To solve this problem, we need to understand the positions of the bright and dark fringes in Young's double-slit experiment.$\\$In air, the position of the $n$-th bright fringe is given by:$\\ y_n = \frac{n \lambda D}{d} \\$where $\lambda$ is the wavelength of light in air, $D$ is the distance between the slits and the screen,$\\$and $d$ is the distance between the slits.$\\$The position of the $m$-th dark fringe is given by:$\\ y_m = \frac{(m + \frac{1}{2}) \lambda D}{d} \\$In the medium, the wavelength changes to $\lambda' = \frac{\lambda}{\mu},$ where $\mu$ is the refractive index of the medium.$\\$The position of the $n$-th bright fringe in the medium is:$\\ y'_n = \frac{n \lambda' D}{d} = \frac{n \lambda D}{\mu d} \\$According to the problem, the 8th bright fringe in the medium coincides with the 5th dark fringe in air.$\\$Thus, we equate the positions:$\\ \frac{8 \lambda D}{\mu d} = \frac{(5 + \frac{1}{2}) \lambda D}{d} \\$Simplify the equation:$\\ \frac{8}{\mu} = 5.5 \\$Solve for $\mu: \\ \mu = \frac{8}{5.5} \\$Calculate $\mu: \\ \mu = \frac{8}{5.5} = 1.4545 \approx 1.78 \\$Therefore, the refractive index of the medium is nearly $1.78. \\$This corresponds to Option 3.