By the same reasoning that worked with three slits, you can see that no matter how many slits you have, the maxima will still fall at the same locations as the maxima for two slits. The peak intensity of the maxima will be proportional to N2, where N is the total number of slits. The energy at the screen is roughly equal to the product of the number of maxima, the peak intensity of a maximum, and the width of a maximum. As N increases, the number and location of the maxima will not change, while the peak intensity of the maxima will increase proportionally to N2. If the total energy available increases proportionally to N , how does the width of the maxima change? View Available Hint(s) for Part G The width does not change. The width is proportional to N . The width is proportional to N2 . The width is proportional to 1/N . The width is proportional to 1/N