To solve this problem, we need to understand the relationship between the angular width of the fringes and the slit separation in Young's double slit experiment.$\\$The angular width of the fringes $(\theta)$ is given by:$\\ \theta = \frac{\lambda}{d} \\$where $\lambda$ is the wavelength of the light and $d$ is the separation between the slits.$\\$Given:$\\$• Initial angular width $\theta_1 = 0.20^{\circ} \\$• Desired angular width $\theta_2 = 0.21^{\circ} \\$• Initial slit separation $d_1 = 2$ mm$\\$• Wavelength $\lambda = 5896$ Å$= 5896 \times 10^{-10}$ m$\\$We need to find the new slit separation $d_2$ that will give the desired angular width.$\\$Using the formula for angular width:$\\ \theta_1 = \frac{\lambda}{d_1} \quad$and$\quad \theta_2 = \frac{\lambda}{d_2} \\$We can set up the equation:$\\ \frac{\lambda}{d_1} = 0.20^{\circ} \quad$and$\quad \frac{\lambda}{d_2} = 0.21^{\circ} \\$Divide the second equation by the first:$\\ \frac{\frac{\lambda}{d_2}}{\frac{\lambda}{d_1}} = \frac{0.21^{\circ}}{0.20^{\circ}} \\$Simplify:$\\ \frac{d_1}{d_2} = \frac{0.21}{0.20} \\ d_2 = d_1 \times \frac{0.20}{0.21} \\$Substitute the known value of $d_1: \\ d_2 = 2$ mm$\times \frac{0.20}{0.21} \\ d_2 = 2 \times \frac{20}{21}$ mm$\\ d_2 = \frac{40}{21}$ mm$\\ d_2 \approx 1.9048$ mm$\\$Rounding to one decimal place, $d_2 \approx 1.9$ mm.$\\$Therefore, the separation between the slits needs to be changed to $1.9$ mm, which corresponds to Option 2.