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(C) Let $I_0$ be the intensity due to a single slit. The resultant intensity $I$ in a double-slit experiment is given by $I = I_0 + I_0 + 2\sqrt{I_0}\

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$\frac{D\lambda}{3d}$

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(C) Let $I_0$ be the intensity due to a single slit. The resultant intensity $I$ in a double-slit experiment is given by $I = I_0 + I_0 + 2\sqrt{I_0}\sqrt{I_0}\cos\phi = 2I_0(1 + \cos\phi)$.Given that the intensity at the point is equal to that of a single slit,we set $I = I_0$.$I_0 = 2I_0(1 + \cos\phi) \Rightarrow 1 = 2(1 + \cos\phi) \Rightarrow \frac{1}{2} = 1 + \cos\phi$.$\cos\phi = -\frac{1}{2} \Rightarrow \phi = \frac{2\pi}{3}$.The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda}\Delta x$.$\frac{2\pi}{3} = \frac{2\pi}{\lambda}\Delta x \Rightarrow \Delta x = \frac{\lambda}{3}$.For small angles,the path difference $\Delta x = \frac{dy}{D}$.Equating the two,$\frac{dy}{D} = \frac{\lambda}{3} \Rightarrow y = \frac{D\lambda}{3d}$.

In a Young's double-slit experiment,$D$ is the distance of the screen from the slits and $d$ is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is the same as that due to a single slit is equal to:

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