Video Transcript
Light with a wavelength of 625 nanometers passes through a sheet in which there are two parallel narrow slits. 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. The angle between 𝐿 and a line that intersects the center of the bright fringe closest to the central bright fringe is 2.67 degrees. What distance separates the slits on the sheet? Give your answer to the nearest micrometer.
So what we’re looking to do here is a way to relate the distance between the slits on the sheet, which we’ll hereafter call 𝑑, to the other variables that we are given. Here, the wavelength is 625 nanometers and the angle we are given is 2.67 degrees. Importantly, this angle is the angle between a line going to the central bright fringe and a line going to the center of a bright fringe closest to the central bright fringe. When there are bright fringes involved, we should always think of constructive interference because that is what causes the bright fringes.
Constructive interference occurs between two waves of wavelength 𝜆 any time their path length difference is 𝑛𝜆, where 𝑛 is an integer. This means that these bright fringes can only occur at integer values of 𝑛 for this path length difference. If we were to draw out the waves coming from the two slits that lead to the central bright fringe, we would find that they have no path length difference, which is to say the 𝑛-value of the central bright fringe is zero. What they’re saying when a problem mentions the central bright fringe is that that is the bright fringe that has a value of 𝑛 equal zero. And because these bright fringes can only exist in whole integer steps of 𝑛, it must mean that the bright fringe closest to the central bright fringe would have an 𝑛-value of one, meaning that the path length difference between the two light waves coming from the slits would be 𝑛𝜆.
Let’s recall the equation that can relate all of these variables together when constructive interference is involved. 𝑑 sin 𝜃 equals 𝑛𝜆, where 𝑑 is the distance between the slits. 𝜃 is the angle between 𝐿 and another line that intersects the center of another bright fringe. And the right side of the equation is the path length difference, which in this case is 𝑛𝜆, since we are looking at where there’s constructive interference at the bright fringes.
Now, in order to solve for the distance between the slits 𝑑, we will have to divide both sides by sin 𝜃. Doing this causes sin 𝜃 to cancel on the left side, leaving behind just 𝑑. Now, let’s look at the other variables we are given and substitute them in to this equation. We’re measuring the angle between the line 𝐿, which is at the central bright fringe, and the first bright fringe from the central bright fringe, which is to say where 𝑛 equals one. And we’ve already been given the value of the angle, 2.67 degrees, and the value of the wavelength, 625 nanometers, which for ease of calculation we’ll convert to scientific notation as 6.25 times 10 to the negative seven meters. Substituting all of these values into the equation and subsequently using our calculators to solve gives a result of 1.342 times 10 to the power of negative five meters.
Now, we want this answer to the nearest micrometer, which in scientific notation is on the scale of 10 to the power of negative six. Since we’re already at the power of negative five, to get this in our answer, all we have to do is move the decimal place one spot to the right, which makes the answer 13.42 times 10 to the power of negative six meters, or micrometers. So, rounding our answer to the nearest micrometer gives us an answer of 13 micrometers. So the distance that separates the slits on this sheet is 13 micrometers.
