Video Transcript
Light with a wavelength of 597 nanometers passes through a sheet in which there are two parallel narrow slits 7.64 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, intersecting the central bright fringe of the pattern on the screen. Two lines, line I and line II, intersect L at the position of the sheet. Line I intersects the center of the dark fringe closest to the central bright fringe, and line II intersects the center of the bright fringe closest to the central bright fringe. Both lines I and II are on the same side with respect to line L. What is the angle between line I and line II? Give your answer to one decimal place.
We’ve constructed this diagram using the information that the question has told us. So let’s clear some space by removing part of the question. Now then, to find the angle between line I and line II, let’s look at just the lines, line I, line II, and L. Line L goes straight across from the center of the slits all the way over to the central bright spot on the screen. This means that when light passes through the two slits, it constructively interferes at this point. Constructive interference occurs where the path length difference of two waves is 𝑛𝜆, where 𝑛 is an integer and 𝜆 is the wavelength of the waves.
In the case of this central bright fringe, there is no path length difference between the two waves. Both waves are traveling the same distance, meaning that the path length difference for the two waves that make up the central bright fringe is zero, which still satisfies the equation for constructive interference, since zero is an integer. So the value of 𝑛 for the central bright fringe is zero. The reason that we care about the path length difference is because we can use an equation that relates the distance between the two slits with this path length difference. This equation 𝑑 sin 𝜃 equals the path length difference contains an angle which is exactly what we need to solve for the angle between line I and line II. We’ll see what angle this 𝜃 is actually referring to later.
For now, let’s focus on finding the path length difference for line I and line II. For now, we’ll just move the equation up here so we don’t forget it. Let’s now look at our line that ends at our other bright spot, which is actually line II. Line II ends at the bright spot nearest to the central bright spot, or the one with line L. Bright fringes occur where there is constructive interference, and constructive interference only occurs when the path length difference is equal to 𝑛𝜆, where 𝑛 is an integer. So since the central bright fringe has an 𝑛 of zero, the next bright fringe up has to have an 𝑛 that is exactly one greater, so 𝑛 is equal to one here. This means that the path length difference for the bright fringe at the end of line II, which is constructive interference, is one 𝜆, or just 𝜆.
Now, let’s look at line I which ends at the first dark fringe. Dark fringes are the spaces of darkness between the bright fringes and occur where there is destructive interference. Destructive interference occurs where the path length difference of two waves is the product of 𝑛 plus one-half and 𝜆, where 𝑛 is an integer and 𝜆 is the wavelength of the two waves, except here 𝑛 refers to the dark fringes instead of the bright fringes. So where 𝑛 equals zero, we would have the first dark fringe of the screen, just like how for a constructive interference, 𝑛 equals zero refer to the first bright fringe of the screen. So since line I is the first dark fringe from the central bright fringe, its 𝑛 is equal to zero. Meaning that its path length difference is zero plus one-half times 𝜆, which is just of course one-half 𝜆, or 𝜆 over two.
Now that we have the path length differences for line I and line II, let’s find out where exactly 𝜃 comes from by taking a closer look at these slits. When we are zoomed-in enough to actually see 𝑑 which, remember, is on the scale of micrometers, then the angle that the light waves make as they pass through the slits is extremely similar, which we can see by drawing lines normal to the slits. The differences in angles between these two waves are so small that we can consider them the same angle, despite knowing that they do eventually converge on the opposite screen.
So these two lines have the same angles between them. But where does 𝜃 come in? It turns out that 𝜃 actually comes from an angle near 𝑑. If we draw a straight line from the top slit down to the first light wave such that it forms a 90-degree angle with the light wave, then this angle here is 𝜃. The base of the triangle here that is then formed expresses the difference in distance that these two light waves travel, which is to say that the length of the base of this triangle is the path length difference between these two waves. So it seems like we have everything that we need to find 𝜃. We have 𝑑, and we have the path length difference.
But this angle 𝜃 seems rather strange, since it is not in the orientation we would expect for measuring the angles between line I and line II. We would actually want one of the angles here which, remember, are basically the same angle, which means if we had a line coming from the very center like, say, line L, then we could treat both of these light waves as just one light wave coming from the center. And we could measure that angle there, and it would be the same as the ones above and below it. This line then is actually line I or II, since it’s going directly to one of the fringes. So we just have to find the angle between L and line I or II, the angle that is again the same as the angles between the top or bottom triangles. So we can find this angle right here as part of the smaller triangle.
For now, let’s call this angle 𝜃 𝐴. When we look at our small triangle right here, we see that we have two other angles, a right angle and another small angle here, which we can call 𝜃 𝐵. And we can see that 𝜃 𝐵 is also an angle present in this larger triangle here. By noting that this larger triangle has a right angle right here, we can see then that the larger triangle is made of three angles 𝜃, 𝜃 𝐵, and this right angle. And similarly the smaller triangle is made of the angles 𝜃 𝐴, 𝜃 𝐵, and the right angle. Side by side, these triangles look something like this, both with a right angle, an angle 𝜃 𝐵, and then the angle 𝜃 or 𝜃 𝐴. And since both of these triangles are triangles, they have to have the same number of degrees within them, which means that 𝜃 must be equal to 𝜃 𝐴.
This means that the angle between line L and line I or II will be the angle 𝜃, which changes depending on whether we’re looking at the rays for line I, line II, or any other line that we want to draw. So all we have to do now to find the angle between line I and line II is solve for 𝜃 in this equation. Let’s start by solving for the angle between line L and the line I, which we’ll denote by having a subscript one for 𝜃. The path length difference for line one is 𝜆 over two. We then want to isolate sin 𝜃 one, which we’ll do by dividing both sides by 𝑑, causing the 𝑑’s to cancel on the left side of the equation. And on the right side, the 𝑑 just goes into the denominator.
We can then take the inverse sine of both sides, causing the original sine to cancel on the left side and leaving behind just 𝜃 one. The value of 𝜆 is 597 nanometers, or written slightly differently 597 times 10 to the power of negative nine meters. And 𝑑, the distance between the two slits, is 7.64 micrometers, which is 7.64 times 10 to the power of negative six meters in scientific notation. Substituting these values into the equation and using our calculators, we find that the value of 𝜃 one, the angle between line L and line I, is equal to about 2.239 degrees.
Now let’s find 𝜃 two, which is the angle between line II and line L. So we’re going to use the path length difference of 𝜆. The equation is reduced the same way. Both sides are first divided by 𝑑. And then the inverse sine is taken of both sides, leaving behind 𝜃 two. Wavelength is 597 times 10 to the power of negative nine meters, and the distance between the slits is 7.64 times 10 to the power of negative six meters. When we put these values into our calculator, we get 4.482 degrees.
The last thing we have to do is find the angle between line I and line II. So this means we just have to find the difference between these two angles. So 𝜃 two minus 𝜃 one equals 4.482 degrees minus 2.239 degrees, which is about equal to 2.24 degrees, which rounded to one decimal place is 2.2 degrees. So the angle between line I and line II to one decimal place is 2.2 degrees.
