Through some dedcution, we can find out that the width of the central maximum can be written as:
[tex]\Delta x=2(\frac{\lambda D}{d})[/tex]Thus, for our case:
[tex]\Delta x=2*(\frac{465.2*10^{-9}*1.54}{0.12*10^{-3}})=0.01194m=11.94mm[/tex]Thus, the width of the central maximum is 11.94mm
