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]D=37.0 cm=0.37 m[/tex]
[tex]\lambda=530 nm=5.3 \cdot 10^{-7} m[/tex]
while the distance between the first and the fifth minima is
[tex]y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m[/tex] (2)
If we use the formula to rewrite [tex]y_5, y_1[/tex], eq.(2) becomes
[tex] \frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m [/tex]
Which we can solve to find a, the width of the slit:
[tex]a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm[/tex]
