Answer:
Δx = 5.82mm
Explanation:
To find the distance between the eight maximum and the third one you use the following formula:
[tex]x_m=\frac{m\lambda D}{d}[/tex] (1)
λ: wavelength = 452*10^-9 m
m: order of the fringes
D: distance to the scree = 0.49m
d: distance between slits = 0.190*10^-3 m
you use for m=8 and m=3, then you calculate x8 - x3:
[tex]x_8=\frac{8(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=9.32*10^{-3}m\\\\x_3=\frac{3(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=3.49*10^{-3}m\\\\\Delta x_{8,3}=5.82*10^{-3}m=5.82mm[/tex]
hence, the distance between these fringes is 5.82mm
