A flake of glass (refractive index mu = 1.5) | vedclass

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(C) The fringe shift $\Delta x$ due to the introduction of a glass plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{D}{

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(C) The fringe shift $\Delta x$ due to the introduction of a glass plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{D}{	ext{d}}(\mu - 1)t$.Since the fringe width is $\beta = \frac{\lambda D}{	ext{d}}$,we can write the shift in terms of the number of fringes $n$ as $\Delta x = n\beta$.Equating the two expressions: $n\beta = \frac{D}{	ext{d}}(\mu - 1)t$.Substituting $\beta = \frac{\lambda D}{	ext{d}}$,we get $n \left( \frac{\lambda D}{	ext{d}} ight) = \frac{D}{	ext{d}}(\mu - 1)t$.This simplifies to $n\lambda = (\mu - 1)t$.Given $n = 7$,$\lambda = 600 \, nm$,and $\mu = 1.5$,we have $7 	imes 600 = (1.5 - 1)t$.$4200 = 0.5 	imes t$.$t = \frac{4200}{0.5} = 8400 \, nm$.

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