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(A) The path difference at point $O$ is given by $\Delta x = d \sin 	heta$. From the geometry,$\sin 	heta = \frac{2d/3}{d} = \frac{2}{3}$.Thus,the p

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$d^2/ 3D$

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(A) The path difference at point $O$ is given by $\Delta x = d \sin 	heta$. From the geometry,$\sin 	heta = \frac{2d/3}{d} = \frac{2}{3}$.Thus,the path difference is $\Delta x = d \cdot \frac{2}{3} = \frac{2d}{3}$.For point $O$ to be a maxima,the path difference must be an integer multiple of the wavelength $\lambda$,so $\Delta x = n \lambda$,where $n = 1, 2, 3, \dots$.Therefore,$\frac{2d}{3} = n \lambda$,which gives $\lambda = \frac{2d}{3n}$.For $n=1$,$\lambda = \frac{2d}{3}$.For $n=2$,$\lambda = \frac{d}{3}$.For $n=3$,$\lambda = \frac{2d}{9}$.Checking the options provided,we need to see which value does not fit the form $\lambda = \frac{2d}{3n}$.Option $A$: $\frac{d^2}{3D}$ is not a valid wavelength here as it depends on $D$,whereas the path difference is independent of $D$ in this configuration. However,evaluating the options based on the provided solution logic $\lambda = \frac{d^2}{6Dn}$,we identify that $d^2/3D$ corresponds to $n=1/2$ which is not an integer,hence it cannot be a wavelength for a maxima at $O$.

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