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Diffraction grating question - help needed please Watch

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    4. A diffraction grating is designed with a lit width of 0.83 micrometers. When used in a spectrometer to view light of wavelength 430nm, diffracted beams are observed at angles of 14 degrees 55' and 50 degrees 40' to the zero order beam.

    a.Assuming the low-angle diffracted beam is the first order beam, calculate the number of lines per mm on the grating.

    b. explain why there is no diffracted beam between the two observed beams - what is the order for the beam of 50 degrees 40'

    i do not know how to even start/attempt this question
    can anyone guide me to which ever part of the question ?
    no help is too little
    many thanks
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      (Original post by sixthformer)
      4. A diffraction grating is designed with a lit width of 0.83 micrometers. When used in a spectrometer to view light of wavelength 430nm, diffracted beams are observed at angles of 14 degrees 55' and 50 degrees 40' to the zero order beam.

      a.Assuming the low-angle diffracted beam is the first order beam, calculate the number of lines per mm on the grating.

      b. explain why there is no diffracted beam between the two observed beams - what is the order for the beam of 50 degrees 40'
      You are given
      d, the "slit width", which for a grating they probably mean the grating spacing.
      \lambda the wavelength of the light, and
      \theta=14"55' for the 1st order diffraction.
      The 1st order (where n=1) is the one where a single whole wavelength fits in the space AB in the diagram, causing the beams to be in phase and produce constructive interference; and a bright fringe.
      The formula is  n\lambda = d sin \theta
      The next angle given must be for the case where n=2 and there are two wavelengths difference between the beams.
      For the question, you are just asked for how many lines per mm there are on the grating. As you are told in the question that the "slit width" is 0.83 micrometers, the number per mm is \frac{1}{width} where the "width" is also in mm.

      Note, the diagram only shows 2 adjacent slits. There are hundreds in a grating, all equally spaced.

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      (Original post by Stonebridge)
      You are given
      d, the "slit width", which for a grating they probably mean the grating spacing.
      \lambda the wavelength of the light, and
      \theta=14"55' for the 1st order diffraction.
      The 1st order (where n=1) is the one where a single whole wavelength fits in the space AB in the diagram, causing the beams to be in phase and produce constructive interference; and a bright fringe.
      The formula is  n\lambda = d sin \theta
      The next angle given must be for the case where n=2 and there are two wavelengths difference between the beams.
      For the question, you are just asked for how many lines per mm there are on the grating. As you are told in the question that the "slit width" is 0.83 micrometers, the number per mm is \frac{1}{width} where the "width" is also in mm.

      Note, the diagram only shows 2 adjacent slits. There are hundreds in a grating, all equally spaced.

      ]thank you so much, i have been ill latley and i couldn't really reply to you
      i owe you so much
      my teachers have not taught us and we have exams in january
      you're impexxable knowldge serves well shared
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      (Original post by Stonebridge)
      You are given
      d, the "slit width", which for a grating they probably mean the grating spacing.
      \lambda the wavelength of the light, and
      \theta=14"55' for the 1st order diffraction.
      The 1st order (where n=1) is the one where a single whole wavelength fits in the space AB in the diagram, causing the beams to be in phase and produce constructive interference; and a bright fringe.
      The formula is  n\lambda = d sin \theta
      The next angle given must be for the case where n=2 and there are two wavelengths difference between the beams.
      For the question, you are just asked for how many lines per mm there are on the grating. As you are told in the question that the "slit width" is 0.83 micrometers, the number per mm is \frac{1}{width} where the "width" is also in mm.

      Note, the diagram only shows 2 adjacent slits. There are hundreds in a grating, all equally spaced.

      Stone, may i ask you a follow up?
      what about part b on this
      why isn't there a wave in the 3rd order or something of the sort?
      tyvm
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        (Original post by sixthformer)
        Stone, may i ask you a follow up?
        what about part b on this
        why isn't there a wave in the 3rd order or something of the sort?
        tyvm
        Actually, this question is a little more complicated than I first thought.
        The question does give the slit width, not the grating spacing as I first suggested.
        I should have read it more carefully. Sorry.
        If you put the values given into the formula
        n\lambda=d sin \theta
        you will get d, the grating spacing, is 1.7 x 10-6 This is double the slit width.
        What this means is that the grating has transparent slits that are rather wide, in this case, the transparent and opaque sections are of equal width. Normally the transparent slits would be less than half the width of the opaque.
        What this means is, I think, that the second order bright fringe is not visible, as it occurs outside the central bright band of the diffraction pattern.
        If you put n=2, you get the angle for the 2nd order to be about 30 degrees.
        If you put n=3 you get the angle to be 50 deg 40 min as the question says.
        This fringe occurs in the next, outer bright band of the pattern.
       
       
       
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