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(B) When a transparent film of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits in a Young's double slit experiment,the

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$14.5$

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(B) When a transparent film of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits in a Young's double slit experiment,the central fringe shifts by a distance $\Delta x = \frac{(\mu - 1)tD}{d}$.This shift is equivalent to a number of fringes $N$ given by $N = \frac{\Delta x}{\beta}$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Substituting the expression for $\Delta x$,we get $N = \frac{(\mu - 1)tD/d}{\lambda D/d} = \frac{(\mu - 1)t}{\lambda}$.Given values: $\mu = 1.45$,$t = 0.02 \ mm = 2 	imes 10^{-5} \ m$,and $\lambda = 620 \ nm = 6.2 	imes 10^{-7} \ m$.Substituting these values: $N = \frac{(1.45 - 1) 	imes 2 	imes 10^{-5}}{6.2 	imes 10^{-7}}$.$N = \frac{0.45 	imes 2 	imes 10^{-5}}{6.2 	imes 10^{-7}} = \frac{0.9 	imes 10^{-5}}{6.2 	imes 10^{-7}} = \frac{0.9}{6.2} 	imes 100 = \frac{90}{6.2} \approx 14.516$.Rounding to the nearest provided option,the number of fringes is $14.5$.

$A$ transparent film $(\mu=1.45)$ of thickness $0.02 \ mm$ is placed 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 center if the film is removed?

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