WEBVTT
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welcome to the first section in our unit on wave
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optics. In this section, we're going to be
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discussing some of the most basic examples and wave optics
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of interference. Now we're going to start, though
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, by looking at the Huygens principle. So Huygens
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principle is Thea underlying principle of all wave optics.
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What it says is that we have a source that
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produces a wave front and we can consider each point
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on this wave front to be producing small wave lits
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. He called them that look like this. So
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each point on the wavefront is actually producing secondary wave
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lits and then to find where the next wave front
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will be, What we do is we draw a
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line that is tangent to the tip of all the
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wave. Let's like this and then to move to
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the next step, we would draw, wave,
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let's all over the second line and then again draw
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the line tangent toe all these wave lits. Okay
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, so hopefully you get the idea. It's a
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pretty simple principle. But as we'll see as we
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look at this interference, um, it is it
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is very powerful in helping us analyze interference. Now
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, looking at, uh, interference. Generically,
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we know that we have a couple of basic rules
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, the first one being that we have constructive interference
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and destructive interference. Remember, given a situation where
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we have two light sources that are producing the same
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wavelength of light, and they are a distance R
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one and R two from a point of interest than
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the difference in path length K. So that's Delta
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R is the difference in path length for the past
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R one and R two. That difference in path
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length must either be a integer multiple of love wavelengths
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in order to obtain construct of interference or a half
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into germ multiple of wavelengths in order to obtain destructive
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interference. And this principle is still true. It's
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still correct, even for light, as it was
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for sound, because this is a mathematical truth about
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waves. Um, but as we'll see, it
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has some different implications for what we'll do here with
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optics. The first optical experiment we want to consider
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is double slit interference. The idea with double slit
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interference is that you have a source of waves and
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that source of waves. The waves hit a double
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slit, so this is a wall with two slits
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in it. And when it does that, the
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two slits themselves now act as sources of light producing
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waves. And when we see how that interferes notice
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here, what's gonna happen is we're gonna have consistent
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positions where the waves overlap and what those are as
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thes our positions. These air going to lead to
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what are known as light bands, K or interference
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bands. So these bands or fringes, as they're
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called in some books are going to represent the different
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positions at which constructive interference is met, where we
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have the requirement that Delta R is equal to mm
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times lambda. So here we have m Z equals
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zero. Then we have to m equals one.
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And to m equals to notice that M equals zero
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is right down the center in the middle between the
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two slits now thinking about thes the screen here,
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the filter and then the wall that the light is
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hitting being separated by a distance. L we can
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consider what's the mathematics of this specific situation? Drawing
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out a large perspective here in thinking l to be
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much larger than d such that we would find the
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angles between r one and R two and the horizontal
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will have to be the same. They look like
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parallel lines as long as l is a far enough
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distance away. Yeah, that is to say much
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larger than D. When we do this, what
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we'll find is that the path difference between R one
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and R two all happens right here at the beginning
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because they're parallel After we go past this segment here
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, what that means is that our one in our
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two will have the same path. So what is
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this segment are two minus. Our one is going
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to be D, which is our high pot news
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here multiplied by the sign of data as we can
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see from our graphic here. So we have Delta
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r equals d sine theta must be equal to M
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times Lambda in order to have constructive interference. Now
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we can simplify this further recognizing that theta is generally
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going to be a very small angle, especially if
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l is very large. Remember, that's the distance
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between the filter and the wall. And if we
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make the small angle approximation, what we find then
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is that for these angles we will have constructive interference
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now. It could be difficult to major angles sometimes
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, but it's not so difficult to measure distances.
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So some people prefer to instead right out the position
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instead of the angle. So again, thinking about
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are the picture we here As we've blown it up
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, we have a distance l and an angle theta
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, which means the vertical distance along the wall is
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going to be l times the tangent of theta.
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Therefore, we come up with y equals l tangent
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data. Applying this small angle approximation, we again
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have tangent of data is approximately Thatta and we confined
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that the positions of our bright fringes are going to
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be m times Lambda Times L over d, where
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all we've done is plug in theano Gle that we
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found in the previous picture. Now again, the
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EMS can be 01 or two or any integer.
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But one thing we should recognize is that these are
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fringes. So it's a certain range over which we
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have, Ah, bright fringe. And then there's
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a space. Before we have another bite fringe.
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In between the two fringes, there is a dark
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fringe, and that dark fringe is going to be
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where the requirement for destructive interference is met. Which
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again, if we just apply the same logic we
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did before. If we just say m plus one
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half instead of em where M is restrained to India
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Cher's, then we'll find that these are the positions
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of our dark fringes. So this is very helpful
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. We have the positions of our bright fringes and
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the positions of our dark fringes. I should note
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that if you're having trouble with any of these images
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, I highly recommend you go and look at your
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textbook where they'll have very nice, uh, color
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images drawn by professionals. Um, that will perhaps
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make a little more sense to you. One note
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I should make here is that the intensity of the
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different fringes can be found using this equation. You'll
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see that there's a lot of variables here that we're
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kind of taking for granted D and lambda and L
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and why all of these different things you need to
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keep track of what they all mean. In this
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case, we've added a nine on which is the
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initial intensity of the light coming in before the filter
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so you can use this actually draw a picture.
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If I were to have an M equals zero here
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, I'd have the highest point and then it would
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come down and then we'd have it coming back up
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on either side. It would be symmetric. So
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where a Thai? We have a bright fringe where
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it's low, we have a dark fringe. So
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here are some bright fringes and here are some dark
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fringes. Okay? The other experiment we should take
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into account is something known as the diffraction grating.
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The diffraction grating is kind of similar to the double
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slit experiment. Except we have ah lot of slits
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in this case. And if we have, say
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, end number of slits that's going toe, all
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all of them are going to contribute to what we
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see on the screen. Now we end up seeing
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more or less the same pattern of light and dark
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fringes. Except we have very narrow, very bright
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fringes compared to the double slit experiment. This makes
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the diffraction grating especially helpful when we're looking working in
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a small lab where we can get very obvious results
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that are easy to measure and easy to see now
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in this case If we want to define our distance
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d, we're going to need to define it as
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millimeters per slits or, in this case, the
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number of slits per millimeters that would be n divided
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by the total length l or not. L but
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we'll call this h the height h of the diffraction
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grating in millimeters. We'll give you the distance in
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millimeters one over the distance and millimeters. So using
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that, we can actually same used the exact same
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formulas that we have before, except recognizing that we're
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going to have a maximum intensity equal to end squared
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, multiplied by our initial intensity high. Not now
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. One thing to keep in mind here for this
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fringe is that we are going to have a rather
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quick fall off with intensity where we're going to have
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a high intensity in the middle, and then it
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will fall off as we move away from the M
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equals one position. The same thing will actually happen
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with the doubles. The experiment, if you go
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back and think about cosine squared as why gets bigger
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here, Cosine squared moves farther and farther away from
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one. So we will have while the positions air
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. Correct. The sighs of these bumps will get
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smaller and smaller like this, so we won't just
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maintain a constant intensity for all fringes. It will
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be a small reduction as we move away from the
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M equals zero position. So these are our first
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two examples of interference using wave optics. Let's do
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a couple of examples really diving into each one.