In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
[tex]y_n= \frac{n \lambda D}{a}[/tex] (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
[tex]\lambda[/tex] is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
In our problem,
[tex]\lambda=520 nm=5.2 \cdot 10^{-7} m[/tex]
[tex]a=0.22 mm=0.22 \cdot 10^{-3} m[/tex]
while the width of the central maximum on the screen corresponds to twice the distance of the first minimum from the center, and it is equal to
[tex]2 y_1 = 8.30 mm=8.3 \cdot 10^{-3} m[/tex]
Therefore the distance of the first minimum from the center is
[tex]y_1 = \frac{8.3 \cdot 10^{-3} m}{2}=4.15 \cdot 10^{-3} m[/tex]
If we plug these numbers into eq.(1), we can find D, the distance of the screen from the slit:
[tex]D= \frac{y_1 a}{ 1 \lambda }= \frac{(4.15 \cdot 10^{-3} m)(0.22 \cdot 10^{-3} m)}{(1)(5.2 \cdot 10^{-7} m)}= 1.76 m[/tex]
