Answer
a. 8.00 mm.
b. 8.00 mm.
Work Step by Step
The width of a bright fringe is the distance between the two adjacent destructive minima. The screen is very far, so assuming the small angle formula for destructive interference, we know that $y_m=R\frac{(m+\frac{1}{2})\lambda}{d}$.
Find the distance between any two successive minima: $\Delta y =R\frac{\lambda}{d}$.
$$\Delta y =\frac{R\lambda}{d}=\frac{ (4.00m)(400\times10^{-9}m)}{0.200\times10^{-3}m}=8.00\times10^{-3}m$$
Thus, the answer to both parts is 8.00 mm.
