Practice: 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?

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Diffraction | 9 mins | 0 completed | Learn |

36.02 (A, n/a): Diffraction with Huygen's Principle | 15 mins | 0 completed | Learn |

Young's Double Slit Experiment | 24 mins | 0 completed | Learn |

Single Slit Diffraction | 28 mins | 0 completed | Learn |

Concept #1: Single Slit Diffraciton

**Transcript**

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 the 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 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 peaks sorry alternating spots of brightness and darkness.The big difference between the double slit experiment and the single slit experiment is concerned with the central bright spot 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 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 of the bright fringes are going to have the same width. 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. Let me minimize myself so we can see this figure, like in a double slit the diffraction pattern is produced due to interference.The big difference between the double slit and the single slit is that in the double slit you actually have two 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 light leaving different parts of the slit leave at different angles so you have all of these different angles that the light can travel at leaving both slits. Sometimes two 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 one wave arrive at a peak and another wave arrives 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 because with destructive interference. Comes a smaller amplitude for the interfered wave smaller amplitude means its darker. 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. 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, in the double slit we have two equations one for the bright fringes one for the dark fringes for the single slit we only have one for the dark fringes. Dark fringes are located at angles given by sine of theta M equals M lambda over D. Where M is our indexing number this time but because M indexes the dark fringes and not the bright fringes it turns out that a requirement is we cannot have M equals zero, this is crucial to remember there is no M equals zero index for dark fringes due to a single slit. 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 fringes on the top side and the bottom side right, and then we could say some arbitrary M. The Mth dark fringe is given by theta sub M where theta follows this equation. Now solving problems with a single slit is going to be exactly the same as solving problems with a double slit 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 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 I physically drawn one slit instead of two now I'm going to draw that central axis and let me read the problem before continuing a 450 nanometer laser is shown through a single slit of width 0.1 millimeters if the screen is at a distance of 140 centimeters away from the slit how wide is the central bright spot? So the screen is a distance of 140 centimeters away which is equivalent to 1.4 meters and what we're looking for is the width of the central bright spot so the central bright spot looks like this right and it will continue and there will 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 theta 1 that first dark fringe angle and this is theta 1 as well and notice. This distance is just that distance that we are looking for this is the width which I'll call W, that's the width of the central bright spot now to make solving this problem easier we can notice that the top triangle the triangle above the horizontal axis and the triangle below the horizontal axis are identical so this height Y1 and this height Y1 are the same.Now all we need to do is find the angle theta 1 using our equation so that we can find Y1, so first remember that our equation for the location of dark fringes in a single slit experiment is M lambda over D where M starts at 1 remember this is different than the location for the bright spots on a single slit experiment where M started at 0. We're looking for a theta 1 so M is just going to be 1. What's lambda its a 450 nanometer laser nano is 10 to the -9.What is the single slit width D. D was also different in the single slit than it is in the double slit in the double slit D represented the separation between the two slits for a single slit D represents the width of the single slit. We're told that the single has a width of 0.1 nanometers sorry millimeters and milli is ten to the power of -3, plug this into a calculator you're going to find that this is 0.0045 or that theta 1 is 0.26 degrees. Let me minimise myself again now we can see we've just found theta 1 with this triangle right here we can then find Y1 so I'm going to redraw that triangle. The angle is theta 1 which we know is 0.26 degrees the base is 1.4 meters and the height we called Y1 and once again the width is not going to be Y1 the width is actually going to be 2 times Y1 so let's solve this triangle notice that we have the opposite edge and the adjacent edge to that angle so we consider the tangent of the angle is equal to the opposite edge divided by the adjacent edge and so all I have to do is multiply 1.4 up to the other side of the equation.Plug this into a calculator and we get an answer of 0.0064 meters which is equivalent to 6.4 millimeters don't forget this is not the final answer, this is simply Y1 the width W which is what we're looking for is 2 times Y1 it's twice this height or the sum of those two heights but they're the same so this is 2 times 6.4 millimeters which is 12.8 millimeters.Now don't forget that the central bright fringe is twice as wide as all the other bright fringes so what would the width of all the other bright fringes be it would just be Y1 which was 6.4 millimeters. Alright guys that wraps up this video on the single slit diffraction. Thanks for watching.

Practice: 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?

Example #1: Number of Dark Fringes on a Screen

**Transcript**

Hey guys, let's do an example. Light from a 500 nanometer laser is shone from a single slit a width 0.5 millimeters, with a screen placed 3 and a half meters from the slit. If the screen is 2 centimeters wide, how many dark fringes can fit on the screen? To illustrate what's going on in this problem I'm actually going to draw an extra big diagram because this problem is a little bit different than problems that we've seen before. Now I'm going to draw my central axis. We're told that this is, the screen is 3 and a half meters behind the sorry the single slit and we're told that the screen itself is 2 centimeters wide and we want to know how many dark fringes fit. So what we're going to have is we're gonna have like 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 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 will go past the screen and so you won't see diffraction pattern. The diffraction pattern ends at the screens width. We haven't really addressed in any problem, either a double slit or single slit problem, what happens when you consider the finite width of a screen. 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. How do we relate this to the number of dark fringes? Well notice that each dark fringe, 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 one dark fringe, this is the M equals two, this is the M equals three, this is the M equals four. Something important to remember is that the dark fringes are evenly spaced angularly. So whatever angle is right here, which I'll call phi. The angle between the central axis in the first dark 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 the same as the angle between the third and the fourth dark fringe. They are evenly spaced angularly alright. Now notice that the angle phi is equal to that angle between the central axis and the first dark fringe, this angle right here. Well that we know by definition has to be theta one so the first bit of information that we know is that phi equals theta one and that's really important. Our equation for the angles for dark fringes in a single slit is that sine of theta sub M is M lambda over D. So that tells us that sine of theta one has to equal just lambda over D when M is one. We know what lambda is, we know what D is, so we can find sine of theta one and then theta one. This is a 500 nanometre laser, nano's 10 to the -9. The width of the slit is half a millimeter, so this is 0.5 milli is 10 to the -3 and if you plug this into your calculator you get an answer of 0.001 which is equivalent to theta one equals 0.057 degrees. And remember that that equals phi, the angle between each successive dark fringe. Now how do we find out how many dark fringes there are? 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 goanna be theta max divided by phi. 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. 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 5. 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. Well notice that theta max is just given by this triangle right here. We know what the base of this triangle is, it's 3 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. So we know the opposite edge, we know the adjacent edge, so we can use the tangent. So the tangent of theta mass is going to be the opposite edge, which is 1 centimeter centi is 10 to the -2 and the adjacent edge is the base or 3 and a half meters if you plug this into your calculator you get 0.0029 which means that theta max is 0.16 degrees. Now we also have theta max. 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. So you can fit, or you can find at least 5 dark fringes in those 2 centimeters. You would also partially have some bright fringes hanging off the edge of both of those ends 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 the screen. It doesn't end at a dark fringe. Alright guys, that wraps up this problem. Thanks for watching.

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Concept #1: Single Slit Diffraciton

Practice #1: Width of an Unknown Single Slit

Example #1: Number of Dark Fringes on a Screen

Coherent laser light passes through a narrow slit that has width 1.50 x 10 -4 m and falls on a screen that is 4.00 m from the slit. In the diffraction pattern on the screen the width of the central maximum is 3.00 x 10-2 m. What is the wavelength of the light?

A single slit that has a width of 0.40 mm is illuminated by coherent light of wavelength 500 nm. The diffraction pattern is observed on a screen that is a large distance from the slit. On the screen the width of the central maximum is 4.0 mm. If the width of the slit is changed to 0.20 mm. the width of the central maximum becomes
(a) 4.0 mm (unchanged)
(b) 2.0 mm
(c) 8.0 mm
(d) none of these

Coherent light with wavelength λ = 600 nm falls on a single thin slit and the resulting diffraction pattern is observed on a screen that is 2.0 m from the slit. The width of the central bright fringe is measured to be 4.0 mm. If the width of the slit is doubled while the wavelength of the light remains the same, the width of the central bright fringe in the diffraction pattern
(a) stays the same
(b) increases
(c) decreases

Coherent light of wavelength λ = 700 nm passes through a single narrow slit that has width a. The interference pattern is observed on a screen a distance 4.0 m from the slit. The central diffraction pattern on the screen has a width of 12.0 mm. What is the width a of the slit?

Light of 540 nm wavelength is shone through a single slit with a width of 10 μm. What is the maximum possible number of dark fringes that could fit on any screen place in front of the slit?

Monochromatic light of wavelength λ = 500 nm passes through a narrow slit of width a and on a screen 2.0 m from the screen the width of the central maximum is 3.0 mm. If the width of the slit in increased, the width of the central maximum is
A) 3.0 mm
B) larger than 3.0 mm
C) smaller than 3.0 mm

Consider the setup of a single slit experiment. Determine the height y3, where the third minimum occurs.
1. y3 = 5λ/a L
2. y3 = 7λ/2a L
3. y3 = λ/a L
4. y3 = 3λ/2a L
5. y3 = 9λ/2a L
6. y3 = 2λ/a L
7. y3 = 4λ/a L
8. y3 = 5λ/2a L
9. y3 = 3λ/a L
10. y3 = λ/2a L

Coherent light of wavelength 600 nm is incident on a narrow slit. The diffraction pattern is observed on a screen that is 4.00 m from the slit. On the screen the width of the central maximum of the diffraction pattern is 3.00 mm. What is the width of the slit?

(Figure 1) shows the light intensity on a viewing screen behind a single slit of width a. The light's wavelength is λ.Is λ < a, λ = a, λ > a, or is it not possible to tell? Answer the question by filling in the terms or equations to the appropriate blanks to complete the sentences.(a) The width of the central maximum in the intensity pattern for a single slit is given by ___________.(b) If λ ≥ a than w will be equal to or ________ 2L.(c) It is obvious that w << 2L, hence λ is _______ a.w = 2Lλ/aw = 2aL/λw = 2La/λgreater thanless thanequal tow = 2aλ/L

Part A: At what angle θ1 to the normal would the first dark ring be observed?Part B: Suppose that the light from the pinhole projects onto a screen 3 meters away. What is the radius of the first dark ring on that screen? Notice that the angle from Part A is small enough that sinθ≈tanθ .Part C: The first dark ring forms the boundary for the bright Airy disk at the center of the diffraction pattern. What is the area A of the Airy disk on the screen from Part B? answer must be in mm2

Infrared light of wavelength 2.5 micrometers illuminates a 0.20-mm-diameter hole.Part A: What is the angle of the first dark fringe in radians?Part B: What is the angle of the first dark fringe in degrees?

Draw the pattern you observed on the screen for the red laser. Which aspect of your drawing is due to double-slit interference, and which is due to single-slit? How can you tell? Indicate these on your drawing.What happens to the pattern when you:a. Increase the wavelength (What happen for double-slit pattern and single-slit pattern ?)b. Slits are moved closer (What happen for double-slit pattern ?) c. Each slit is widened (What happen for single-slit pattern?)

Coherent microwave light with a frequency f = 2.0 × 1010 Hz is incident on a d = 5.0 cm double slit barrier, producing an interference pattern of a number of maxima and minima. A detector is free to swing around the full 180 degrees in order to find the presence of interference maxima and minima. How many different minima will this detector detect, as it is allowed to swing around the full 180 degrees? ( include minima on both sides of the centerline in your count.)

An experiment is conducted in which red light is diffracted through a single slit.Then, each of the following alterations to the original experiment is made, one at a time and the experiment is repeated. After each alteration, the experiment is returned to its original configuration.A. The slit width is halved.B. The distance between the slits and the screen is halved.C. The slit width is doubled.D. A green, rather than red, light source is used.E. The experiment is conducted in a water-filled tank.F. The distance between the slits and the screen is doubled.A. Which of these alterations decreases the angles at which the diffraction minima appear?

You have been asked to measure the width of a slit in a piece of paper. You mount the paper 80.0 centimeters from a screen and illuminate it from behind with laser light of wavelength 633 nanometers (in air). You mark two of the intensity minima as shown in the figure, and measure the distance between them to be 17.9millimeters. (Figure 1)Part AWhat is the width, a, of the slit?Express your answer in micrometers, to three significant figures.Part BIf the entire apparatus were submerged in water, would the width of the central peak change?a) The width would increase.b) The width would decrease.c) The width would not change.

What happens to a pattern of diffraction when you:a. Increase the wavelength (for double slit and single slit pattern)?b. Slits are moved apart (for double-slit pattern)?c. Each slit is widened (for single slit pattern)?

A 0.50-mm-wide slit is illuminated by light of wavelength 500 nm.What is the width of the central maximum on a screen 2.0 m behind the slit?

A 0.52 mm-diameter hole is illuminated by light of wavelength 510 nm. What is the width of the central maximum on a screen 2.2 m behind the slit?

The spreading of waves behind an aperture isa) more for long wavelengths and less for short wavelengths.b) less for long wavelengths and more for short wavelengths.c) the same for long and short wavelengths.d) not discussed in this chapter.

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