To solve the problem, apply the concepts related to constructive and destructive interference. For this purpose, we will start by defining the distance from the dark fringe from the central maximum in terms of the slit separation, the number of fringe the wavelength and the distance of the screen to the slit. Mathematically this is,
[tex]y_n = \frac{(2n-1)\lambda D}{2d}[/tex]
[tex]D = \frac{2dy_n}{(2n-1)\lambda}[/tex]
Here,
d = Slit separation
n = Number of fringe
[tex]y_n[/tex] = Distance of the dark fringe from the central maximum
D = Distance of the screen to the slit
[tex]\lambda[/tex] = Wavelength
Our values are given as,
[tex]\lambda = 679nm = 679*10^{-9}m[/tex]
[tex]d = 1.1mm = 1.1*10^{-3}m[/tex]
[tex]y_{14} = 8.01cm = 8.01*10^{-2} m[/tex]
[tex]n = 14[/tex]
Replacing,
[tex]D = \frac{2(1.1*10^{-3})(8.01*10^{-2})}{(2(14)-1)(679*10^{-9})}[/tex]
[tex]D = 9.61m[/tex]
Therefore the screen is located from the slit to 9.61m
