CBSE Class 12 · 2020 · Wave OpticsShort Answer

The fringe width is $\beta =\frac{\lambda D}{d}$.
(a) $\text{Screen moved away}$ ($D$ increases): $\beta \propto D$, so the fringe width $\text{increases}$.
(b) $\text{Slit separation increased}$ ($d$ increases): $\beta \propto \frac{1}{d}$, so the fringe width $\text{decreases}$.
(c) $\text{Source slit moved closer}$ to the double slit: the condition for distinct fringes is $\frac{s}{S}<\frac{\lambda }{d}$, where $s$ is the source-slit width and $S$ its distance. Moving the source closer decreases $S$, increasing $\frac{s}{S}$; once $\frac{s}{S}\ge \frac{\lambda }{d}$ the fringes lose contrast and eventually $\text{disappear}$.