Light Waves and Colors Legacy Problem #22 Guided Solution

You go to the museum and you notice an exhibit on Young's experiment and in the experiment red light is being used and it has a wavelength of 648 nanometers and passes through a double slit and produces an interference pattern consisting of alternating bright, red, and dark spots. This interference pattern is projected onto a screen 16.8 meters from the slits. A metric ruler located above the interference pattern clearly shows that on this interference pattern of bright, red, and dark bands that the fifth dark band is 42.0 centimeters from the central bright spot. We wish to calculate the slit separation distance. This is clearly a Young's equation problem in which we wish to solve for d, the slit separation distance. So I'm going to take Young's equation, lambda equal yd divided by ml, and I'm going to rearrange it to solve for d. Doing proper algebra yields d equal m times l times lambda divided by wavelength. Now if all I can do, if I can find the ml wavelength and the y value, I can substitute it and I can solve for my unknown quantity d. Now it's advisable that you first get all of the quantities in the same unit so that when you, so that when you substitute in and solve, you're going to get a clean single unit for the d value. So here we go. I'm going to identify the known values beginning with wavelength. Lambda equals 648 nanometers, and I'm going to convert that to meters by dividing by ten to the ninth. After all, there's ten to the ninth little nanometers in one meter. What I do, I get lambda equals 6.48 times ten to the negative seventh meters. L is the distance from the slits to the screen, and that's stated as 16.8 meters. L equals 16.8 meters, no need to convert there, it's already in meters. Y equals 42.0 centimeters. That's the distance from the central bright spot to the fifth dark spot on one side of the pattern. I use m equals 4.5 for that since the dark fringes or dark bands are always assigned a half number to them. So if the central one is m equals zero, the first dark fringe is m equals a half. That means that the fifth dark fringe is a half less than five, thus m equals 4.5. And y equals 42 centimeters, I can convert that to meters, 0.42 meters. Now I'm ready to substitute into my equation. D equals m times L times lambda divided by y. And when I do, I can solve for D, and it comes out to be 1.1664 times ten to the negative fourth meters. I can round that to three significant digits, so that's 1.17 times ten to the negative fourth meters. And if needed, I can convert that to centimeters and millimeters. In millimeters it would be 0.117 millimeters.