620 \times {10}^{- 9}}\]
\[ = 14 . 5\]
Hence, 14.5 fringes will cross through the centre if the paper is removed.
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