Phys 4910 Spectroscopy
Spring 2016 Assignment 2


  1. A square diffraction grating is 10 cm on a side, and is ruled with 600 lines per millimeter.
    1. Assuming that light falls on the whole grating, how many lines are illuminated? 600 lines/mm * 100 mm = 60,000
    2. What is the maximum resolving power of the grating in 2nd order? R = pN = 2 * 60,000 = 120,000
    3. What is the minimum wavelength difference of two green lines (wavelength ~ 500 nm) which can be seen as two separate lines? = /R = 500 nm/120,000 = 0.004 nm
  2. Light falls on the grating from question 1 at normal incidence. Diffracted light is to be detected with photomultiplier with a sensitivity which runs from 300 nm to 800 nm.
    1. How many full orders are seen? (A full order is one for which the whole range of wavelengths can be seen for a given value of the order number). For each of them, what is the range of angles covered by that order?
      1. for a full order sinθ = p / d must be no bigger than 1 for the largest wavelength to be detected
      2. d = 1 mm / 600 lines = 1.67 µ = 1667 nm
      3. pfull = d sinθ / = 1667 nm * 1 / 800 nm = 2.08
      4. since p must be an integer the last full order is p = 2
    2. How many partial orders are seen? (A partial order is one for which only a portion of the range of wavelengths can be seen. Other wavelengths are forbidden if the sine of the angle becomes greater than one). For each of them, what is the range of wavelengths that can be seen?
      1. for a partial order sinθ = p / d must be no bigger than 1 for the smallest wavelength to be detected
      2. ppartial = d sin / = 1667 nm * 1 / 300 nm = 5.55
      3. since p must be an integer the last partial order is p = 5
  3. The grating is now placed inside a monochromator, with exit slits which are 75 cm away. What is the spectral width of the exit slits if they are physically
    1. Layout of solution
      1. λ = d sin(θ)/p
      2. sin(θ) = 2 * 500 nm / 1667 nm = 0.6
      3. θ = 36.9º
      4. dλ/dθ = d cos(θ) /p
      5. dλ/dl = 1/R*dλ/dθ = dcos(θ)/pR = 1.67 µ *cos(36.9º) / 2*75 cm = 8.9 x 10-7
    2. 5 µm wide?
      1. with dl = 5 µ, dλ = 8.9 x 10-7 * 5 µm = 4.5 x 10-5 µm = 0.0045 nm
    3. 100 µm wide? (As in question 1, take the wavelength to be 500 nm in 2nd order)
      1. with dl = 100 µ, dλ = 8.9 x 10-7 * 100 µm = 8.9 x 10-4 µm = 0.089 nm
  4. Comment on the relative size of the answer from question 1.3, and the answer you got for each part of question 3.
    1. With the narrow slit width (5 µm) the resolution of the monochromator approaches that of the grating. Reducing the slit width further would not improve the resolution appreciably. However, with the wider slit width (100 µm) the resolution of the monochromator is much larger than that of the grating. A large improvement in resolution (at the expense of signal intensity) could be achieved by reducing the slit with.

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