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What is the concept behind Diffraction, n , Finding Dark/Bright Spots

  1. Mar 17, 2014 #1
    What is the concept behind Diffraction, "n", Finding Dark/Bright Spots

    I used the following youtube video to help me learn about the concept of diffraction but still could not make sense of my lecture notes:

    (All 4 parts)

    SINGLE SLIT DIFFRACTION


    In order to find the minima the following formula is used, a*sinθ=n*λ, what does "n" exactly mean and how does the order relate to the dark/bright spots on the screen, (a=slit width,λ=wavelength).

    The extra distance travelled by the rays = a/2*sinθ, while the phase difference is always λ/2 (learned via youtube video above)

    So, a/2*sinθ=λ/2 from which we find the formula a*sinθ=n*λ

    Sinθ can be replaced by x/D, where x is the distance from the central max to the desired Max/Min and "D" is the distance from the screen to the slit.

    1st Dark Spot on the screen: a/2*x/D=λ/2, what does "n" equal here, is it 1? and does a*sinθ=n*λ apply?

    1st Max Spot: a/3*x/D=λ/2 (found from video), what does "n" equal here and does a*sinθ=n*λ also apply here.

    How will i know in a question if i need to find a Max or a Min? As i can't seem to find the relationship between the order and the dark/bright spots on the screen.

    Are the following correct:

    Central Max: n=0
    1st Dark: n=1
    1st Max: n=2
    2nd Dark: n=3

    and then does a*sinθ=n*λ need to be applied to find the desired solution? but i can't apply it to the Max or the Min, for example 1st Dark Spot is a/2*x/D=λ/2 (learned from the video), how do i get this from a*sinθ=n*λ and what does "n" equal in this case for dark spot and the same for the 1st Max, 2nd Dark, 2nd Max etc.

    Also in my notes i see a graph of intensity(y-axis) and sinθ(x-axis) with values such as λ/a, 2λ/a, what do they mean? i know that a*sinθ=λ where sinθ=λ/a is from the 1st Dark Spot, but then i have 2λ/a where sinθ=2λ/a but what does that represent?

    I see it may come from a*sinθ=n*λ where the "n" value is number 2 but how do i know what the "n" number represents because i can't figure out if they are dark or bright spots. Also if λ/a is a bright spot? then 2λ/a would represent the 1st max? but how can that be when it is a/3*x/D=λ/2

    As you can see i have a basic understanding of the science behind what is happening but i need to understand the thinking behind the use of the formulas to find the dark/bright spots and their relationship with "n" which is my main issue. Since their is a set formula for minima the one which i have shown above i want to understand how to use it to find the dark spots but since it applys to minima i assume there must be one for maxima and i hope to understand how the formula can be used to find the bright spots and the relationship with "n" in all of this.

    DOUBLE SLIT DIFFRACTION

    With double slit difraction the extra distance travelled for the following is:

    1st Max: 1λ
    2nd Max: 2λ
    3rd Max: 3λ

    1st Dark: λ/2
    2nd Dark 3/2*λ

    where for example to find 1st dark you have d*sinθ=λ/2, is there a general formula in this case to find max or min which also involves the use of "n" to find the bright or dark spots.

    CIRCULAR APETURE AND DIFFRACTION GRATING

    For circular apeture, is there anything beyond sinθ=1.22*λ/a(diameter of opening) and R(radius)=1.22*λ*D/a and do these formulas help me find the max and minima? use of "n"? etc

    Diffraction grating, d*sinθ=m*λ what is "m" and how do i know which value for "m" corresponds to a dark or bright spot when i am trying to find one. For example:

    extra distance travelled for 1st max: λ and 2nd max: 2λ in this case does that mean for the 1st max m=1 and second max m=2 and then i apply it to the formula d*sinθ=m*λ to find the solution i require?

    Sorry for the lengthy question wanted to make sure i fully understand what is happening, will appreciate your time and patience.

    Thank you very much for any help guys.
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Mar 17, 2014 #2
    Hi there!
    For future reference, I recommend you to numerate and group together your questions which will them easier to answer. This, I would say from personal experience, boost the probability for your inquiry to get an answer.

    Nevertheless, let "path length difference between interfering waves" = ψ, and you write

    "...
    Are the following correct:

    Central Max: n=0
    1st Dark: n=1
    1st Max: n=2
    2nd Dark: n=3
    ..."

    which corresponds to

    ψ = n/2 * λ; n = 0 - constructive interference
    ψ = n/2 * λ; n = 1 - destructive interference
    ψ = n/2 * λ; n = 2 - constructive interference
    ψ = n/2 * λ; n = 3 - destructive interference

    ... is correct. If I understand your question correct, you want to know the physical interpretation of n? Diffraction patterns are mostly about (constructive and destructive) interference. Furthermore, constructive interference gives maxima and destructive gives minima in the diffraction pattern.

    Now, let two light waves of same frequency/wavelength start from the same point in space, meaning the single slit, but not necessarily at the same time. Then they meet up to interfere at some other space point for example the screen showing the diffraction pattern, at the same time, where they interfere. The two waves will interfere depending on the different path lengths they have traveled (as you have written) and here is where the n comes in:

    (1) If the difference in path length is exactly zero, they interfere constructively giving a maxim
    (2) If the difference in path length is exactly ONE wavelength, they interfere constructively giving a maxim
    (3) If the difference in path length is exactly TWO wavelengths, they interfere constructively giving a maxim

    Or in general, if the difference in path length is exactly , where n=0,1,2..., they interfere constructively giving a maxim, of order n. Hence, in this example, n gives the number of wavelengths one wave have traveled longer than the other as they meet to interfere.

    Then what happens if we turn it around? Say, one wave traveled /2 longer than the other. This implies destructive interference, because now one wave's top meets the other's bottom.

    I believe the thing about diffraction, is not to get stuck at the n. Think more like,

    → Constructive interference when top/bottom meets top/bottom: this occurs when one the waves have traveled an integer multiple of its wavelength () relative to the other. This creates diffraction pattern maxima.

    → Destructive interference when top meets bottom: this occurs when one the waves have traveled a half integer multiple of its wavelength (n/2 λ) relative to the other. This creates diffraction pattern minima.

    And here is where its easy to mess up, because some leave the half in the formula with the sine itself (as you did) which is okay, but then of course, constructive interference comes for half integers of n, and destructive for whole integers. However, read the arrows (-->) carefully, as this argument concerning the wavelengths should always hold true!

    =======================================

    I hope that should clarify most of things, but I'll wrap in some other questions for short.

    You write: "Also in my notes i see a graph of intensity(y-axis) and sinθ(x-axis) with values such as λ/a, 2λ/a .. ". Here you need to be careful, because the intensity(!) is given by this formula:

    [itex]
    I(\theta) = I_0 sinc^2 \left(\frac{d \pi}{\lambda} \sin(\theta) \right)
    [/itex]

    Which you should find in your textbook or on Wikipedia. This behaves the same way, obviously, but it's a little bit more complex to show where maxima are found. To show this, you may apply that sinc(x) is zero for
    [itex]x=n \cdot \pi [/itex],

    for any
    [itex]n=0,1,2,...[/itex],
    meaning you get a minim at

    [itex]
    I(\theta) = I_0 sinc^2 \left(\frac{d \pi}{\lambda} \sin(\theta) \right) = 0
    [/itex]
    for,
    [itex]
    \left(\frac{d \pi}{\lambda} \sin(\theta) \right) = n \pi
    [/itex]

    and solve! Of course you ca also replace [itex] \sin \theta [/itex] for an expression with x and y using some trigonometry if you want to.

    Double slits are just the same as single slit, but consider the waves which interfere to come from two different points in space (different slits). Find the wavelength path difference between the two slits to a generalized point on the screen. This is derived explicitly on Wikipedia. When you have the path length difference, find where on the screen one of the waves from two different slits have traveled exactly n wavelengths longer/shorter than the other. These will be maxima!

    There exists a general method for finding different (far field) diffraction patterns from any type of slit, using Fourrier transformations etc, but I do not know advanced optics you are into?

    If not all enlightening, I hope some of this makes at least something clearer or motivates you to find out more. Please, do not hesitate to ask further.
     
  4. Mar 17, 2014 #3

    BvU

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    Personally I like sections of this page and and this one but there must be lots of even nicer displays.

    Google Airy disk for the circular aperture.
     
  5. Mar 19, 2014 #4
    thanks for the reply, will keep that in mind regarding grouping questions. I am just working through your post now, thanks for taking your time out to help my understanding
     
  6. Mar 19, 2014 #5
    I have a better understanding of constructive and destructive interference thanks to your post, just have some confusion with the math and formulas.

    ψ = n/2 * λ; n = 0 - constructive interference, so this is the Central Max
    ψ = n/2 * λ; n = 1 - destructive interference, 1st Dark
    ψ = n/2 * λ; n = 2 - constructive interference, 1st Max
    ψ = n/2 * λ; n = 3 - destructive interference, 2nd Dark

    From the video link in the OP, to find the 1st Max (Bright spot/constructive interference) on the screen:

    a/3*sinθ=λ/2, which gives x=3λD/a2. Is that correct? I was trying to find it by what you had said for constructive interference, " difference in path length is exactly nλ, where n=0,1,2..., they interfere constructively giving a maxim"

    So this is what i was doing, a/3*sinθ=nλ where n=2 due to the 1st Max but was unable to get x=3λD/a2 which is the correct answer for finding the 1st Max on the screen

    And also for the 1st Dark, we have a/2*x/D=λ/2 which becomes x=Dλ/a

    For this 1st dark spot(destructive), i done the following:

    a/2*sinθ= nλ/2 where n=1 due to the 1st Dark spot but was unable to get x=Dλ/a

    I want to be able to find the position of the required Max/Min from the central max on the screen but don't know which formula to apply


    Also for constructive interference you said that the difference in path length was nλ but also said the path difference is ψ = n/2 * λ for constructive interference when n=0 and n=2, shouldn't ψ = n/2 * λ just be for destructive interference where:

    to find a minimum its: a/2*sinθ= nλ/2
    to find a maximum its: a/2*sinθ= nλ

    I was trying to apply those formulas to find where max and min are when:

    n = 0 - constructive interference, so this is the Central Max
    n = 1 - destructive interference, 1st Dark,
    n = 2 - constructive interference, 1st Max
    n = 3 - destructive interference, 2nd Dark

    but i can't do that when the extra distance travelled by the rays is:

    n = 1 - destructive interference, 1st Dark, = a/2*x/D
    n = 2 - constructive interference, 1st Max, = a/3*x/D
    n = 3 - destructive interference, 2nd Dark, =a/4*x/D

    I've been trying to find the correct position of the Min/Max on the screen by usng the extra distance travelled by the rays and making them = to "nλ/2" or "nλ" where appropriate but could not get the correct answer as shown above.

    In the video i watched all the extra distances travelled a/2*x/D, a/3*x/D etc for the different spots on the screen always equalled to "λ/2" (Is this the phase difference, in the video it said that it never changes) and then the formula is rearranged to find "x" the distance on the screen from the central max but i failed to find the same answer for "x" when i make the extra distances travelled for the different Max/Min equal to nλ/2 or nλ where the following "n" values were used

    n = 0 - constructive interference, so this is the Central Max
    n = 1 - destructive interference, 1st Dark,
    n = 2 - constructive interference, 1st Max
    n = 3 - destructive interference, 2nd Dark

    I think my problem with "n" is related to the extra distance travelled, a/2, a/3, a/4 etc for the different Max/Min?

    Thanks for the previous post again and any help greatly appreciated
     
  7. Mar 23, 2014 #6
    Bump, any help with my above post please? Thanks.
     
  8. Mar 26, 2014 #7
    I have cut down my previous post and wish to know the answers to the following questions regarding the formula in finding the bright/dark spots for single and double slit diffraction please

    (a=slit width,λ=wavelength, D=Distance from screen, x=distance of max/min from central max and n=order)

    SINGLE SLIT


    In order to find the min is the following formula used, a/2*x/D=n*λ/2

    and to find the max the following is used, a/2*x/D=n*λ (i think this one might be wrong?)

    where n= 0.,1,2,3 or 4 etc

    are the above formulas correct?

    DOUBLE SLIT

    Bright spot forumla, d*x/D= n*λ

    Dark spot formula, d*x/D=(n+1/2)*λ

    where n may = 0,1,2,3 or 4 etc



    are the above formulas correct to find the bright/Dark spots for the double slit as well or am i wrong?



    And finally For the double/single slit will the "n" be given in a specific question to find where the bright/dark spots are and does the "n" have a relationship with the Dark/Bright spots?


    Thanks
     
  9. Mar 27, 2014 #8
    Hi again!
    I'm sorry I haven't been able to answer before now as I've been struggling with a project of my own. Anyways, lets see..

    First, this MIT-paper answers all your questions in a really good way. I highly recommend you just read from the start of chapter 14.4 (it is not very long!)
    MIT-paper: http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide14.pdf

    THE SINGLE SLIT
    It's a little difficult to say your formula is wrong as I do not know where you got it from. Nevertheless, look at the picture given from this HyperPhysics site: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html, as seen here:

    sinslit.gif

    Then study each of the three rays. With "Fraunhofer diffraction", we mean that the distance (D) between the slit and the diffraction patter (screen) is so large that the angle between each ray and the D-line is essentially equal, namely [itex]\theta [/itex].

    That is, by convention: [itex]\theta [/itex] is the angle between mid-ray and D-line. The Angle between top ray and D-line is [itex]\theta - \Delta_1[/itex] and angle between bottom ray and D-line is [itex]\theta + \Delta_2[/itex] from some angles Δ, and as D becomes laaaaarge (Fraunhofer diffraction): [itex]\Delta_1 = \Delta_2 \approx 0[/itex].

    Using the picture: Now, we study only the two rays, bottom and top, and only use the middle as reference. Convince yourself that the bottom rays travels δ longer than the top ray!! You see that δ changes with [itex]\theta [/itex]? For example is δ zero for [itex]\theta =0[/itex], giving diffraction max of order zero (constructive interference).

    The whole point from here is to find the relation between δ and the y-coordinate on the screen, given by [itex]\theta [/itex]. We do know, that diffraction max(!) occurs for [itex]\delta = n\lambda [/itex], [itex]n=0,1,2,... [/itex]. But by trigonometry (see the picture), you see that,
    [itex]
    sin(\theta ')= \frac{ \delta }{ a }, \theta ' \approx \theta
    [/itex]
    as D is large. Here δ is (still) the path length difference between top and bottom ray.

    And here comes the hard part, where I personally thinks the picture fails. Until now we have studied the path length difference between top and bottom! For interference, we actually want light from the end of the slit to have traveled [itex]\delta = n\lambda [/itex] longer than the middle, reference ray!! Not the bottom ray! Think about it before you continue to read. This is not trivial: ask your lecturer/teacher, and read the MIT-paper.

    When we accept this, we may write instead (a → a/2), and use different order for max and min:

    [itex]
    \sin(\theta)= \frac{ \delta }{ a/2 }
    [/itex]
    Here δ is now the path length difference between top and middle rays! Furthermore we write

    [1] Constructive interference (diffraction max) for [itex]\delta = n\lambda [/itex]
    [itex]
    a \sin(\theta)= 2 n \lambda, \quad n=0,1,2,3,...
    [/itex]

    [2] Destructive interference (diffraction min) for [itex]\delta = \frac{m}{2}\lambda [/itex]
    [itex]
    a \sin(\theta)= m \lambda, \quad m=1,2,3,...
    [/itex]

    Now is the time to relate [itex]\theta[/itex] to the screen. In the picture, they use y-coordinates so I'll do the same, with origin as indicated in the picture (where D-line cross y-axis). Then with [itex]\theta[/itex] as the angle between reference ray (mid ray) and D-line, then y-coord. is given by,

    [itex]
    \tan (\theta) = \frac{y}{D}
    [/itex]

    Now, since D is very large we say that [itex]D >> y[/itex] meaning [itex]\theta[/itex] becomes small, which means we can write [itex]\tan(\theta) \approx \sin(\theta) \approx \theta[/itex], which makes us able to connect with the other equation yielding (for destructive interference),

    [itex]
    a \frac{y}{D}= m \lambda
    [/itex]
    and finally,

    =============================

    [itex]
    y = \frac{m \lambda D}{a}
    [/itex]
    For destructive interference (diffraction min)


    and

    [itex]
    y = \frac{2n \lambda D}{a}
    [/itex]
    For constructive interference (diffraction max)


    Where only n may be zero..
    which are identical to what you write (x → y)??
    ================================

    Now I use quite some time answering as I cross check my sources to give solid answers, and this is what I managed right now. I'll come back for more as soon as I got the time, but in the meantime look at this. For multiple slits, it isn't really much different than for a single: you might figure it out without more help, but I or someone else will be back.

    Good luck!
     
    Last edited: Mar 27, 2014
  10. Mar 30, 2014 #9
    Thanks a million! I have a better idea in-regards to find the Max/Min. Just have some questions if you don't mind.

    I can see that the condition to find the minimum can be generalized to the following formula:

    a*sin(θ)=mλ, m=1,2,3


    where to find the 2nd min spot m=2 and then when we use "a*sin(θ)=mλ" we get y=2λD/a which is correct.

    However to find the first max we use the following formula which you have designated, a*sin(θ)=2nλ. And say i need to find the 1st Max, would n=1? as its the first Max. Then Since n=1, y=2λD/a but this is wrong i think?

    In the video, to find the 1st Max we have the following: a/3*sin(θ)=λ/2 where y=3λD/2a how can i apply a*sin(θ)=2nλ in this case to find the 1st Max?


    Also for Double slit interference i worked out that to find the Max the a*sin(θ)=mλ, m=1,2,3 formula can be applied, but which formula can be used to find the minimum?

    Thanks again for any help
     
  11. Apr 15, 2014 #10
    Hi again! How are you doing, figured it out yet?

    Reading my own answers made me realize I could have formulated myself much better. Besides, I have made a rather crucial mistake which you actually point out! Creds to you!:)

    You write that we can se from the condition to find the minimun can be generalized to the following formula:

    [itex]a \sin (\theta) = m \lambda, \quad m = 1,2,3 [/itex]
    (I'll call this the destructive interference eq. [DIE])

    which seems be correct, and then you quote me who find the maximum bu coming up with some answer which you believe is incorrect, and here you're right as well. I did the math making some pretty bad assumptions, but won't take this further here now.

    The thing is, and I quote wikipedia, which after finding our (correct) expression for the minima, says: "There is no such simple argument to enable us to find the maxima of the diffraction pattern.". And why is that actually?

    One way to think is that you are solving an equation describing the intensity (of light or E-field if you'd like) of the diffraction pattern. Solving this single equation for finding the minimum is obviously done by finding y (i.e. the screen coordinate) which gives zero intensity (i.e. one side of the eq. is zero). Finding the max is not possible this way as we have no idea what the intensity actually is at the max: is it 1, 10, 143652? You get my point..

    So, how do we find the max? You'll have to assume Fraunhofer diffraction and do some magical math which makes you able to show that,

    [itex]
    I (\theta) = I_0 sinc^2 \left( \frac{a \pi}{\lambda} \sin(\theta) \right)
    [/itex]
    (and I'll denote this the intensity eq. [IE])

    This is the "intensity-function" with respect to Θ (which may be translated to y-coord.), and is easily deduced for it's maxs and mins. Here you could for instance check if the mins of DIE correponds with IE, concerning coordinates Θ or y.

    I haven't derived this [IE] for quite some time, but I may look it up and show you if you haven't figured this all out by now.

    =================================================

    A side comment to my earlier answer concerning the reference line from the picture from HyperPhysics

    I wrote that you need to find the difference in path length for the top and middle ray: this is rather inconsistent and confusing (at least i think so). Fact is, that the middle ref. line/ray is just a mean of the two ends. In reality, we should not only look at waves emitted from the ends of the slit but from all of it! Let's look into it,

    The Physics (or rather Huygens' Principle) state that any point on a wave wront may be treated as a new single point source emitting new waves. What does it mean? Look at this picture,
    http://upload.wikimedia.org/wikipedia/commons/6/60/Refraction_on_an_aperture_-_Huygens-Fresnel_principle.svg

    As the (assumed plain) wave enters the slit to be diffracted, we may discretize the wave's front as a set of new point sources, all emitting new waves. Such point sources are indicated as yellow dots in the picture. Of course, in reality, any point is a new point source, meaning the yellow dots are actually continuous along the slit opening. However, it illustrates my point, that each of these yellow dots emits waves, which will interfere constructively or destructively creating a diffraction pattern depening on each yellow dot's distance to the point on the screen.
     
    Last edited: Apr 15, 2014
  12. Apr 21, 2014 #11
    Thanks, that makes thing much clearer in my mind now. I had researched a bit too and found that there was no way to find the max that way while it was easier to find the minimum via the designated formula.

    I managed to find the 1st Max via the following formula:

    sin(θ) = (2m + 1)* λ / 2a where m= 1,2.

    Cheers for all the help!
     
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