Answer:
The wavelength of incident light is [tex]1.38x10^{-6}m[/tex]
Explanation:
The physicist Thomas Young established, through his double slit experiment, a relation 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 = 2.60m[/tex]
[tex]d = 0.230mm[/tex]
[tex]\Lambda x = 1.57cm[/tex]
Then, [tex]\lambda[/tex] can be isolated from equation 1
[tex]\lambda = \frac{d \Lambda x}{L}[/tex] (2)
However, before equation 2 can be used, it is necessary to express [tex]\Lambda x[/tex] and d in units of meters.
[tex]\Lambda x= 1.57cm \cdot \frac{1m}{100cm}[/tex] ⇒ [tex]0.0157m[/tex]
[tex]d = 0.230mm \cdot \frac{1m}{1000mm}[/tex] ⇒ [tex]2.3x10^{-4}m[/tex]
Finally, equation 2 can be used.
[tex]\lambda = \frac{(2.3x10^{-4}m)(0.0157m}{(2.60m)}[/tex]
[tex]\lambda = 1.38x10^{-6}m[/tex]
Hence, the wavelength of incident light is [tex]1.38x10^{-6}m[/tex]
