Video Transcript
Which of the following formulas correctly relates the angle 𝜃 at which light with a wavelength 𝜆 emerges from a pair of narrow slits separated by a distance 𝑑 to the order 𝑛 of a bright fringe of an interference pattern produced by the light on a screen? Is it (A) 𝑑 equals 𝑛𝜆 sin 𝜃, (B) 𝑑𝜆 equals 𝑛 sin 𝜃, (C) 𝑑 sin 𝜃 equals 𝑛 over 𝜆, (D) sin 𝜃 equals 𝑑𝑛𝜆, or (E) sin 𝜃 equals 𝑛𝜆 over 𝑑?
All of these answers have the variables 𝑑, 𝜃, 𝜆, and 𝑛. In order to relate all of these variables together, we’re going to use some trigonometry. But before we do that, let’s just make sure we really know what our variables are. 𝜆 is the wavelength of light before and after it passes through the slits; it does not change at any point in the problem. Similarly, 𝑑 is the distance between the two slits in the screen and will also not change throughout the problem. 𝜃 is the angle that the light waves make as they come out of the slits with lines normal to or at 90 degrees with the screen with the slits in it. This angle will change depending on which bright fringe these two light waves will converge at.
And yes, even though these two light waves converge, we still say that they have the same angle despite knowing that they’re not actually parallel. And the reason that we do this is because the top 𝜃 value is so close to the bottom 𝜃 value that we can treat them as basically the same. They only differ by a few tiny, tiny fractions of a degree. So, it’s okay. Now, when these light waves converge and start to form these bright fringes, this is where we see the variable 𝑛. And this is because these bright fringes only occur at very specific points, where constructive interference occurs.
Constructive interference is a phenomenon in which two waves which are in phase with each other are incident at the same point. “In phase” means that these two waves, if we were to stack them on top of each other, would look the same at about the same moment in time. An easy way to check for this is to see if the peaks and the valleys of the waves match up at the same moments in time. If they do, then the waves are in phase, causing them to combine at the point where they converge, which in the case of light waves creates a more intense bright area, a bright fringe.
But to create this constructive interference, these light waves don’t have to line up exactly. There can be a gap and yet they can still constructively interfere at certain points. This gap length can vary, but in order for there to be constructive interference, the gap length must be exact integer multiples of 𝜆, which we represent as 𝑛𝜆, where 𝑛 is an integer.
Because this interference only occurs at very specific gap lengths or path length differences, it means that we can represent each of these bright fringes on the screen over here as having a separate value of 𝑛. So, this fringe in the center with an 𝑛 of zero has no path length difference. This one has a path length difference of one 𝜆, so just 𝜆. This one is two 𝜆, and so on. So, there is constructive interference whenever the path length differences between these two light waves is 𝑛𝜆, which we can represent with our diagram on the left like this.
In the case where these light waves are angled upwards, the bottom one will be longer. And if we draw a line down from the top of the slit down to the end of the path length difference such that it forms a 90-degree angle, then we have successfully formed a triangle that relates 𝑑 and the path length difference 𝑛𝜆. Now all we need is to have the angle 𝜃 somewhere in this triangle so we can relate all four of these variables. And it turns out it is.
To see how, let’s say that this angle up here is 𝜃 one. If we look at this large triangle right here, which forms a 90-degree angle between the screen and a line normal to the screen, then we would have a triangle that looks something like this. Its other angle, which is this angle right here, we’ll call 𝜃 two. So, this triangle has three angles, 𝜃 one, 𝜃 two, and a 90-degree angle.
Now, there is actually a smaller triangle right next to this triangle as well. And we see that it has an angle here of 𝜃, a right angle right here, and this angle right here is the same angle that it shares with the large triangle, 𝜃 two. So, the angles that the smaller triangle has are 𝜃, 𝜃 two, and 90 degrees. We see that both of these triangles share 𝜃 two and 90 degrees as angles. This means that since all triangles have the same number of degrees, 180, the third angles must be shared too, meaning that 𝜃 one and 𝜃 are actually the same angle, which means that this angle is actually just 𝜃, meaning that we now have a triangle that relates all four variables of 𝑑, 𝜃, and 𝑛𝜆. The triangle that contains these variables is a right triangle.
If we were to take the sin of this angle 𝜃, then we know it would be equal to the length of the opposite side, 𝑛𝜆, over the length of the hypotenuse 𝑑. Substituting in these values gives us the equation sin 𝜃 equals 𝑛𝜆 over 𝑑. So, the formula that correctly relates the angle 𝜃, wavelength 𝜆, slit distance 𝑑, and order 𝑛 of a bright fringe of an interference pattern produced by light on a screen is (E) sin 𝜃 equals 𝑛𝜆 over 𝑑.
