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}{1×10^{6}nm}[/tex] ⇒ [tex]6.33×10^{-4}mm[/tex]

Finally, equation 1 can be used:

[tex]\Lambda x = (2000mm)\frac{(6.33×10^{-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.