In Young's double slit experiment, the fringe width $(\beta)$ is given by the formula:$\\ \beta = \frac{\lambda D}{d} \\$where:$\\$• $\lambda$ is the wavelength of light used.$\\$• $D$ is the distance between the slits and the screen.$\\$• $d$ is the separation between the slits.$\\$Initially, the fringe width is:$\\ \beta_1 = \frac{\lambda D}{d} \\$According to the problem, the separation between the slits $(d)$ is halved,$\\$so the new separation is $\frac{d}{2}. \\$Also, the distance to the screen $(D)$ is doubled,$\\$so the new distance is $2D. \\$The new fringe width $(\beta_2)$ is:$\\ \beta_2 = \frac{\lambda (2D)}{\frac{d}{2}} \\$Simplify the expression:$\\ \beta_2 = \frac{2\lambda D}{\frac{d}{2}} \\ \beta_2 = \frac{2\lambda D \cdot 2}{d} \\ \beta_2 = \frac{4\lambda D}{d} \\$Thus, the new fringe width is four times the original fringe width:$\\ \beta_2 = 4 \beta_1 \\$Therefore, the fringe width becomes four times, which corresponds to Option 2.