Video Transcript
Which of the following formulas correctly relates 𝜆, the wavelength of light that emerges from a pair of narrow slits; 𝑠, the separation between the slits; lowercase 𝑑, the distance from the center of an interference pattern produced by the light on a screen; and uppercase 𝐷, the distance from the slits to a bright fringe in the pattern of order 𝑛?
Let’s start by determining where each of these variables will appear on a diagram that shows the interference pattern produced by a pair of narrow slits. The interference pattern comes from this light wave as it passes through or is incident upon the pair of narrow slits. The separation between the slits forms our variable 𝑠. The wavelength of the light wave before it passes through the slits is 𝜆, and after it passes through the slits, it splits into two. And both of these light waves also have the wavelength of 𝜆. It is unchanged. The difference then between these two light waves on their way to a single point is the distance that they travel.
One of these light waves is traveling a slightly longer distance than the other one. This distance is called the path length difference. Finding the path length difference of every point on the screen by manually measuring every single line possible from the slits can get tedious pretty quickly. So instead, we solve for this trigonometrically. The lengths of these two lines combined with the slit distance forms a triangle. We can draw a line from the shorter path length to the longer path length, forming a 90-degree angle. This then allows us to find this angle here, which we’ll call 𝜃. We know the length of the separation between the slits 𝑠, which would be this length right here.
Now we can use our understanding of sin 𝜃. For a right triangle, sin 𝜃 is equal to the length of the opposite side of 𝜃 divided by the length of the hypotenuse. The opposite side here is the path length difference, which is what we want to find, and the hypotenuse is just 𝑠. This means that for this triangle, sin 𝜃 is equal to the path length difference divided by 𝑠. If we multiply both sides by 𝑠, the 𝑠’s cancel on the right-hand side of the equation, meaning that the final equation is the product of 𝑠 and sin 𝜃 is equal to the path length difference.
Now, the reason that we did all of this is because we’re looking for the bright fringes in the pattern. When this light wave passes through the slits, bright fringes occur at regular intervals on the screen and are created by constructive interference by the two waves as they go through the slit. Constructive interference is when two waves, in this case the first wave from the first line and the second wave from the second line, happen to line up, which is to say the waves are doing the same things at the same times. We also call this being in phase. When these two in-phase waves are incident at the same point, they constructively interfere, making a wave with greater amplitude, which in the case of light waves makes it brighter.
So in order for these waves to constructively interfere and form these bright spots, they must be in phase with each other. And it turns out that these waves are in phase with each other when the path length difference between them is 𝜆 or any integer multiple of 𝜆, which we’ll represent with an 𝑛. So for there to be a bright fringe at a point, the path length difference must be equal to 𝑛𝜆, which is also equal to 𝑠 sin 𝜃. Looking back at the question text, we have the variables 𝜆, 𝑠, and 𝑛. But now we need to find the lowercase 𝑑, the distance from the center of an interference pattern, and uppercase 𝐷, the distance from the slits to a bright fringe.
The way that we find these is by doing an interesting trick. We assume that these two light waves, despite matching up at one point, are parallel to each other. We can do this pretty safely actually because the distance that these light waves travel is very large compared to 𝜆. When we assume that these lines are parallel, we can draw out horizontal lines from the narrow slits and find angles with the original light wave lines. This angle is the same for both lines, and it happens to be 𝜃 because this is the same angle. And because we’re assuming that these lines are parallel, we now no longer need both lines, but rather just one coming from the center of the slits, forming an angle 𝜃 with a line horizontal from the center of the slits.
This single line here now represents the distance from the slits to a bright fringe in the pattern, which means that it represents uppercase 𝐷. And we also have the distance lowercase 𝑑, which is from the center to an interference pattern. So we have a right triangle with an angle 𝜃, a known side opposite this 𝜃, and the hypotenuse. Recall that sin 𝜃 is equal to the length of the opposite side over the length of the hypotenuse, which is just lowercase 𝑑 and uppercase 𝐷, respectively. This means that sin 𝜃 is equal to lowercase 𝑑 over uppercase 𝐷, which we can substitute into this equation here.
Looking at the available answers, they all have 𝑠 on one side of the equation, so we should start doing the same thing. In order to get rid of this uppercase 𝐷 in the denominator, we multiply both sides by uppercase 𝐷, causing the uppercase 𝐷s on the left side to cancel out. This leaves 𝑠 and lowercase 𝑑 on the left side. By dividing both sides by lowercase 𝑑, they can cancel on the left side, leaving behind this equation: 𝑠 equals the product of 𝑛𝜆 capital 𝐷 divided by lowercase 𝑑.
So the formula that correctly relates 𝜆, the wavelength of light; 𝑠, the separation between the slits; lowercase 𝑑, the distance from the center of an interference pattern produced by the light on a screen; and uppercase 𝐷, the distance from the slits to a bright fringe in the pattern of order 𝑛, is (C).
