On placing a thin film of mica of thickness 1 | vedclass

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(D) The path difference introduced by the mica film is given by $\Delta x = (\mu - 1)t$. In Young's double slit experiment,the shift in the fringe pat

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$1.5$

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(D) The path difference introduced by the mica film is given by $\Delta x = (\mu - 1)t$. In Young's double slit experiment,the shift in the fringe pattern $\Delta y$ is related to the path difference by $\Delta y = \frac{D}{d} \Delta x = \frac{D}{d} (\mu - 1)t$. Since the fringe width is $\beta = \frac{\lambda D}{d}$,we can write the shift as $\Delta y = \frac{\beta}{\lambda} (\mu - 1)t$. Given that the shift is equal to the fringe width,$\Delta y = \beta$,we have $\beta = \frac{\beta}{\lambda} (\mu - 1)t$. This simplifies to $(\mu - 1)t = \lambda$. Substituting the given values,$\mu - 1 = \frac{\lambda}{t} = \frac{6 	imes 10^{-5} 	ext{ cm}}{12 	imes 10^{-5} 	ext{ cm}} = 0.5$. Therefore,$\mu = 0.5 + 1 = 1.5$.

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