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Diffraction Grating and White Light Problem

  1. Dec 6, 2017 #1

    Abu

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    1. The problem statement, all variables and given/known data
    White light containing wavelengths from 400 nm to 750 nm falls on a grating with 6000 lines/cm. How wide is the first-order spectrum on a screen 2.0 meters away

    2. Relevant equations
    dsinθ=mλ for constructive interference
    dsinθ=(m+1/2)λ for destructive interference
    Δx = λL/d

    3. The attempt at a solution
    I am having trouble deciding which formula to use. From my understanding of the question, we are not talking about the width of the central spectrum, but the first order, which is either the first constructive/destructive point above or below the central (lets just say above).

    Different wavelengths will go to different spots on the screen, right? I think my main problem is attempting to picture this scenario.

    If we were imagining that the first order spectrum was constructive interference, it would be surrounded by a destructive point above and below it, correct? So my first thought was to find the angles of each wavelength to those points but I have a feeling that is just ridiculous and hard to comprehend.

    If I were to mindlessly use what I have in the equations, I would probably do:

    dsinθ=mλ
    (1/6000 *10^-2)sinθ = 1(400*10^-9(
    θ = 13.8 degrees

    dsinθ=mλ
    (1/6000 *10^-2)sinθ = 1(750*10^-9)
    θ = 26.7 degrees

    For one, I don't even know what these angles mean, even if they are correct.
    Two, I don't know if my idea to find the destructive points around the first order is correct or not
    Three, I don't know how to picture this example. Is it saying that one end of the first order is the 750 wavelength and the other end of the first order is the 400 wavelength?

    Sorry if my attempt seems lackluster, I'm trying my best.
    Thank you.
     
  2. jcsd
  3. Dec 6, 2017 #2

    Charles Link

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    It's called a "diffraction" grating, but it's perhaps better referred to, at least when you are first learning it, as an "interference grating of many equally spaced slits or lines" that have constructive interference at ## m \lambda=d \sin{\theta} ## for integers ## m ##. (The equation is the same as for the two-slit interference pattern). With many lines on the grating, the interference maxima become very narrow bright spots, as opposed to the wider maxima that you get for a 2 slit interference pattern. Also, the lines on a grating are usually much closer than what is typically used in a two-slit experiment. In a grating where there are many equally spaced sources, the region between maxima, e.g. berween the ## m=1 ## and ## m=2 ## is dark. You don't have a dark spot that is centered between two wide bright spots like you get in a two-slit interference pattern. (So you don't have an equation for dark spots/destructive interference ## (m+1/2) \lambda=d \sin{\theta} ## etc... that you have in the 2-slit case). The diffraction grating is useful, just like a prism to sort out the spectrum. In the case of the diffraction grating, you can get multiple "rainbows" from the different maxima=e.g. m=1 makes one rainbow for the different wavelengths, m=2 makes another, etc. ## \\ ## Meanwhile, I think you computed the angles correctly. If the screen is 2 meters away from the grating, it is simple trigonometry to compute the distance on the screen that the pattern will cover for ## m=1 ##.
     
    Last edited: Dec 6, 2017
  4. Dec 6, 2017 #3

    Abu

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    Thank you very much. It makes sense that in between the constructive interference spots there is not simply a dark spot that is centered, but rather that complete area is pretty much entirely dark. But now what I don't understand is where the destructive interference is? Because we have the formula dsinθ=(m+1/2)λ for destructive interference. If the entire space in between constructive points is dark, where is the destructive point?
     
  5. Dec 6, 2017 #4

    Charles Link

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    (Please see also my edited post 2). For 3 slits equally spaced, the peaks get higher and narrower. For 4 slits equally spaced, even more narrow, but you still get constructive interference at the same angles because all the slits constructively interfere at the same angle. With a diffraction grating, typically you use ## N=100 ## or more lines/slits on the grating at the same time (the light beam incident on the grating will cover 100 or more lines/slits, even if it is a narrow beam), so the interference peaks get very narrow and bright=the interference is very sharp, and if you move off of the maximum even a slight amount, the 100 sources effectively combine in a non-constructive way and very little energy goes anywhere except to the angles for which there are narrow (and very bright) primary maxima. There is still a finite width (## \Delta x ## (on a screen) which corresponds to a ## \Delta \theta )## to the maximum, but it is quite narrow. ## \\ ## Incidentally, you can get a similar effect that you get from a diffraction grating by shining light onto a CD. I believe CD's have closely spaced tracks on them, so they behave very much like a diffraction grating. You should get a rainbow of colors=perhaps 2 or 3 rainbows with different m's by shining white light on a CD. A somewhat narrow beam of light works best.
     
    Last edited: Dec 6, 2017
  6. Dec 6, 2017 #5

    Abu

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    It is good to know that I was on the right track with those angles, but I am not entirely sure what they mean. The only reason why I did that was because it seemed to fit into the formula. And regarding your most recent post, I realize that the more slits that are used, the narrower and brighter the interference pattern is. However, my problem with that idea is where exactly the destructive point lies.

    Thank you very much for your patience, sir.
     
  7. Dec 6, 2017 #6

    Charles Link

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    If you want to know the complete details, there is an interference intensity formula that can be derived for ## N ## equally spaced slits (It works for any ## N ##): ## I(\theta)=I_o \frac{sin^2( N \phi/2)}{sin^2 ( \phi/2)} ## where ## \phi=(2 \pi/ \lambda ) d \sin{\theta} ##. To analyze the formula mathematically takes a little practice: The primary maxima occur when both numerator and denominator are zero at the same time. When that happens ## I=N^2 I_o ## (in the limit as the numerator and denominator go to zero), and that happens when ## \phi/2=m \pi ## which is the same thing as ## m \lambda=d \sin{\theta} ##. If you get a computer to graph it for ## N=100 ##, it might be helpful. Otherwise, you can look at it and see the denominator is no longer zero, and the numerator is less than 1, so in general, it won't have a value like ## I=N^2 I_o ## except at the maxima. ## \\ ## You can also compute the minima (zeros of intensity) by looking at the numerator. If the numerator is zero and the denominator is non-zero, you get minima. For two slits (N=2), you get one minimum between primary maxima. For N=3, you get 2, and for N=4, you get 3. Between these zero points/minima are located what are called secondary maxima. When ## N=100 ##, these 98 secondary maxima, except for the first couple very near the primary maximum, contain very little energy.
     
    Last edited: Dec 6, 2017
  8. Dec 6, 2017 #7

    Abu

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    Thank you for your reply and effort. I will have to tackle the question again tomorrow and I will get back to you then when I have either made some progress or perhaps have another question about how to solve it. Side note: I realized in your edited post (#2), you said that my equation dsinθ=(m+1/2)λ did not apply for destructive interference in diffraction grating. I will have to ask my teacher for clarification on that tomorrow as well because he is the one that told me otherwise.

    Thank you once again.
     
  9. Dec 7, 2017 #8

    Abu

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    Okay so I made some progress on the question and ended up getting the right answer. After I drew a diagram and found out that the first order (constructive) would have had a variety of colors in it, and at one end is the 400 nm wavelength and the 750 is at the other end, presumably. I will very roughly replicate the drawing I did so I can ask my other question regarding this problem.
    Capture.PNG
    So what I did was find theta 1 (angle between 400 nm wavelength and the center) and theta 2 (angle between 750 nm wavelength and the center)
    dsinθ=mλ
    (1/6000 *10^-2)sinθ = 1(400*10^-9(
    θ = 13.8 degrees for theta 1

    dsinθ=mλ
    (1/6000 *10^-2)sinθ = 1(750*10^-9)
    θ = 26.7 degrees for theta 2

    Then I used trigonometry to find the distance X (from the 400 nm wavelength to the center), then subtracted that distance by Y (the entire distance, from 750 to the center). The length is 2 meters.
    tan13.8 = x/2
    x = 0.49 m
    tan26.7 = y/2
    y = 1.005
    1.005 - 0.49 = 0.513 m, which is the correct answer.

    Now it is obviously great that I solved it, but I also had some different methods along the way that did not work, and I am not sure why.
    We have the formula: Δx = λL/d which is the distance between two constructive fringes or two destructive fringes.
    To get that formula, we have x = mλL/d for the distance between a constructive fringe and the center.
    So.. what I attempted to do before I solved this question was find X and Y, as shown in the diagram, with that formula instead, and then subtract the difference between Y and X to, hopefully, get 0.513 meters.

    But instead I got:
    x = mλL/d
    x = (1)(400*10^-9)(2)/(1/6000 * 10^-2) = 0.48 meters
    and for Y:
    y = mλL/d
    y = 1(750*10^-9)(2)/(1/6000 * 10^-2) = 0.9 meters

    y - x = 0.9 - 0.48 = 0.42 meters
    Obviously 0.42 meters is a completely different answer, but I cannot figure out why.
    Thank you.
     
  10. Dec 7, 2017 #9

    Charles Link

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    What you did was very good. :) Your first answers are the correct ones. For small angles, the approximation can often be made that ## \sin{\theta} \approx \tan{\theta} \approx \theta ## with the last approximation valid when ## \theta ## is measured in radians. In the case you have here, the two ## \theta 's ## that you computed were not small angles, so that is why your first answer is the correct one. The other formulas, which are approximations, are not very accurate because the angles ## \theta_1 ## and ## \theta_2 ## are not small angles.
     
  11. Dec 7, 2017 #10

    Abu

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    Thank you so much! I am afraid that I will make the same error again in the future on an exam, by using the formula x = mλL/d that you clarified only applies for small angles. For problems such as these, do you suggest finding the angles first, and if they are small, use x = mλL/d, but if they are large, use the first method that worked? Also, what is considered a small angle; less than 10 degrees?
     
  12. Dec 7, 2017 #11

    Charles Link

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    Small angles is usually ## \theta <.1 ## radians (which is about 6 degrees) or even smaller. ## \\ ## Editing: You should always start with ## m \lambda= d \sin{\theta} ## if you are working with primary maxima of an interference pattern of equally spaced sources, and take it from there. Whether ## \sin{\theta} \approx \tan{\theta}=x/L ## is something you can assess after getting the exact answer.
     
    Last edited: Dec 7, 2017
  13. Dec 7, 2017 #12

    Abu

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    Alright. And would you agree with my plan of finding the angles first in problems such as these so I can decide whether or not the angle is small enough to use x = mλL/d?
     
  14. Dec 7, 2017 #13

    Charles Link

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    See my edited post above.
     
  15. Dec 7, 2017 #14

    Abu

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    Okay, well that concludes it for this problem. Thank you very much for your patience, help and effort! Sorry for my intense questioning, I was very confused before this thread.
     
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