The slits in a double-slit interference exper | vedclass

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Question

(D) The shift in the fringe pattern due to the introduction of a thin film of thickness $t$ and refractive index $\mu$ is given by the formula: $\Delt

Vedclass · Accepted Answer

$24 \ \mu m$

Vedclass · Answer

(D) The shift in the fringe pattern due to the introduction of a thin film of thickness $t$ and refractive index $\mu$ is given by the formula: $\Delta x = \frac{(\mu - 1)tD}{d}$.The shift in terms of the number of fringes $N$ is given by $\Delta x = N \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Equating the two expressions: $N \left( \frac{\lambda D}{d} ight) = (\mu - 1)t \left( \frac{D}{d} ight)$.Simplifying,we get $N = \frac{(\mu - 1)t}{\lambda}$.From the given graph,when $\mu = 2.00$,the number of fringes shifted is $N = 40$.Substituting these values into the equation: $40 = \frac{(2.00 - 1)t}{600 	imes 10^{-9} \ m}$.$40 = \frac{1 	imes t}{600 	imes 10^{-9}}$.$t = 40 	imes 600 	imes 10^{-9} \ m = 24000 	imes 10^{-9} \ m = 24 	imes 10^{-6} \ m = 24 \ \mu m$.Therefore,the correct option is $D$.

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