To solve this problem, we need to understand the intensity distribution in a Young's double-slit experiment.$\\$In a Young's double-slit experiment, the intensity at any point on the screen is given by:$\\ I = I_0 \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) \\$where $I_0$ is the maximum intensity, $d$ is the distance between the slits, $\lambda$ is the wavelength,$\\$and $\theta$ is the angle of the point on the screen with respect to the central maximum.$\\$Given:$\\ d = 5\lambda \\ D = 10d = 50\lambda \\$We need to find the intensity in front of one of the slits. In this case, $\theta = 0$ because the point is directly in front of the slit.$\\$Substituting $\theta = 0$ into the intensity formula:$\\ I = I_0 \cos^2\left(\frac{\pi \cdot 5\lambda \cdot \sin 0}{\lambda}\right) \\$Since $\sin 0 = 0,$ we have:$\\ I = I_0 \cos^2(0) \\$We know that $\cos(0) = 1,$ so:$\\ I = I_0 \cdot 1^2 \\ I = I_0 \\$Therefore, the intensity in front of one of the slits is $I_0. \\$This corresponds to Option 3.