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Diffraction Grating homework

  1. Jan 17, 2008 #1
    Firstly, after thinking wether this should go in the Homework and Coursework section, I can to the conclusion, that was not the right place, as I am just wanting to expand on what I have done in a lesson.

    Well, firstly we just touched on it, and I didn't really get the concept. I know that it separates light of different wavelengths, and then the angles at which these different colour s are recorded, but really I am not to sure of...

    A) What it actually is, I have seen the apparatus used, and I would know how to carry the experiment out a bit, as we talked a bit about it, but I'm not sure what it actually is, which I guess is probably the most important things.

    B) I would be able to record the angles, but I am not sure what these angles represent in terms of wavelength. I have the formula "n lambda = d sin( theta )" and know/think that D is the distance from the selected region to the origin or the from the straight line, and "n" is the the number of the line left or right.

    I've looked on Wikipedia, and Physics.smu but I struggle to comprehend what they are saying.

    Any help would be great, and if this is in the wrong forum, I will move it straight away. Thanks =]
  2. jcsd
  3. Jan 17, 2008 #2
    Infact, I might need that formula checking out, I dont know if It is correct, and neither what the angles actually represent in terms of wavelength?
  4. Jan 18, 2008 #3
    You'd probably get more response in the quantum mechanics section. But I'd watch the Feynman videos found here http://vega.org.uk/video/subseries/8 for a clear explanation. Or check out Feynman's easy to read QED.
  5. Jan 18, 2008 #4


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    firstly, from you have said I am not sure whether your title of diffraction grating matches what you are actually intended to ask. But judging from the formula you have provided, namely,
    [tex]n \lambda = d \sin \theta[/tex]

    it appears that what you are actually asking is how to use the diffraction pattern you seen on the screen to deduce the wavelegth of the source. am I right?

    firstly, the formula means

    [tex]\text{path difference of two light rays} = n \lambda[/tex]
    where n is an integer and [tex]\lambda[/tex] is the wavelength of a coherent source. Path difference = n times wavelength of light in this case ensure constructive interference, giving you the bright spots on the screen. the path difference is between two rays scattered/reflected off two adjacent gratings (on a piece of metal or something)

    so the typical experimental set up is you have a laser of some fixed but unknown wavelength, you shine the beam on an object with gratings (should be spaced uniformly with fixed distance between gratings of d); record location of the bright spots and use trigonometry and the formula above to help you work out the wavelength of the light.

    but I don't think d sin (theta) = n lambda should be the formula for doing this. So may be you were actually asking just diffraction not diffraction grating. But until you can clarify that that's all I can say at this point.
  6. Jan 18, 2008 #5

    Andy Resnick

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    It's tough to figure out what you are asking; you apparently want to know what diffraction or a diffraction grating *is*, and how the scattering properties depend on wavelength.

    Diffraction is a fancy word for scattering: light diffracts at interfaces of different refractive indices. Lots of objects diffract light- a window screen, clouds, ruled gratings, etc. Any type of wave can undergo diffraction as well. The diffraction can be mostly transmission or mostly reflection, or an intrmediate combination.

    A diffraction grating is a series of closely spaced parallel grooves: the goove profile is a variable, as is the specific spacing. A CD/DVD is a common example or a reflection grating. I have some diffraction covers for a laser pointer to generate different shapes: an arrow, a hand, a smiley face; these are transmission diffractive elements- they are not gratings, but they work by diffraction as well.

    The formulas you have is a simple expression for a simple diffractive grating, and gives the angles of the first diffracted order- the primary rainbow, if you will. The Richardson Grating Laboratory has a short book "Diffraction grating handbook" that (at the time I got one) is free and is fairly comprehensive. Try www.gratinglab.com to see if you can get one.
  7. Jan 20, 2008 #6

    Claude Bile

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    Diffraction gratings are devices that spatially modulate a light source in some way, typically gratings that either reflect or transmit an incoming light wave through the use of a series of ridges or slits placed at regular intervals.

    The main characteristic all gratings share is that they are periodic in some way.
    For diffraction gratings where the incident beam is at normal incidence;

    [tex]dsin(\theta) = n\lambda[/itex]

    [itex]d[/itex] is the period of the grating (i.e. the distance between slits).
    [itex]\theta[/itex] is the angle of the reflected beam.
    [itex]n[/itex] is the diffraction order (equal to some integer that can be +ve or -ve)
    [itex]\lambda[/itex] is the wavelength.

    You can see fairly plainly from this equation that the angle varies with wavelength. This is the source of the dispersion that allows a diffraction grating to separate an incident beam into its constituent wavelengths.
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