Answer
$w = 4.0~mm$
Work Step by Step
We can find the width of the central maximum:
$w = \frac{2~\lambda~L}{d}$
$w = \frac{(2)(500\times 10^{-9}~m)(2.0~m)}{0.50\times 10^{-3}~m}$
$w = 4.0\times 10^{-3}~m$
$w = 4.0~mm$
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