To solve this problem, we need to understand how the immersion in water affects the angular width of the first minima in a double slit experiment.$\\$The angular position of the first minima in a double slit experiment is given by:$\\ \theta = \frac{\lambda}{d} \\$where $\lambda$ is the wavelength of light in the medium, and $d$ is the slit separation.$\\$Initially, the wavelength of light in air is $\lambda_{air} = 400$ nm.$\\$The angular width of the first minima in air is given as $0.2^\circ. \\$When the apparatus is immersed in water, the wavelength of light changes to:$\\ \lambda_{water} = \frac{\lambda_{air}}{\mu_{water}} \\$Given $\mu_{water} = \frac{4}{3},$ we have:$\newline \lambda_{water} = \frac{400 nm}{\frac{4}{3}} = 300 nm \\$The angular position of the first minima in water is then:$\\ \theta_{water} = \frac{\lambda_{water}}{d} = \frac{300 nm}{d} \\$The angular width of the first minima is proportional to the wavelength, so:$\\ \frac{\theta_{water}}{\theta_{air}} = \frac{\lambda_{water}}{\lambda_{air}} \\$Substitute the known values:$\\ \frac{\theta_{water}}{0.2^\circ} = \frac{300 nm}{400 nm} \\$Simplify to find $\theta_{water}: \\ \theta_{water} = 0.2^\circ \times \frac{3}{4} = 0.15^\circ \\$Therefore, the angular width of the first minima when the apparatus is immersed in water is $0.15^\circ. \\$This corresponds to Option 2.