Explanation:

A. Angular separation of the fringes remain constant \(\left( {\sin \theta = n\lambda {\rm{/}}d} \right)\). The separation between fringes increases as the screen moves away from the plane of the slits.
B. The separation of the fringes decreases with decrease in wavelength.
C. When medium is water, \(\lambda ' = \frac{{{\lambda _{air}}}}{{4{\rm{/}}3}} = \frac{3}{4}{\lambda _{air}}\)
\(\therefore \;\beta ' = \frac{{\lambda 'D}}{d} = \frac{3}{4}\left( {\frac{{\lambda D}}{d}} \right) = \frac{3}{4}\beta \)
D. When \(d\) is reduced, \(\beta \propto \frac{1}{d}\)
So, \(\beta \) is increased. Option (3) is correct.