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Resolving Power of a collimated Littrow diffracting telescope

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose a simplified telescope and grating spectrometer system:

    diffractiontelescope.jpg

    D is the telescope's primary diameter, W is the width of the collimated beam, β is the grating angle in a Littrow configuration, and [itex]\alpha[/itex] is the aperture diameter in angular units on the sky.

    The goal is to find an expression of the resolving power R that depends only on those 4 quantities. We can assume that the detector size in the exit focal plane is small compared to the projected aperture size.


    2. Relevant equations

    R=[itex]\frac{\lambda}{\Delta\lambda}[/itex]
    R=mN where N is the number of grating slits or ridges and m is the diffraction order.

    Since the grating is in Littrow configuration, we know that 2dsinβ=mλ, so λ=[itex]\frac{2d\sin{\beta}}{m}[/itex] where d is the distance between ridges/slits.

    D and W are related to [itex]\alpha[/itex] by D=[itex]2f_1\sin{\alpha}[/itex] and W=[itex]2f_2\sin{\alpha}[/itex], where [itex]f_1[/itex] and [itex]f_2[/itex] are the focal lengths from the primary lens to the aperture and from the collimating lens to the aperture, respectively.

    We know the number of slits/ridges illuminated is given by N=[itex]\frac{W}{d\cos{\beta}}[/itex]

    If I'm not mistaken, the reflected light is normal to the grating.

    3. The attempt at a solution


    [itex]\frac{\delta\lambda}{\delta\beta}=\frac{2d\cos{β}}{m}[/itex]

    Therefore, approximating [itex]\delta\lambda[/itex] to Δλ, R=[itex]\frac{\tan{\beta}}{\delta\beta}[/itex]

    So from this point, we need [itex]\delta\beta[/itex] in terms of [itex]\alpha[/itex], D, and W. But here is where I'm stuck. I can't for the life of me figure out what to do next.
     
  2. jcsd
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