Consider a Young\'s two-slit experiment in which the wavelength of light is 37%

Question

Consider a Young's two-slit experiment in which the wavelength of light is 37% smaller than the distance between the slits.
How many complete dark fringes would be seen on a distant screen? (Assume for all these questions that the screen is as large as it needs to be.)

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How many total interference maxima would be seen on a distant screen?

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Now suppose that the wavelength of light is 143% larger than the slit separation. How many complete dark fringes would be seen on a distant screen?

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Explanation / Answer

I'm assuming a "complete" dark fringe means all the way to the minimum, not all the way to the next maximum. Also I hope 48% "smaller than" means 52% as large. If not repeat calcs below using correct multiplicative factor. A. For an infinite-width screen, theta = 90 deg. Then the maximum path-length difference is ?L = d = 1.923 lambda. The total number of dark fringes to either side = 2 (one at ?L = 0.5 lambda and one at ?L = 1.5 lambda), for a total of 4. B. The total number of bright fringes to either side = 1, at ?L = lambda, for a total of 3 including the center maximum. C. 145% "larger than" means 245% times as large, I hope. If not repeat calcs below using correct multiplicative factor. Now we have max ?L = d = 0.408 lambda

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