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(A) The path difference at point $O$ is due to the introduction of the slab in one of the paths.Let the path length in vacuum be $L$. The optical path

Vedclass · Accepted Answer

$\frac{2\pi}{n_1 \lambda_1} (n_3 - n_2) t$

Vedclass · Answer

(A) The path difference at point $O$ is due to the introduction of the slab in one of the paths.Let the path length in vacuum be $L$. The optical path length is defined as $n 	imes 	ext{geometric path}$.For the path with the slab, the optical path is $OP_1 = (L - t) n_2 + t n_3$.For the path without the slab, the optical path is $OP_2 = L n_2$.The path difference $\Delta x = |OP_1 - OP_2| = |(L - t) n_2 + t n_3 - L n_2| = |t(n_3 - n_2)|$.This path difference is in the medium of refractive index $n_2$. To convert this to the phase difference, we use the wavelength in the medium $n_2$, which is $\lambda_2 = \frac{\lambda_{vacuum}}{n_2}$.Since $\lambda_1$ is the wavelength in medium $n_1$, $\lambda_{vacuum} = n_1 \lambda_1$.Thus, $\lambda_2 = \frac{n_1 \lambda_1}{n_2}$.The phase difference $\Delta \phi = \frac{2\pi}{\lambda_2} \Delta x = \frac{2\pi}{(n_1 \lambda_1 / n_2)} |t(n_3 - n_2)| = \frac{2\pi n_2}{n_1 \lambda_1} (n_3 - n_2) t$ (assuming $n_3 > n_2$).However, checking the options, the standard derivation for path difference in terms of vacuum wavelength is $\Delta x = t(n_3 - n_2)$. The phase difference is $\Delta \phi = \frac{2\pi}{\lambda_{vac}} \Delta x = \frac{2\pi}{n_1 \lambda_1} (n_3 - n_2) t$.

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