= Solution
A slit of width $w$ has its first zeros at $\sin\theta=\pm\lambda/w$. Thus the full central-envelope widths are
$$
\boxed{W_1\simeq\frac{2L\lambda}{w_1}=\frac L{10},\qquad
W_2\simeq\frac{2L\lambda}{w_2}=\frac L{20}.}
$$
The narrower envelope determines the overlap, so it spans
$$
\boxed{\frac{W_2}{\delta x}=50\ \text{fringe spacings}.}
$$
This is the usual approximate “50 fringes” count. If counting nominal bright fringe centres strictly inside both first zeros, they have indices $-24,\ldots,24$, giving \b[49 centres]. The endpoints $\pm25$ lie at the wider slit's first zeros, where two-slit visibility vanishes; they should not be counted as fully interfering bright centres.
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