Single Slit Diffraction - Online Tutor, Practice Problems & Exam Prep

Single slit diffraction results in a unique pattern of alternating bright and dark fringes due to the interference of light waves. The central bright spot is notably larger and brighter than the others, which are uniform in width. The dark fringes are calculated using the equation $\sin (\theta$_m)=mλ/d, where m indexes the dark fringes. Understanding this concept is crucial for grasping wave behavior and interference patterns in physics.

Single Slit Diffraciton

Hey, guys. In this video, we're going to talk about single slit diffraction. So what happens to the light as it passes through a single slit as opposed to what we saw before with a double slit. Alright. Let's get to it. Now light shown through a double slit had unexpected results as we talked about if you don't consider diffraction. Okay? If you do not consider diffraction, then it's an unobvious result. And, obviously, back before they understood diffraction, they had certain expectations for the experiment, and the experiment turned out differently. Likewise, light shown through a single slit also displays the same sort of unexpected result, which we call a diffraction pattern, which is alternating spots of brightness and darkness. Right? The big difference between the double slit experiment and the single slit experiment is concerned with the central bright spot. Okay? In a double slit, the central bright spot is just as wide as all of the others. It's the same width as all the others. So every single bright spot across the entire screen is going to be of uniform width. But in a single slit, the central bright spot is actually twice as large as all of the other ones. It's also considerably brighter. So that central one is definitely going to be larger than any of the other bright ones. But all the other bright spots, all the other bright fringes are going to have the same width. Okay? That only applies to the central bright fringe. All of the dark fringes have the same width in the single slit just as they did in the double slit experiment. Okay?

Let me minimize myself so we can see this figure. Like in a double slit, the diffraction pattern is produced due to interference. Okay? The big difference between the double slit and the single slit is that in the double slit you actually have 2 sources of light that are interfering. In the single slit, you have one source of light. It's just that light leaving at the top part of the slit and light leaving at the bottom part of the slit does not leave at the same angle. Okay? Light leaving different parts of the slit leaves at different angles. Okay? So you have all of these different angles that the light can travel at leaving both slits. Okay? Sometimes 2 beams of light will arrange themselves so that when they arrive they're both at a peak. Right? When they arrive on the screen they're both at a peak. This produces constructive interference, and light that constructively interferes produces bright fringes. The amplitude of the light increases under constructive interference. Other times, you can have a wave arrive at a peak and another wave arrive at a trough. And when you have a peak and a trough meeting, you have destructive interference, and light that destructively interferes produces a dark spot or a dark fringe. Okay? Because with destructive interference comes a smaller amplitude for the interfered wave. Smaller amplitude means it's darker. Okay? Just like we did for the double slit experiment, we talked about the single slit conceptually, and now we want to actually talk about the mathematics of solving single slit problems. Where are these fringes actually located? Okay.

Now the key difference in the math between the single slit and the double slit experiment is that in the single slit experiment you only have an equation for dark fringes. Okay? Dark fringes are located at angles given by sinθ_m=mλd. Okay? Where m is our indexing number this time. Okay? But because m indexes the dark fringes and not crucial to remember there's no m equals 0 index for dark fringes due to a single slit. Okay? So what we have here is we have the first dark fringe or the m equals 1 dark fringe. We have a corresponding m equals 1 dark fringe on the other side and then we would have the m equals 2 dark fringe on the top side and the bottom side. Right? And then we could say some arbitrary m, the nth dark fringe, is given by θ_m where theta follows this equation. Okay?

Now solving problems with a single slit is going to be exactly the same as solving problems with a double slit. Okay? The first thing that we're going to do is we're going to draw the figure that I have above me in the green box for a single slit, and it's going to look identical to the figure for a double slit. Okay? Let me minimize myself so I can draw this off to the side. Alright. So here's our single slit. Right? This drawing is going to look absolutely identical to that of a double slit. The only difference is that I've physically drawn 1 slit instead of 2. Now I'm going to draw that central axis. Okay?

And, well, let me read the problem before continuing. A 450nm laser is shown through a single slit of width 0.1mm. If the screen is at a distance of 1.4m away from the slit, how wide is the central bright spot? Okay. So the screen is a distance of 1.4m away which is equivalent to 1.4m. Okay? And what we are looking for is the width of the central bright spot. So the central bright spot looks like this, right? And it'll continue and there'll be a second bright spot and a third bright spot, etc. The equation that we have for the single slit tells us the locations of the dark fringes. So we know this is the m equals 1 dark fringe, This is the equivalent m equals 1 dark fringe on the bottom side. And this is theta1, that first dark fringe angle, and this is theta 1 as well. And notice this distance is just that distance that we are looking of the central bright spot. Okay?

Now to make solving this problem...

Light from a 600 nm laser is shown through a single slit of unknown width. If a screen is placed 4.5 m behind the slit captures a diffraction patter with a central bright fringe of width 20 mm, what is the width of the single slit?

• A. 0.13 mm
• B. 0.11 mm
• C. 0.44 mm
• D. 0.26 mm

Number of Dark Fringes on a Screen

Hey guys. Let's do an example. Light from a 500 nanometer laser is shown through a single slit of width 0.5 millimeters with the screen placed three and a half meters from the slit. If the screen is 2 centimeters wide, how many dark fringes can fit on the screen? Okay. To illustrate what's going on in this problem, I'm actually going to draw an extra big diagram. Okay? Because this problem is a little bit different than problems that we've seen before. Now I'm going to draw my central axis. Okay. We're told that the screen is three and a half meters behind the single slit, and we are told that the screen itself is 2 centimeters wide. Okay? And we want to know how many dark fringes fit. Okay? So we're going to have a dark fringe and a bright fringe and a dark and a bright. Right? These are alternating. The huge central bright fringe, etc. Something like that. Okay?

Now there's going to be a maximum angle that light can leave the single slit and still reach the screen. Right? That's this angle right here to the edge of the screen, and I'll call this theta max. Clearly, if light leaves at a larger angle than that, it will not reach the screen. It'll go past the screen, and so you won't see a diffraction pattern. The diffraction pattern ends at the screen's width. Okay? We haven't really addressed in any problem, either a double slit or a single slit problem, what happens when you consider the finite width of the screen. Okay? So this is the first problem that we're actually talking about that. But if you exceed this maximum angle, then obviously the light just goes past the edge of the screen and it's not captured and you don't see anything. Okay? How do we relate this to the number of dark fringes? Well, notice that each dark fringe I I actually skipped one here. Hold on. Let me reposition myself. There's one, there's two, there's three, there's four, etc. However many fit, right? This is the m equals 1 dark fringe. This is the m equals 2. This is the m equals 3. This is the m equals 4. Something important to remember is that the dark fringes are evenly spaced angularly. Okay? So whatever angle is right here, which I'll call phi, the angle between the central axis and the first ark fringe is the same angle between the first dark fringe and the second dark fringe. And that's going to be the same as the angle between the second and the third dark fringe which is going to be the same as the angle between the third and the fourth dark fringe. Okay? They are evenly spaced angularly. Alright? Now notice that the angle phi is equal to the angle between the central axis and the first ark fringe. This angle right here. Well that we know by definition has to be theta 1. So the first bit of information that we know is that phi equals theta 1, and that's really important. Our equation for the angles for dark fringes in a single slit is that sin(θm)=mλd. So that tells us that sin(θ1)=λd, right, when m is 1. Okay? We know what lambda is. We know what d is, so we can find sin(θ1) and then theta 1. This is a 500 nanometer laser. Nano is 10 to the negative 9. The width of the slit is half a millimeter so this is 0.5 millis 10 to the negative 3. And if you plug this into your calculator you get an answer of 0.001 which is equivalent to theta1 equals 0.057 degrees. Okay? And remember that that equals phi, the angle between each successive dark fringe. Okay?

Now how do we find out how many dark fringes are there? Well if each dark fringe is spaced by phi, and we have some total angle theta max, then the number of dark fringes is just going to be theta max divided by phi. Okay? So the number of dark, I'm just going to call it number dark, is the maximum angle that light can still reach the screen divided by phi. Okay? This isn't quite right though. The reason it's not quite right is because look at theta max. Theta max is actually the angle between the central axis and the furthest ray. But that's not going to tell us how many dark fringes we have. That's only going to tell us how many dark fringes we have on this side. You need the total angle, this angle, to figure out how many dark fringes you have across the entire screen. So technically this should actually be 2 times theta max over phi. Okay? So this is going to be the equation that we're going to use to figure out what the number of dark fringes are. We already know phi. Now we need to find theta max. Okay? Well notice that theta max is just given by this triangle right here. We know what the base of this triangle is. It's three and a half meters, right? What's the height of this triangle? Well the width of the entire screen is 2 centimeters so this should just be half that or 1 centimeter. So now I can just redraw that triangle. Alright? And that should give us theta max. The adjacent edge, right, the base is 3.5 meters and the height I said was 1 centimeter. Okay. So we know the opposite edge, we know the adjacent edge, so we can use the tangent. So the tangent of theta max is going to be the opposite edge which is 1 centimeter, since he's 10 to the -two and the adjacent edge is the base or three and a half meters. If you plug this into your calculator you get 0.0029 which means that theta max is 0.16 degrees. Okay? Now we also have theta max. Okay? So all we need to do is plug theta max and phi into the above equation to find the number of dark fringes. That's 2 theta max over phi. So that's 2 times 0.16 degrees. It's fine to leave this in degrees. You don't need to convert it to radians. And this is 0.057 degrees. The width, the angular width, or the angular distance between each dark fringe, and plugging this into your calculator you get 5.6. Okay, so you can fit, or you can find at least 5 dark fringes in those 2 centimeters. Okay? You would also partially have some bright fringes hanging off the edge of both of those ends. Okay? Because you actually have 5.6, right? If this was 5 on the nose then you would have a dark fringe at the edge at each edge of the screen. But since it's slightly above 5 you have some sort of brightness at the edge of each screen. It doesn't end at a dark fringe. Alright guys. That wraps up this problem. Thanks for watching.

Here’s what students ask on this topic:

What is single slit diffraction and how does it differ from double slit diffraction?

Single slit diffraction occurs when light passes through a narrow slit and produces a pattern of alternating bright and dark fringes due to interference. The central bright spot in single slit diffraction is notably larger and brighter than the other fringes, which are uniform in width. In contrast, double slit diffraction involves two slits, resulting in multiple sources of light that interfere. The central bright spot in double slit diffraction is the same width as the other bright spots. The key difference lies in the central bright spot's size and brightness, and the number of light sources causing interference.

How is the location of dark fringes in single slit diffraction calculated?

The location of dark fringes in single slit diffraction is calculated using the equation:

$\theta =\frac{\mathrm{m\lambda }}{d}$

where $m$ is the fringe order (1, 2, 3, ...), $\lambda$ is the wavelength of the light, and $d$ is the width of the slit. This equation gives the angle $\theta$ at which the dark fringes occur. Note that there is no $\mathrm{m=0}$ dark fringe in single slit diffraction.

Why is the central bright spot in single slit diffraction larger and brighter than the other fringes?

The central bright spot in single slit diffraction is larger and brighter because it results from the constructive interference of light waves that travel nearly parallel paths through the slit. This central maximum is twice as wide as the other bright fringes and has a higher intensity due to the greater number of light waves contributing to the constructive interference at this central point. The other bright fringes are formed by light waves that interfere constructively at specific angles, resulting in uniform but narrower and less intense bright spots.

What is the equation for the width of the central bright spot in single slit diffraction?

The width of the central bright spot in single slit diffraction can be determined by first finding the angle $\theta$ for the first dark fringe using the equation:

$\theta =\frac{\lambda }{d}$

Then, the width $w$ of the central bright spot on a screen at distance $L$ from the slit is given by:

$w=2L\cdot \theta$

This accounts for the distance from the central maximum to the first dark fringe on either side.

How does interference cause the bright and dark fringes in single slit diffraction?

In single slit diffraction, interference occurs because light waves leaving different parts of the slit travel at different angles. When these waves meet, they can interfere constructively or destructively. Constructive interference happens when waves align with their peaks and troughs matching, resulting in bright fringes. Destructive interference occurs when a peak of one wave meets the trough of another, canceling each other out and creating dark fringes. This alternating pattern of bright and dark fringes is the diffraction pattern observed on the screen.