Answer:
The distance between first-order and second-order bright fringes is 12.66mm.
Explanation:
The physicist Thomas Young establishes through its double slit experiment a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.
[tex]\Lambda x = L\frac{\lambda}{d} [/tex] (1)
Where [tex]\Lambda x[/tex] is the distance between two adjacent maxima, L is the distance of the screen from the slits, [tex]\lambda[/tex] is the wavelength and d is the separation between the slits.
The values for this particular case are:
[tex]L = 2.0m[/tex]
[tex]\lambda = 633nm[/tex]
[tex]d = 0.100mm[/tex]
Notice that is necessary to express L and [tex]\lambda[/tex] in units of milimeters.
[tex]L = 2.0m \cdot \frac{1000mm}{1m}[/tex] ⇒ [tex]2000mm[/tex]
[tex]\lambda = 633nm \cdot \frac{1mm}{1x10^{6}nm}[/tex] ⇒ [tex]6.33x10^{-4}mm[/tex]
Finally, equation 1 can be used:
[tex]\Lambda x = (2000mm)\frac{(6.33x10^{-4}mm)}{(0.100mm)} [/tex]
[tex]\Lambda x = 12.66mm[/tex]
Hence, the distance between first-order and second-order bright fringes is 12.66mm.
