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Grating Spectrometer - Bandpass

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  1. Apr 19, 2014 #1
    1. The problem statement, all variables and given/known data

    2zxpbhx.png

    Part (a): Find intensity distribution of N-slit grating
    Part (b): Find resolving power
    Part (c): Find an expression for Bandpass, and estimate its value. What value of slit width gives same bandpass as resolving power theoretically?

    2. Relevant equations



    3. The attempt at a solution

    I've solved most parts of the question, except the very last part in part (c). I thought the bandpass is independent of wavelength?


    Part(a)

    [tex]I = I_0 \frac{sin^2(\frac{Nkd sin \theta}{2})}{sin^2 \frac{kd sin \theta}{2}} sinc^2 (\frac{ka sin \theta}{2}) [/tex]

    Part (b)

    Width from central maxima to first minimum ## \Delta \theta = \frac{1}{N}(\frac{\lambda}{d})##

    For pth order fringe maximum: ## sin \theta_{max} = \frac{p\lambda}{d}##, I get angular dispersion as ##\frac{d\theta}{d\lambda} = \frac{p}{d cos \theta} \approx \frac{p}{d}##.

    Thus, Resolving Power ##\frac{\lambda}{\Delta \lambda} = \frac{\lambda}{\Delta \theta / \frac{d\theta}{d\lambda}} = Np = N## for p = 1.

    Part (c)

    Bandpass is defined as the amount of lambda it lets through, so Bandpass is:
    [tex] \frac{d\lambda}{d\theta} \Delta \theta[/tex]
    [tex] = \frac{d\lambda}{d\theta} \frac{w}{f}[/tex]

    Where w is exit path width and f is focal length.

    I found bandpass = ##5.56 * 10^{-5} m##

    Now for the last part, theoretically for first order resolving power is independent of λ?

    Resolving power = N

    [tex]\frac{wd}{f} = N[/tex]
    [tex] w = 3*10^{11} m [/tex]

    This answer seems ridiculous.
     
  2. jcsd
  3. Apr 21, 2014 #2
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