This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
[tex]y= \frac{m\lambda D}{a} [/tex]
from the center of the pattern. In the formula, m is the order of the minimum, [tex]\lambda[/tex] the wavelenght, [tex]D[/tex] the distance of the screen from the slit and [tex]a [/tex] the width of the slit.
In our problem, the distance of the first-order band (m=1) is [tex]y=0.22 cm[/tex]. The distance of the screen is D=86 cm while the wavelength is [tex]\lambda = 384 nm=384 \cdot 10^{-7}cm[/tex]. Using these data and re-arranging the formula, we can find a, the width of the slit:
[tex]a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm [/tex]
