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(D) The optical path length is defined as the product of the refractive index of the medium and the geometric path length.For the path $S_1O$,the ligh

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$\left| {\left( {{\mu _2} - {\mu _1}} \right)t} \right|$

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(D) The optical path length is defined as the product of the refractive index of the medium and the geometric path length.For the path $S_1O$,the light travels through the liquid of refractive index $\mu_1$ for the entire distance $S_1O$. Thus,$(S_1O)_{optical} = \mu_1(S_1O)$.For the path $S_2O$,the light travels through the slab of thickness $t$ with refractive index $\mu_2$,and the remaining distance $(S_2O - t)$ through the liquid of refractive index $\mu_1$. Thus,$(S_2O)_{optical} = \mu_2 t + \mu_1(S_2O - t)$.Given that $S_1O = S_2O$,the optical path difference $\Delta x$ at point $O$ is:$\Delta x = |(S_2O)_{optical} - (S_1O)_{optical}|$$\Delta x = |\mu_2 t + \mu_1(S_2O - t) - \mu_1(S_1O)|$Since $S_1O = S_2O$,the terms $\mu_1(S_2O)$ and $\mu_1(S_1O)$ cancel out:$\Delta x = |\mu_2 t - \mu_1 t| = |(\mu_2 - \mu_1)t|$.

Young's double slit experiment is conducted in a liquid of refractive index $\mu_1$ as shown in the figure. $A$ thin transparent slab of refractive index $\mu_2$ and thickness $t$ is placed in front of the slit $S_2$. The magnitude of the optical path difference at point $O$ is

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In the $YDSE$ shown,the two slits are covered with thin sheets having thickness $t$ and $2t$,and refractive index $2\mu$ and $\mu$ respectively. Find the position $(y)$ of the central maxima.

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