3. [25 points]
Point B is the third minimum, corresponding to a D Dadj of 4/3 l ,. Solving the equation dsinq =4/3l ,.for q gives 19.5° .
d sin q = nl , where n=0,1,2 etc. for the principal maxima. If d remains constant, sin q increase if the wavelength increases. Therefore, the angles to each of the principal maxima will increase and the adjacent principal maxima will move farther apart.
The location of the principal maxima does not depend on the width of the slit. The principal maxima will remain in the same location.

Was the change in intensity caused by covering slit 1, slit 2, or either slit 1 or 2? Explain your reasoning.
Since point A was originally a secondary maxima, D Dadj for each of the three slits is 1/2 wavelength. The path length difference for slits 1 and 3 is then a full wavelength. If slit two is covered, the light from slits 1 and 3 will interfere constructively, increasing the intensity at point A. If slit 1 is covered, the two sources would be out of phase at point A. Therefore, the intensity would increase at point A only if slit 2 were covered.