Young's Double Slit Experiment: Study with Video Lessons, Practice Problems & Examples

Young's Double Slit Experiment

Guys, in this video, we're going to talk about Young's double slit experiment, which is just an analysis of how light rays are going to interfere with one another after initially collimated light passes through a double slit. Okay? Let's get to it. A beam of light that's shown onto a double slit, that just means it's aimed at a double slit. Okay? Was thought initially before diffraction was understood to produce a single spot of brightness. Okay? However, when the experiment was performed, that wasn't the case. So in this image here on the left, we have the expected outcome of the experiment. If we shine this initially collimated light through a double slit, we expect to get just uniform brightness at a single point on the screen, the width of the slit. Okay? But what actually happens, what was demonstrated, is that there are multiple bright spots on the screen. That this initially collimated light, when it passes through the double slit, does not produce a uniform brightness, a constant brightness at one spot. It actually produces a varying brightness that alternates between bright and dark along the screen, and it produces a diffraction pattern. That's what we call this. This diffraction pattern is due to the lights interfering, so this is due to interference. That as these waves pass through these slits, they come out at different directions. So maybe here, you get constructive interference, and you get a bright spot. Maybe down here, you get destructive interference, and you get a dark spot. But that pattern of alternating brightness and darkness, that diffraction pattern is produced entirely by those two light waves interfering with one another when they reach the screen. Okay? And this is entirely due to diffraction, because if without diffraction, sorry, that collimated light encountering the slit goes through undisturbed. It goes through collimated. But with diffraction, when that collimated light encounters the slit, it spreads apart and comes out isotropic. So you get all those different angles for the light coming through. Okay? Now each slit produces a ray in every direction. Alright? Isotropic, right? The same in all directions. This means that some rays are going to interact constructively. You can see in the figure that here we have a maximum, and here on the top right we have a minimum. And when you have a maximum and a minimum, or a peak and a trough, you get destructive interference. Right? Down here you can see that both waves are at a maximum. When both waves are at maximum or both waves are at a minimum, you get this constructively interfere, and this depends upon where along the screen you want to look at the lights. If the light happens to constructively interfere, that means that the amplitude is going to get larger, you get bright, what are called, fringes. Up until this point, I've been calling them spots, but they're technically called fringes. Okay? That's the physics term for this. Some of those light rays are going to also destructively interfere, and they're going to produce a light with less amplitude or a dark fringe. And we're going to get alternating spots or alternating dark and light fringes, alternating spots of brightness and darkness. Okay? Now that's all conceptually, but how do we actually apply math to figure out where these bright spots are or the bright fringes and where the dark fringes are? Okay? The bright fringes are located at angles given by this one formula, that sinθ=mλd. Okay? Now what's m? M is what we would call an indexing number, and it's allowed to be 0, 1, 2, 3, etcetera. All positive integers and 0. The reason we have this so-called indexing number is because there is not a single angle where brightness occurs, a bright fringe, but there are a multitude of angles where this occurs. Okay? I have on this figure here the diagram that you're always going to draw when analyzing a double slit problem. Okay? Here we have the double slit where d is the width oh, sorry, the separation between the two slits. And we're going to draw a dashed center line that lines up with the central brightness peak, the m=0 peak. And if we choose an arbitrary bright spot, which is given by an arbitrary index, θmm is the angle that the light has to travel at to reach that bright spot. Okay? Likewise, we can also find the angles that dark spots occur at or dark fringes occur at. This is indexed by a different letter, n. Okay? It really doesn't matter that we use m and n. Maybe your book uses I and j. Maybe it uses m twice. That's also popular. The indexes are different for dark fringes and bright fringes. Okay? Because they occur at different positions, and the equation is slightly different. This is n+12λ/d. Okay? Where d once again is the separation between the two slits, n is the indexing number for the dark fringes, and λ is just the wavelength, what lambda has always been. Okay? And you want to memorize this figure because you're always going to draw this figure when trying to solve double slit problems. Alright? Let's do a quick example. A 650 nanometer laser is shown through a double slit of 10 10 millimeter separation. What angle is the 4th brightest fringe located at? If this double slit is 2.8 meters from the screen, how far from the brightest fringe is the 4th brightest? Okay? So let's draw our little diagram of the double slit problem, which is always what you want to start with. Okay? Now you're going to have a central bright fringe, and then you're going to have a less bright one, a less bright one, and a less bright one. Okay? And this is the first brightest. This is the second brightest. This is the third brightest, and this is the fourth brightest. Now remember that the equation for the angle for bright fringes is indexed by m, we called it. So the question is, what m does the fourth brightest fringe occur at? Well, m can be 0, 1, 2, 3, 4, etc. 0 is that first brightest fringe, that central fringe. The second brightest is m equals 1. The third brightest is m equals 2, and that fourth brightest is actually m equals 3. Okay? So this problem was written in a very specific way to teach you that just because this is the fourth brightest fringe, doesn't mean you have to assume that m equals 4. In this case, because m starts at 0, m actually equals 3. Okay? So we want to find that angle θm3 that the n equals 3 fringe occurs at or the fourth brightest fringe occurs at. So sinθ3=3λd, right? Which is going to be 3 lambda, the wavelength, we're told to 650 nanometer laser. Okay? So 650 nanos times 10 to the negative 9, and d is that separation distance between the slits. We're told that there's 10 millimeters of separation or 10 times 10 to the negative 3. And so plugging this into a calculator, you get 0.000195. Okay? Or you get that λθ3 equals 0.011 degrees. Okay? Something to note here is that this is an incredibly small angle. The angles are always really going to be very, very small. Alright? And that's because actually to get this equation that we use, we actually have to assume that the angles are always small. This might be something that you read in your book. This might be something your professor points out, but that's the reason why the angle should be small. Okay. They're always going to be this point 011 degrees. Alright. It's not even it's barely over a hundredth of a degree. It's a very, very small angle. Okay? Now the next part of the question is, if the double slit is 2.8 meters away from the screen, how far from the brightest fringe is the fourth brightest? We want to know this distance. How far is it? And I'll call that distance y. Okay? Look at this. This is just a triangle. So I'm just going to draw this triangle. This angle is our θm3 angle, which is 0.011 degrees. The base of the triangle is 2.8 meters, right? That's how far the screen is from the double slit, and the vertical distance sorry, let me actually call that y sub 3. That's just that vertical distance of that third, that m equals 3 peak. Okay? And the height of this triangle is going to be y. So clearly I can use trigonometry. I have the opposite edge and the adjacent edge, so I can use tangent, and say tangent of our angle 0.011 degrees equals the opposite edge y 3, sorry, over the adjacent edge, which is 2.8 meters. Or we have that y 3 if I just multiply 2.8 meters up, we have y 3 equals 2.8 meters times tangent 0.011 degrees, which plugging into our calculator is just 0 point 00054 meters or 0.54 millimeters. Okay? So you see how small that angle is? Even though this horizontal distance is 2.8 meters, because that angle is so small, because it's this tiny angle, the height y 3 is only half a millimeter roughly. Alright, guys. That wraps up this talk about Young's double slit experiment. Thanks for watching.

A 450 nm laser shines light through a double slit of 0.2 mm separation. If a screen is placed 4 m behind the double slit, how wide are the bright fringes of the diffraction pattern?

• A. 9.0 mm
• B. 4.5 mm
• C. 2.2 mm
• D. 0.055 mm

Unknown Wavelength of Laser through Double Slit

Hey guys, let's do an example about Young's double slit experiment. A laser of unknown wavelength shines monochromatic light through a double slit of 0.2 millimeter separation. If a screen is 5.5 meters behind the double slit, you find the angular separation of each bright fringe to be 0.15 degrees. What is the wavelength of the laser? Okay. First, I want to discuss this big word right here, monochromatic. Okay? This means single-colored. Mono is Latin for 1. Chromo is Latin for color. Monochromatic, single-colored. This just means light at a single wavelength. Okay? And lasers are most typically monochromatic. Okay? But there are multichromatic lasers. Okay? So it's specifically monochromatic. The light is only emitted at a single wavelength and that wavelength is unknown. That's what we want to find. So, the first thing we're gonna do is we're gonna draw the situation because that's what we always do with these double slit problems. We're told that the screen is 5 and a half meters behind the double slit. And what we are told is that the angular separation of each bright fringe is 0.15 degrees. What does this mean? Well, we have our central bright fringe, right, and then the second bright fringe and the third bright fringe etcetera. What this is saying is the angle separating each bright fringe is 0.15 degrees. If I were to draw a line through the next bright fringe, that angle would also be 0.15 degrees. If I were to draw a line through this next bright fringe right here, this would also be 0.15 degrees. The separation between every single bright fringe, the angular separation, is 0.15 degrees. Notice that first ray that I drew, this blue one right here, this has to be our theta₁ angle because that's the angle for m equals 1. We're talking about the angle between the m equals 0 fringe and the m equals 1 fringe.

To find the angular location of bright fringes, our equation is:

sinθ_m=mλd

We know that the angle we're looking for is of that second bright fringe, the one just after the central bright fringe which occurs when m equals 1. So we have m equals 1 and so sin( theta₁ ), that angle that we want to find, is 1λ/d. λ is our unknown. We know what theta₁ is. Right? It's 0.15 degrees. And for theta₂, the angle at the location of the m equals 2, do you think it'd be 0.15 degrees? No. It's the entire sweep of this angle. So it's 0.15 + 0.15 which is 0.3. And you could also solve this problem by finding the theta₂ angle. That would also tell you the same wavelength. Length. Alright? And you guys can double-check what I'm saying at the end of the problem if you want for the theta₂ or the theta₃ or the theta₄. Okay? But for theta₁ we know what d is. The separation is 0.2 millimeters. We know what theta₁ is. It's 0.15 degrees. Wavelength is our unknown. So I'm gonna multiply the d up to the other side and this lambda is just d*sin(theta₁) which is 0.2 millimeters, so times 10 to the negative 3 meters and theta is 0.15 degrees. So our wavelength is 5.24 times 10 to the negative 7 meters. Okay?

Now, you can leave it like this and be done. That's the answer. But I'm gonna rearrange this slightly. What I'm gonna do is I'm gonna increase the order of magnitude by 2. I move the decimal place over 2 points which is going to increase my order of magnitude by 2. So I need to decrease my exponent by 2. Alright, if I'm gaining 2 in the decimal place, I have to lose 2 in the power. Alright, this times 10 to the negative 9 is nanometer. So this is 524 nanometers. And you're gonna see most of your problems are gonna describe wavelength in nanometers because on the 100 of nanometer, about 450 nanometers to 750 nanometers, if I remember correctly. That is visible light. That's light that you can see. 450 nanometers is purple light, the lowest wavelength light, and 750 nanometers is red light, the highest wavelength light. Alright guys. That wraps up this problem. Thanks for watching.

Here’s what students ask on this topic:

What is Young's double slit experiment and what does it demonstrate?

Young's double slit experiment is a fundamental demonstration of the wave nature of light. In this experiment, collimated light passes through two closely spaced slits and creates a pattern of alternating bright and dark fringes on a screen. This pattern, known as a diffraction pattern, results from the interference of light waves. Constructive interference occurs where the light waves are in phase, producing bright fringes, while destructive interference occurs where the waves are out of phase, producing dark fringes. The experiment provides strong evidence for the wave theory of light, showing that light can exhibit properties of both waves and particles.

How do you calculate the positions of bright and dark fringes in Young's double slit experiment?

The positions of bright and dark fringes in Young's double slit experiment can be calculated using specific equations. For bright fringes (constructive interference), the equation is:

$\sin \left(\theta \right)=m\lambda /d$

where m is an integer (0, 1, 2, ...), λ is the wavelength of the light, and d is the separation between the slits. For dark fringes (destructive interference), the equation is:

$\sin \left(\theta \right)=(n+1/2)\lambda /d$

where n is an integer (0, 1, 2, ...). These equations help determine the angles at which the bright and dark fringes appear on the screen.

What is the significance of the indexing number in Young's double slit experiment?

The indexing number in Young's double slit experiment is crucial for identifying the positions of the bright and dark fringes. For bright fringes, the indexing number m (0, 1, 2, ...) indicates the order of the fringe, with m = 0 representing the central bright fringe. For dark fringes, the indexing number n (0, 1, 2, ...) helps locate the dark fringes between the bright ones. These numbers are essential for applying the equations:

$\sin \left(\theta \right)=m\lambda /d$

for bright fringes and

$\sin \left(\theta \right)=(n+1/2)\lambda /d$

for dark fringes, where λ is the wavelength and d is the slit separation.

Why are the angles in Young's double slit experiment typically very small?

The angles in Young's double slit experiment are typically very small because the separation between the slits (d) is usually much larger than the wavelength of the light (λ). This results in small values for the sine of the angle (sin(θ)) in the equations:

$\sin \left(\theta \right)=m\lambda /d$

for bright fringes and

$\sin \left(\theta \right)=(n+1/2)\lambda /d$

for dark fringes. Since sin(θ) is small, the angles θ themselves are also small. This small-angle approximation simplifies the mathematical analysis and is a common assumption in the study of diffraction and interference patterns.

How does the separation between the slits affect the diffraction pattern in Young's double slit experiment?

The separation between the slits (d) in Young's double slit experiment significantly affects the diffraction pattern. A larger slit separation results in fringes that are closer together, while a smaller slit separation causes the fringes to spread further apart. This relationship is evident in the equations:

$\sin \left(\theta \right)=m\lambda /d$

for bright fringes and

$\sin \left(\theta \right)=(n+1/2)\lambda /d$

for dark fringes. As d increases, the angles θ for both bright and dark fringes decrease, leading to a more compact pattern. Conversely, decreasing d increases the angles θ, resulting in a more spread-out pattern. This demonstrates the inverse relationship between slit separation and fringe spacing.