A transparent paper (refractive index = 1.45) of thickness 0.02mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620nm. How many fringes will cross through the centre if the paper is removed?
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Given that, $\mu=1.45,$ t = 0.02mm $= 0.02 × 10^{-3}m$ and $\lambda=620\text{nm}=620\times10^{-9}\text{m}$
We know, when the transparent paper is pasted in one of the slits, the optical path changes by $(\mu-1)\text{t}.$
Again, for shift of one fringe, the optical path should be changed by $\lambda.$
So, no. of fringes crossing through the centre is given by,
$\text{n}=\frac{(\mu-1)\text{t}}{\lambda}=\frac{0.45\times0.02\times10^{-3}}{620\times10^{-9}}=14.5$
We know, when the transparent paper is pasted in one of the slits, the optical path changes by $(\mu-1)\text{t}.$
Again, for shift of one fringe, the optical path should be changed by $\lambda.$
So, no. of fringes crossing through the centre is given by,
$\text{n}=\frac{(\mu-1)\text{t}}{\lambda}=\frac{0.45\times0.02\times10^{-3}}{620\times10^{-9}}=14.5$
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