NEET-XII-Physics
17: Light Waves
- #14A 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?Ans : Given:
Refractive index of the paper, μ = 1.45
The thickness of the plate, `` t=0.02\,\mathrm{\,mm\,}=0.02\times {10}^{-3}\,\mathrm{\,m\,}``
Wavelength of the light, `` \lambda =620\,\mathrm{\,nm\,}=620\times {10}^{-9}\,\mathrm{\,m\,}``
`` ``
We know that when we paste a transparent paper in front of one of the slits, then the optical path changes by `` \left(\,\mathrm{\,\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(\,\mathrm{\,\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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