Video Transcript
Light with a wavelength of 675 nanometers passes through a sheet in which there are two parallel narrow slits 10.5 micrometers apart. The light from the slits is incident on a screen parallel to the sheet, where a pattern of light and dark fringes is observed. A line L runs perpendicular to the surface of the sheet and the direction of the slits. The line L intersects the central bright fringe of the pattern on the screen. What is the angle between L and a line that intersects the center of the bright fringe closest to the central bright fringe? Give your answer to two decimal places.
Now, when this question talks about the line that is not line L, this one right here, the one that intersects the center of the bright fringe closest to the central bright fringe, what it’s actually talking about is the line that we’ve already drawn here or a line that’s like this below line L. This line that the question is giving us can be either one of these, since both of these bright spots are the same distance away from the central bright fringe, meaning that they are both the bright fringe that is closest to the central bright fringe. This also means that the angle is the same too. So, for fun, let’s just stick with this configuration where the line is below L.
Now then, we want to find a way to relate the distance between the slits 𝑑 and the wavelength of the light passing through the sheet 𝜆 to this angle between the lines, which we’ll just call 𝜃. Let’s recall the equation that can relate all of these together. Whenever there are two light waves coming from the slits on a sheet, the path length difference between these two light waves is equal to 𝑑 sin 𝜃.
Now then, the value that we should use for the path length difference depends on if we’re looking at a bright fringe or a dark fringe, the area in between the bright fringes. If we are looking at a bright fringe, then it means that we are looking at areas where there is constructive interference, which occurs when the path length difference between the two light waves is equal to 𝑛𝜆, where 𝑛 is an integer. In this equation, 𝜆 is just the value of our wavelength, which is 675 nanometers. If it is not a bright fringe we’re looking at but a dark fringe, the areas in between the bright fringes, then it means we’re looking at areas of destructive interference, which occur when the path length difference between the two light waves is equal to the product of 𝑛 plus one-half and 𝜆, where again 𝑛 is an integer and 𝜆 is just the same 𝜆 that we’re given.
Now, we’re looking at the angle between L and a line that intersects the center of the bright fringe closest to the central bright fringe, meaning that we’re going to be looking at the bright fringes in this problem and have our path length difference be equal to 𝑛𝜆, thus making the equation 𝑛𝜆 equals 𝑑 sin 𝜃. We already know 𝑑, the distance between the slits, as 10.5 micrometers, and we already know 𝜆, which is 675 nanometers. So, in order to solve for 𝜃, we just have to find out what 𝑛 is equal to, which, remember, must be an integer. Because these bright fringes can only occur where the path length difference between the two light waves is 𝑛𝜆, then it must mean that each fringe has its own 𝑛-value.
To determine this value, let’s start by looking at our central bright fringe and considering what path length difference really means. The path length difference, which is to say the difference in distance the two light waves travel, is, in the case of the central bright fringe formed by these two light waves, zero. They both travel the same distance. And because we know that this distance has to be equal to 𝑛𝜆 — 𝑛𝜆 certainly isn’t zero, since we already know that it is equal to 675 nanometers — it means that 𝑛 has to be equal to zero.
This is very helpful now for determining the other values of 𝑛 for the other bright fringes, since we know that these values of 𝑛 have to occur in whole integer steps. So the bright fringes nearest to the area where 𝑛 is equal to zero must have 𝑛-values of one and the ones out from there, 𝑛-values of two, three, and so on. But we’re only interested in the bright fringes closest to the central bright fringe, which is where 𝑛 is equal to one. As previously noted, this could be the bright fringe that is directly above or below the central bright fringe. The path length difference is the same, and thus the angle that we would get 𝜃 is also the same.
So, now that we know our value of 𝑛, which is one, we can start solving this equation for 𝜃. Starting with 𝑑 sin 𝜃 is equal to 𝑛𝜆, we can divide both sides by 𝑑, canceling the 𝑑’s on the left side, leaving behind just sin 𝜃. And then we can take the inverse sine of both sides, which leaves behind just 𝜃 on the left side. Now, all we have to do is substitute in our variables. 𝑛 is one. 𝑑 is 10.5 micrometers. But we want all of our units to cancel. So we’re going to put both 𝑑 and 𝜆 in terms of just meters. So 10.5 micrometers in scientific notation is 1.05 times 10 to the power of negative five meters. Similarly, in scientific notation, 𝜆 is 6.75 times 10 to the power of negative seven meters.
Now then, substituting these values into the equation shows us that the meters cancel within it. And plugging the rest into our calculators and rounding the result to two decimal places, we find that the angle of 𝜃 is equal to 3.68 degrees. So the angle between L and a line that intersects the center of the bright fringe closest to the central bright fringe to two decimal places is 3.68 degrees.
