In a Young's double-slit experiment,let A and | vedclass

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(D) The path difference $\Delta x$ introduced by the films at the central position is given by:$\Delta x = (\mu_A - 1)t_A - (\mu_B - 1)t_B$Expanding t

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shift towards $A$ if $t_B > t_A$; shift towards $B$ if $t_B < t_A$

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(D) The path difference $\Delta x$ introduced by the films at the central position is given by:$\Delta x = (\mu_A - 1)t_A - (\mu_B - 1)t_B$Expanding this expression:$\Delta x = \mu_A t_A - t_A - \mu_B t_B + t_B$Given the condition $\mu_A t_A = \mu_B t_B$,the terms $\mu_A t_A$ and $\mu_B t_B$ cancel out:$\Delta x = t_B - t_A$If $t_A = t_B$,then $\Delta x = 0$,and the central maximum does not shift.However,if $t_A eq t_B$,the path difference is non-zero,causing a shift.Specifically,if $t_B > t_A$,$\Delta x > 0$,the shift is towards $A$.If $t_B < t_A$,$\Delta x < 0$,the shift is towards $B$.

In a Young's double-slit experiment,let $A$ and $B$ be the two slits. Thin films of thicknesses $t_A$ and $t_B$ and refractive indices $\mu_A$ and $\mu_B$ are placed in front of slits $A$ and $B$ respectively. If $\mu_A t_A = \mu_B t_B$,the central maximum will:

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