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Homework Help: Diffraction limit on resolution

  1. Apr 10, 2004 #1
    Q:Spy planes fly at extremely high altitudes (25.3 km) to avoid interception. Their cameras are reportedly able to discern features as small as 5.20 cm. What must be the minimum aperture of the camera lens to afford this resolution? (Use lambda = 550 nm.)

    I first found theta.. .052m/2 = .026m (since .026 above and below the center = .052m)
    tan theta = .026m/25300m, and theta = 4.578x10^-5

    Then plugging into the rayleigh criterion for resolution limit of theta = 1.22lamda/D, where D is the diameter of the objective lens (same as aperture of camera??)... I get
    theta = 1.22lamda/D
    4.578x10^-5 = 1.22(550x10^-9)/D
    D= .0147m, which is incorrect..

    anyone know where I went wrong?
  2. jcsd
  3. Apr 10, 2004 #2
    I get d = .326m using
    [tex] sin \theta = 1.22\frac{\lambda}{d}[/tex]
    and letting [tex] \theta = arctan \left( \frac{.052}{25300}\right)[/tex]
    [tex] d = \frac{1.22x5.5x10^{-7}}{sin({arctan \left( \frac{.052}{25300}\right)})} = .326m[/tex]

    Not sure though. Do you know what the answer is supposed to be?
  4. Apr 11, 2004 #3
    Your answer was the correct one. I tried your process earlier, but I think I was confusing the theta since it was given in radians as opposed to doing sin theta for the degrees..

  5. Apr 12, 2004 #4
    You can get the same answer using the [tex]\theta=sin\theta[/tex] small angle approximation, but then you have to be consistent & do the arctan calculation in radians also.

    So [tex]\theta = arctan \left(\frac{.026}{25300}\right) = 2.0553x10^{-6}[/tex]

    and [tex]d = \frac{1.22x5.5x10^{-7}}{\theta} = .326 m [/tex]
  6. Apr 12, 2004 #5


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    Gnome's answer in slightly more legible LaTeX:

    Last edited: Apr 12, 2004
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