Light passes through a sheet in which there are two parallel narrow slits 12.8 micrometers apart. The light from the slits is incident on a screen parallel to the sheet, where a pattern of bright and dark fringes is observed. A line 𝐿 runs perpendicular to the surface of the sheet in 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 3.09 degrees. What is the wavelength of the light? Give your answer to the nearest nanometer.
So we want to find the wavelength of light 𝜆. And we are given the distance between the two slits, 12.8 micrometers, and the angle between line 𝐿 and another line coming from the center of the slits to the first bright fringe closest to the central bright fringe, 3.09 degrees. We are also being given another variable through the bright fringes. Any time that you see lines intersecting these bright spots or hear the phrase “bright fringe,” you should immediately think of constructive interference and how when the path length difference between two waves is 𝑛𝜆, where 𝑛 is an integer, there is always constructive interference.
For instance, for this central bright fringe here, there is no path length difference between the two waves coming from the slits that interfere at that point. The path length difference here is still 𝑛𝜆. It’s just that 𝑛 in this case is zero. And because this constructive interference which creates these bright fringes can only occur where 𝑛 is an integer, it means that all of these other bright fringes have an integer value of 𝑛 as well, moving in full integer steps up or down from the central bright fringe where we know that 𝑛 is equal to zero. So when the question tells us that there is a line that intersects the center of the bright fringe closest to the central bright fringe, this one right here, it’s telling us that the value of 𝑛 for that bright fringe is equal to one.
So we have the distance between the two slits, 12.8 micrometers, the angle between 𝐿 and this line, 3.09 degrees, and the value of 𝑛 for the path length difference between two light waves meeting up at this point being equal to one. We now just have to relate all of these variables to the wavelength 𝜆. To do this, we’re going to look closer here at the distance between the two slits and do some trigonometry. Here, we have the two slits, the distance between them being 12.8 micrometers. From these slits, two light waves will come out to meet at the screen opposite this sheet, meeting up and forming the bright fringe corresponding to 𝑛 equals one.
If we were to draw lines that were perpendicular to the sheet, then we could use those lines to find the angle that the light waves make as they exit the slits. Because the screen that these light waves are traveling to is so far away, the difference in the angles between these two light waves is extremely small, meaning that we can treat these two angles as basically the same angle. If these angles from the top and bottom slits are the same, then an angle made between two lines in the center of the slits would be the same as well, say, between a line 𝐿 and a line that intersects the center of the bright fringe closest to the central bright fringe where 𝑛 is equal to one. And since we already know this angle as 𝜃, then the other angles are 𝜃 as well, since these are all the same angle.
With this knowledge of the angle 𝜃, let’s relate it to the slit distance 𝑑 and the path length difference between the two light waves. Recall that the path length difference between two waves that eventually constructively interfere, or form a bright fringe, is 𝑛𝜆, which in this case, where 𝑛 is equal to one, the path length difference is just 𝜆. This means that the bottom light wave here is traveling exactly one 𝜆 further than the top light wave here, hence the phrase “path length difference.” If we were to draw a line from the top slit of the sheet down to the end of the path length difference such that we have a 90-degree angle, then we would have a right triangle with a hypotenuse 𝑑 and a known side length of 𝜆. But we still have two unknown angles and an unknown side, or do we?
Let’s look back at this triangle over here and see what happens when we extend this line down a little bit. That’s right. We just made more triangles, which can help us find our unknown angles. Let’s look at this large triangle that’s formed by this line right here. It has a distance 𝑑 here as one of its sides. Its bottom side is a portion of the line perpendicular to the sheet, which forms a 90-degree angle, making this a right triangle. And of course the hypotenuse is just this line extended downwards. Because it uses this same line and the distance 𝑑, both this triangle and the triangle that uses 𝜆 as one of its sides have the same angle here, which we’ll just call 𝜃 one. The other angle that this triangle has, located here, we’ll call 𝜃 two.
Now let’s compare this triangle with the very small triangle right here. One of its sides is the path length difference 𝜆, but it shares its two other sides with this line and this line, both of which make up the larger triangle. Because these lines and these lines are the same, they must share the same angle, meaning that this angle here is 𝜃 two. And because the angle between the line that was drawn down and the light wave was 90 degrees, so too must it be 90 degrees on the other side as well. Meaning this angle is 90 degrees.
Now when two triangles share two angles that are the same, in this case, the 90 degrees and the 𝜃 twos, it means that the third angle must also be the same, since all triangles are made up of the same number of degrees, 180. Meaning that this last angle on the small triangle must be 𝜃 one. And we actually already know the angle between the light wave and the line perpendicular to the sheet. It’s just 𝜃, meaning that 𝜃 one is actually just equal to 𝜃. They’re the same angle.
With this knowledge, we can now get a triangle that relates 𝑑, 𝜃, and 𝜆 together, which would look like this. Since this is a right triangle, we can use the relation sin 𝜃 equals the length of the opposite side over the length of the hypotenuse. The opposite side of angle 𝜃 is 𝜆, and the hypotenuse of this right triangle is 𝑑, giving us the relationship sin 𝜃 equals 𝜆 over 𝑑. We want just 𝜆, so let’s multiply both sides by 𝑑, causing the 𝑑’s on the right side to cancel out, leaving behind 𝑑 sin 𝜃 equals 𝜆. The value of 𝑑 is 12.8 micrometers, or in scientific notation 1.28 times 10 to the negative five meters. And 𝜃 is of course 3.09 degrees.
Substituting these values into the equation and subsequently calculating gives us an answer of 6.899 times 10 to the negative seven meters. But we want this in the nearest nanometer, which are on the order of 10 to the negative nine meters. So we can move the decimal place here two spaces to the right to make this to the power of negative nine, which is nanometers. Now rounding to the nearest nanometer, the value that we’ve found for this wavelength of light is 690 nanometer.