Step 1: Recall the fringe width formula.
In a biprism (two-source) interference pattern, the spacing between consecutive bright or dark fringes is $$W = \frac{\lambda D}{d},$$ where $D$ is the slit-to-screen distance and $d$ is the separation between the two coherent sources.
Step 2: Spot the proportionality.
With wavelength $\lambda$ fixed, $W$ grows in proportion to $D$ and shrinks in proportion to $d$. So instead of recomputing from scratch, we can simply scale the original $W$.
Step 3: Apply the change in $D$.
Increasing $D$ by $25\%$ means $D' = 1.25\,D$, which multiplies $W$ by $1.25$.
Step 4: Apply the change in $d$.
Halving the slit separation gives $d' = d/2$. Since $W$ is inversely proportional to $d$, halving $d$ doubles $W$, a factor of $2$.
Step 5: Combine the two scaling factors.
$$W' = W \times (1.25) \times (2) = 2.5\,W.$$
Step 6: State the result.
The new fringe width is $2.5$ times the original, because the two effects multiply rather than cancel.
\[ \boxed{W' = 2.5\,W} \]