Answer: the width of central maxima decrease a factor 0.5
Explanation: To explain this results we have to consider the relationship given for the locations of the first dark fringes asides to central maximum ofone lit difraction as:
sin θmin= +/- λ/a where a is the slit width so the width on the screen is approximately given by:
y1= Lsin θmin where L is the distance to the screen
Finally we have the full width on the screen is: 2y1= Lλ/a
if we double the width, we have 0.5 of the full width of the central maximun given by the initial slit