In a biprism experiment, the fifth dark fring | vedclass

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(D) The path difference introduced by the film of thickness '$t$' and refractive index '$\mu$' is given by $\Delta x = (\mu - 1)t$.Initially, the $5^{

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$\frac{2.5 \lambda}{(\mu - 1)}$

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(D) The path difference introduced by the film of thickness '$t$' and refractive index '$\mu$' is given by $\Delta x = (\mu - 1)t$.Initially, the $5^{th}$ dark fringe is at the point, so the path difference is $\Delta x_1 = (5 - 0.5)\lambda = 4.5\lambda$.After placing the film, the $7^{th}$ bright fringe is at the same point, so the path difference is $\Delta x_2 = 7\lambda$.The change in path difference is equal to the path difference introduced by the film: $\Delta x = \Delta x_2 - \Delta x_1$.$(\mu - 1)t = 7\lambda - 4.5\lambda = 2.5\lambda$.Therefore, the thickness of the film is $t = \frac{2.5\lambda}{(\mu - 1)}$.

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