In Young's double-slit experiment, the maximum intensity is $I_{max} = I_0$. The point on the screen directly in front of one of the slits is at a distance $y = \frac{d}{2}$ from the central maximum. The path difference $\Delta x$ at this point on the screen is given by: $\Delta x = \frac{y d}{D}$ Substituting $y = \frac{d}{2}$ and $D = 10d$, we get: $\Delta x = \frac{(d/2) d}{10d} = \frac{d^2}{20d} = \frac{d}{20}$ Given that the slit distance $d = 5\lambda$, we substitute this into the path difference equation: $\Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4}$ The corresponding phase difference $\phi$ is related to the path difference by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$: $\phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{4}\right) = \frac{\pi}{2}$ The resultant intensity $I$ at any point with phase difference $\phi$ is given by: $I = I_{max} \cos^2\left(\frac{\phi}{2}\right)$ Substituting $\phi = \frac{\pi}{2}$ and $I_{max} = I_0$: $I = I_0 \cos^2\left(\frac{\pi/2}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right)$ $I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2}$