# Question Video: Determining the Total Number of Bright Fringes on a Screen Physics • 9th Grade

Light with a wavelength of 588 nm passes through a sheet in which there are two parallel narrow slits, 12.6 𝜋m apart. The light from the slits is incident on a screen parallel to the sheet, 1.52 m away, where a pattern of light and dark fringes is observed. The screen is 2.24 m long, and its center is aligned with the midpoint between the slits in the sheet. How many bright fringes are observed on the screen?

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### Video Transcript

Light with a wavelength of 588 nanometers passes through a sheet in which there are two parallel narrow slits, 12.6 micrometers apart. The light from the slits is incident on a screen parallel to the sheet, 1.52 meters away, where a pattern of light and dark fringes is observed. The screen is 2.24 meters long, and its center is aligned with the midpoint between the slits in the sheet. How many bright fringes are observed on the screen?

So what this question wants to know is how many bright fringes are observed on the limited screen space of 2.24 meters, since it is possible for the light waves coming from the slits to go beyond the screen, meaning of course that those particular waves won’t form a bright fringe on the screen. So, to see how many total will fit on the screen, let’s consider comparing the total length of the screen with the individual distance between the fringes, the basic idea being that we can take this total length of the screen and divide it by the individual length between the fringes, which will hopefully give us the number of fringes present.

This may seem nice and simple, since we already know the total length of the screen. But the length between the bright fringes depends on many factors: the wavelength of the light coming through the slits, the distance between the slits, and the distance from the slits to the screen.

To see exactly how all of these relate to each other, let’s begin by looking closer at these slits. When two light waves come out of these slits to meet on the screen and form a bright fringe, they are coming out of the slits at some angle, which while not technically the same angle, since they do eventually converge to form one of these bright fringes rather than staying parallel forever, the difference is so small that they’re basically the same angle. This is because the distance between the slits is very small, compared to the distance between the sheet and the screen. Therefore, we can say that the differences in angles that these waves have will be extremely small, meaning we can basically say they’re the same.

The angles we see here may look a bit different, but remember that’s because they are not to scale. These angles over here are a better representation. Now, by this same logic of having the differences between these angles be very small, if we were to include a line coming from the very center of the slits and similarly have a light wave come from the center of these slits and end at a bright fringe, then the angle between this line and this light wave would also be the same. When referring to any three of these angles, we would always use just the same angle of 𝜃.

Now, the reason that we’ve put this middle line here is that now we’ve created a triangle with a right angle. If we adjust our triangle so that the length of this side of it over here is equal to the distance between two bright fringes, then we can start doing some trigonometry. This triangle here is the same as this triangle over here. Let’s look at what we know about it. It is a right triangle with another angle here that we know is 𝜃. The bottom length we already know as 1.52 meters. But so as not to clutter any equations, we’ll just call it 𝐿 for now. For top length, the hypotenuse we’ll just call 𝐻. And the far side we’ll say has a length of 𝑌 𝑛, which represents the distance between the two bright fringes, in this case the central bright fringe and the bright fringe directly above it.

Now, the reason that we’re using the subscript 𝑛 is because we usually represent specific bright fringes with specific values of 𝑛, denoting their distance from the central bright fringe, which has an 𝑛 of zero. Each bright fringe you have either above or below the central bright fringe has an 𝑛-value one higher than the previous bright fringe. 𝑌 𝑛 here represents the distance between any two of these bright fringes. But for convenience, we’re doing it between 𝑛 equals zero and 𝑛 equals one because that gives us a right triangle. The distance between the bright fringes, that is to say, the value 𝑌 𝑛, is the same between any two adjacent fringes. But the resultant triangle wouldn’t be a right triangle, which would make things way harder. So, we’ll be sticking with 𝑛 equals zero and 𝑛 equals one.

Now then, let’s start relating this angle 𝜃 to the other parts of this right triangle. Recalling some trigonometric rules, we know that sin 𝜃 is equal to the length of the opposite side from the angle over the length of the hypotenuse, which in this case would be 𝑌 𝑛 over 𝐻. The reason we wanted to find this sin 𝜃 relation is because sin 𝜃 is used in this equation, which gives us the path length difference for constructively interfering points, which is to say the bright fringes. We’ll go over what the variables are in the equation in just a moment. But for now, let’s focus on the sine function.

We want to substitute in this value of sin 𝜃, but we can’t do it in its current form, since we don’t know the length of 𝐻, the hypotenuse. Because of this, we’re going to have to do a fancy trick involving the tan of 𝜃. For any right triangle, tan 𝜃 is equal to the length of the opposite side over the length of the adjacent side, which in this case becomes 𝑌 𝑛 over 𝐿. This is handy because we now have a known variable here with 𝐿, which is much better than sin 𝜃, which has two unknown variables 𝑌 𝑛 and 𝐻.

And it turns out there’s a way to relate these two together using something called the small angle approximation. The small angle approximation is basically any time you have a very small angle, like the one we have in this triangle, the sin 𝜃 and tan 𝜃 of that angle are essentially the same, which means for this triangle sin 𝜃 is equal to 𝑌 𝑛 over 𝐻, which is about equal to 𝑌 𝑛 over 𝐿. So when substituting in the value of sin 𝜃 for this equation, we can use 𝑌 𝑛 over 𝐿. So we can start substituting things in here now. But first we have to consider, why this equation? How is this equation going to help us find how many bright fringes are observed on the screen? Well, let’s consider the variables at play here.

𝑑 is the distance between the slits, 12.6 micrometers. 𝜆 on the far side over here is the wavelength of light, 588 nanometers. But this variable 𝑛 here is what we’re really after. The variable 𝑛 represents the 𝑛th bright fringe that a pair of waves is converging to from the central bright fringe. So if we know all of the other variables, we can solve for the maximum 𝑛th bright fringe on the screen. And if we know the maximum 𝑛th bright fringe, then we know that all of the other bright fringes under it must exist too, which means that 𝑛 can be used to find the total number of bright fringes on the screen. So let’s go about solving for it.

The first thing we’re going to want to do is isolate this value 𝑛, which we’ll do by dividing both sides by 𝜆, causing the 𝜆’s on the right side to cancel, leaving behind just 𝑛. Next, we can substitute in sin 𝜃 with 𝑌 𝑛 over 𝐿, which gives us 𝑑 𝑌 𝑛 over 𝐿 over 𝜆, which can be simplified to 𝑑 𝑌 𝑛 over 𝜆𝐿. Now that we have the equation completely in terms of 𝑛 with variables that we know, let’s start looking at those variables.

𝑑 is the distance between the slits and is equal to 12.6 micrometers. But let’s put this in scientific notation for ease of calculation later. 12.6 micrometers in scientific notation is 1.26 times 10 to the power of negative five meters. Next, the wavelength of light coming through the slits is 588 nanometers or 5.88 times 10 to the power of negative seven meters in scientific notation. Next is 𝐿, the length from the screen to the sheet, and is 1.52 meters. This leaves us now with 𝑌 𝑛, the distance between the bright fringes, which we don’t seem to have. But 𝑌 𝑛 does not have to be between two directly adjacent bright fringes. It could be from, say, the central bright fringe all the way up to the maximum allowable bright fringe, in which case the triangle that would be formed would have a 𝑌 𝑛 equal to half of the screen length of 2.24 meters, which would be 1.12 meters.

But there are not just bright fringes above the central bright fringe, but below it too, which means measuring from the central bright fringe, we would have two values of 𝑌 𝑛, one for those above and one for those below the central bright fringe. Meaning we would add these two together to find the total area where the bright fringes would be, which is just the original length of the screen, 2.24 meters. Since we have seen that we can take 𝑌 𝑛 to be the entire length of the screen, the variable 𝑛, which counts the number of bright fringes within the length of 𝑌 𝑛, will tell us the total number of bright fringes on the screen.

Now, when we do this, it may seem like there is a problem, specifically because of the way that we took the sin 𝜃 in this equation. In order to assume that sin 𝜃 was about equal to tan 𝜃 in order to get the variable 𝐿 instead of 𝐻, we used the small angle approximation, which only works if 𝜃 is very small. And we can clearly see that the angles given when we’re measuring over the entire screen are very much not small. But this is actually not a problem because what this equation is actually doing is only measuring between any two bright fringes directly next to each other at a time and simply summing them all up together, which means that each individual summation always maintains a small angle, which means that the small angle approximation that we made is still fine.

Alright then. Now, with that out of the way, let’s start substituting in our variables, which should look something like this. And since we have meters twice on the top and meters twice on the bottom, all of the meters should cancel out. So plugging all of this into our calculators, we should get about 31.579. But we’re looking for how many full bright fringes are on the screen. You can’t have part of a bright fringe. Yet, we would not round up in this case, even though the number after the decimal point is a five. This is because again you can’t have half of a bright fringe, and we’re only looking for how many total full bright fringes there are on the screen. Even if this was 31.99, we would still say that there are only 31 bright fringes observed on the screen. So this specific setup, with light coming through the slits onto the opposite screen, will produce 31 bright fringes.