The simulation (linked below) is a head-on view of the screen that displays a diffraction pattern from a single slit of width a. The slit is incident normally with monochromatic light of wavelength .1 = 600 mm. The perpendicular distance between the plane of the slit and the viewing screen is D = 2.50 m. The diffraction pattern is shown with a y axis directed along the viewing screen perpendicular to the length of the slit. The red dashed line shows the 3; location of a detector. The simulation shows plots of the light intensity (scaled by the maximum intensity at the center of the central maximum) I/Im versus the location )2 of the detector. The plot on the right is a zoomed-in version of the plot of the left, showing the secondary maxima more clearly. The simulation also displays the value of III," for the given detector location )2. Simulation Intensity of Single-Slit Diffraction Pattern 8 Did you refer to the result from Module 36.2 for the intensity of a single-slit diffraction pattern, I = 14%)? Can you use this expression to deduce the value of a for the rst intensity minimum? In Module 36.2, it was useful to dene a new parameter a : W2, where 45 is the phase difference (in radians) between rays from the top and bottom of the slit. What is the value of a (in radians) at the center of the rst dark fringe? (1:0X rad Detector location (cm) v = 0.075 1/m 0.8 0.08L 0.06L 9.04 902 Ad - Le y (cm) -0.4 -0.2 0.2 0.4 MADA y (CM) 0.4 -0.2 0.2 0.4 0.4 1 / Im = 2.02e-15 0.2 y (cm) 0.0 -0.2 -0.4