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Signal Intensity

  1. Feb 19, 2007 #1
    1. The problem statement, all variables and given/known data

    You need to design a diffraction grating that will disperse the visible spectrum (400-700nm) over 30.0 degree in first order.
    a) How many lines per millimeter does your grating need?
    b) What is the first-order diffraction angle of light from a sodium lamp (wavelength = 589nm)?

    2. Relevant equations

    dsin(theta) = m(wavelength)

    3. The attempt at a solution

    m= 1 (as it is first order) wavelength = (400*10^-9)m
    angle = 30 degree wavelength = (700*10^-9) m

    I'm confused because there are two wavelengths given. Do I find the d value for both wavelength and then find difference between them ... d= 0.0000014-0.0000008 = 0.0000006 = 6*10^-7
    then 1667lines/mmm

    for part 2) do i use the d value as 6*10^-7
    because other values are given ... wavelength = 589 *10^-9 and m=1

    If I take the same d value I'm getting as 79 degree ... is it right ?

    Please somebody check the calculation and tell if its right ... I've only one attempt left.

    Can someone also tell me how to put equation on these kind of questions. I'll be very grateful.
  2. jcsd
  3. Feb 19, 2007 #2
    This may help

    Howdy, I'm not 100% sure, but I think what you need to do is look at it from the point of view of rates of change.

    So, from dsin(theta) = m(wavelength), you first need to find the formula for the derivitive d(wavelength)/d(theta). Then set it equal to (change in wave/change in theta) from the given info and solve for d in mm. Then take the reciprical like you did to see how many d in 1 mm. For part 2 you just plug in 589, after taking account of units of course, and solve for theta.

    AI hope that helps.
  4. Feb 19, 2007 #3
    which value of d should take in part 2
  5. Feb 19, 2007 #4
    After completing step 1, and imputing the initial data, there is only 1 d for the system. So, use the d "from the given info and solve for d in mm" (from my 1st post). Strategically, looking at the written problem and supporting equation, and seeing the 400nm and 700 nm givens, makes some people just want to rush ahead and plug them in and solve for each d and go from there, rather, in this case, we are interested in how a change in wavelength (their difference) is related to a change in spread (theta).
  6. Feb 20, 2007 #5
    The derivative works for small changes, but to be more accurate you'll have to solve the full equations. Neither wavelength1 nor wavelength2 have theta = 30 degrees, the question asks what value of d gives a difference in theta values equal to 30 degrees. You know d*sin(theta1) = wavelength1, and you know d*sin(theta1 + 30 degrees) = wavelength2, so you can solve for theta1 and d in part a. Given d, you can then find the value of theta corresponding to 589 nm in part b.
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