The class is about heat, waves, and light. The website wants exactly the perfect answer and in the same metric system units that are stated on the calculators that are on the picture for the question, if not, it'll mark it wrong.

Set 18 Problem 2 How wide is the central diffraction peak on a screen 3.49 m behind a 0.067 -mm-wide slit illuminated by 400 -nm light? VO 0. Width of peak = cm abc sin(x) Ox10 Submit Reset Reset All The distance from the center of the diffraction pattern to the first minimum is half the width of the central peak