 Early 19th century, everyone is convinced that light consists of a stream of particles because that is what the famous and unquestionable Isaac Newton thought. And this, although Newton, who allegedly had an apple fall on his head, could not explain how diffraction works or how Newton's rings, named after him, are formed, which you can observe for example on the surface of a soap bubble. He was firmly convinced that light propagates in rays consisting of many spherical light particles. An ophthalmologist and physicist named Thomas Young was tired of Newton's assertions. He wanted to show that light is a wave. So to prove this theory, Young had to come up with an experiment that would make properties of light visible that no ordinary particle could ever have. And one of those properties is diffraction. Light must be able to propagate around a corner just like a sound wave if the corner is small enough. A particle could never do that. It would hit an obstacle and bounce back. A big enough wave on the other hand would simply pass the obstacle and continue to propagate unhindered. A light wave should behave like a sound wave that is bent around a corner and is audible to a person standing around the corner. But light should also show interference like for example water waves. So if two light waves interfere constructively with each other, then the resulting wave should be brighter. It would be easy to think up such an experiment with water, but not with light. Otherwise, Newton would certainly have already tried it. The double slit experiment was born. Probably one of the most important experiments in physics with which you can directly show that light has a wave character as if it were like sound or water. And the double slit experiment setup is simple. All you need is a monochromatic light source, a screen and two narrow slits closely together. That's why it's called the double slit experiment. The setup is ready. Now what do you observe on the screen when you switch on the light source? The light moves to the double slit and passes through the two slits. Then it lands on the screen and creates alternating bright and dark fringes of light. It produces the same pattern as a water wave passing through two slits. The stripe in the middle is a bright light fringe of zero order. All the following bright fringes are first order bright fringes, second order bright fringes, etc. And dark stripes are called first dark fringes, second dark fringes, and so on. What you see here on the screen is a typical behavior of waves. If you would bombard the double slit with any kind of particles, you would never get such a pattern. You would simply see two stripes on the screen where the particles have landed. Exactly what you would normally expect from particles, one stripe because of one slit and the other stripe because of the other slit. This is why the classical double slit experiment is so fascinating. It shows a property of light that Newton and others didn't believe in. It seems that light does not behave like a particle but rather like a wave. In order to understand how this light pattern comes about, you need to know what happens when waves hit each other. They interfere. The waves can interfere constructively or destructively. Constructive interference occurs exactly when the two identical waves have a path difference delta s that corresponds to a multiple of the wavelength lambda. Delta s is equal to m times lambda. m is an integer. For a destructive interference, we shift the wave by half the wavelength and add a multiple of the wavelength. So the path difference is now one half plus m times lambda. Since the interference fringes occur symmetrically on both sides of the middle fringe, it is sufficient to look at only one side of the screen. For this purpose we only use positive integers for m. Also we want to reserve m equal to zero for the main fringe. Thus m ranges from one to infinity. However, there is a small problem if we insert one for m. The path difference delta s of the first order fringe would be then three halves of the wavelength. But this should be one half of the wavelength. To correct this, we rewrite the relation by subtracting one and at the same time adding one to m. The condition for destructive interference is then m minus one half times lambda. Let's now send such plane waves to the double slit. The red lines represent the maxima of the waves seen from above. As soon as the light waves arrive at the double slit, we use Huygens principle. This states that from every point on the wave front, a spherical wave is created. If the slit is very thin, then it appears like a point source of light that emits a spherical wave. This can propagate in the whole space behind the slit. Hopefully nobody has switched off the light, so the spherical waves pass constantly through the slit. As you can see, the movement of light waves is not restricted by the slits as expected, unlike particles that would mainly move straight through the slits. But how does the pattern of bright and dark fringes on the screen come about? These red lines mark the wave maxima. Exactly in between are thus the wave minima. Let's mark them with dotted lines so that we can follow them better. When the waves from the two slits meet, they interfere. Mathematically, interference means that the amplitudes of the waves add up. Amplitude of one wave plus the amplitude of the other wave gives the amplitude of the resulting wave. So we know where two solid lines intersect, a wave maximum meets another wave maximum, and constructive interference takes place. And where two dashed lines meet, the minima of the waves amplify to form a resulting wave minimum twice as large. This is also constructive interference, but the amplitude is just negative. And do you know where the waves cancel each other out completely? Exactly, where a solid line intersects a dashed line. These are points where a wave maximum meets a wave minimum. And everything in between is partial interference. There is partly cancellation and partly amplification. That's why you don't see any sharp lines on the screen, but rather the bright fringes merge evenly into dark fringes. If you follow the points of constructive interference up to the screen, you will reach the maxima. The zero order fringe occurs in waves with a path difference of zero, that is m equals zero. The first order fringe occurs at the path difference of exactly one wavelength, that is m equals one, and so on. Now you understand where this numbering of the fringes comes from, namely from the number m, which indicates how large the path difference of the interfering waves is. So what about the dark fringes? Follow the points of destructive interference up to the screen and you will see that they lead to the dark fringes. These dark fringes are described by the integer m, which runs from one to infinity. The first dark fringe with a path difference of half lambda, you get when m equals to one, and so on. In this way Mr. Young has demonstrated with his double slit experiment that light shows interference, which clearly indicates the wave character of light. You can go even further by using the double slit experiment to find out the wavelength of the light. For this, let's use a monochromatic light source that emits plain light waves. A laser is perfectly suited for this. Since we have assumed that the waves are plain, every point of the wave front arrives at the double slit at the same time. This means the waves have not experienced any path difference on the wave from the light source to the double slit. So if a wave maximum arrives at the upper slit, at the same moment a wave maximum arrives at the lower slit. Because of the symmetry, the waves from the two slits must arrive simultaneously in the middle of the screen. There is no path difference there. It's clear because the waves from both the upper and the lower slit travel the same distance to the middle of the screen. Therefore an interference maximum occurs in the middle of the screen. That is a bright fringe. For delta S equals zero, obviously M must be zero. This is where the term zero order maximum comes from. If you go up a little bit from the middle, the wave from the lower slit has to travel a longer distance to the screen than the wave from the upper slit. The waves going to the first bright fringe must have a path difference that corresponds exactly to one wavelength. The path difference of the waves going to the second bright fringe must be exactly two wavelengths and so on. When you go in the opposite direction, the path difference behaves the same way. So if you follow the waves up to the bright fringes, then you use the condition for constructive interference. Because there, the path difference is exactly multiple of the wavelength. On the other hand, if you follow the waves to the dark fringes, you obviously use the condition for destructive interference. In this case, the path difference for the first dark fringe is one-half lambda. For the second dark fringe, three-halves lambda. For the third dark fringe, five-halves lambda, and so on. Unfortunately, the value for path difference is unknown, which is why you cannot determine the wavelength from it. Are we now supposed to measure the wavelength with a ruler? No, that's impossible. Therefore, you have to find other experimentally accessible physical quantities and try to find out the wavelength of light with their help. What can you easily measure in this double slit setup here? For example, you can measure the distance from the double slit to the screen with a folding ruler. Let us call this distance A. But not only that. You can also measure the distance X from the main maximum to a bright fringe, for example, with a ruler. Note, however, that you must always measure exactly from the middle of one fringe to the middle of the other fringe. Because, only exactly in the middle of the bright fringe, the condition for constructive interference is fulfilled. The same applies to dark fringes. Always measure from the middle. You almost see a right angle triangle here. So let us complete the triangle by drawing the hypotenuse of the triangle. The hypotenuse shows you under which angle theta, in this case, the second bright fringe lies. So that it doesn't get too complicated, we'll make two approximations. They are justified if we consider a few things. The first approximation is as follows. The distance A between the screen and the double slit is much larger than the distance X of two fringes on the screen. So in order for this approximation to be applicable, we have to place the screen as far away as possible from the double slit in the experiment. This in turn means that the angle theta becomes very small. Then the distance A becomes approximately as large as the hypotenuse. This results in the geometrical relation sin theta is approximately equal to X over A, because sin theta is defined as the ratio of the opposite side to the hypotenuse. For the second approximation, we use a double slit with slits lying as closely together as possible. By this, we want to achieve that the slit distance G is much smaller than the screen distance A. If this is the case, then the paths of the two waves close to the double slit run more or less in parallel. By the way, since the screen distance A is very large, the whole long picture wouldn't fit in. Therefore, a large piece was cut out here. The cut position is indicated by the dotted gray line. Just look at this smaller right angled triangle. Here we call the angle beta. The opposite side of this angle is the path difference. And the hypotenuse is the slit distance G. From this triangle, you can read off. By a simple geometrical consideration, you can find out that the angles theta and beta are equal. Since the screen distance A is very large, this angle here is 90 degrees. And since the sum of angles in a triangle is 180 degrees, this angle must be 180 degrees minus 90 degrees minus beta. So the sum is 90 degrees minus beta. The angle between A and G is, as you can see, 90 degrees. These 90 degrees are composed of the angle theta and 90 degrees minus beta. From this follows, theta is equal to beta. With the help of the second approximation, you know that the angle beta corresponds to the angle theta and you can replace it. Just equate the two approximations and you get a very important equation. If you measure the distance from the center of the screen to a bright fringe, then you insert in Delta S the condition for constructive interference. And if you measure the distance from the center of the screen to a dark fringe, then you insert the condition for destructive interference in Delta S. Know that now all physical quantities are known. X, A and G. Also M is known. To determine M, you count in this case all dark fringes to the dark fringe that is at distance x from the center. Just rearrange the formula for the wavelength and voila, you can use a double slit experiment to determine it. So, that's it. Just imagine. With a double slit experiment, you have not only shown that light has a wave character, but now you are also able to find an important property of light. It's wavelength. From now on, you are the ruler of light. With this in mind, bye and see you next time.