If you were to double the width of a single slit, by how much would you change the width of the central maxima?

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

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