Answer:
a) 2.4 mm
b) 1.2 mm
c) 1.2 mm
Explanation:
To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:
[tex]m\lambda=asin\theta[/tex]
a: width of the slit
λ: wavelength
m: order of the minimum
for little angles you have:
[tex]y=\frac{m\lambda D}{a}[/tex]
y: height of the mth minimum
a) the width of the central maximum is 2*y for m=1:
[tex]w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm[/tex]
b) the width of first maximum is y2-y1:
[tex]w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm[/tex]
c) and for the second maximum:
[tex]w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm[/tex]
