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Homework Help: Diffraction lens diameter Problem

  1. Dec 7, 2003 #1
    Tried this problem for about 2 hrs and no dice...

    A spy satellite orbiting at 150 km above the Earth's surface has a lens with a focal length of 3.5 m and can resolve objects on the ground as small as 36 cm; it can easily measure the size of an aircraft's air intake. What is the effective lens diameter, determined by diffraction consideration alone? Assume = 527 nm.

    I dont think the 150km matters...
    i just did .36 = f*theta and i got theta = .36/3.5


    then i did theta = wavelength /d (diameter) and i get like 5.12e-6M

    totally wrong...help

    btw as soon as someone replies i will prolly reply back since its kinda urgent
     
    Last edited: Dec 7, 2003
  2. jcsd
  3. Dec 8, 2003 #2

    chroot

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    Staff Emeritus
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    Consider the Rayleigh criterion for a circular aperture:

    [tex]\sin \theta_R = 1.22 \frac{\lambda}{d}[/tex]

    In this case, [itex]\theta_R[/itex] is the angle subtended by a 36 cm feature from 150 km away. Hint: draw a right triangle. The angle subtended is

    [tex]\begin{align*}
    \theta_R &= 2 \tan^{-1} \left( \frac{0.18}{150 \cdot 10^3} \right)\\
    & \approx \tan^{-1} \left( \frac{0.36}{150 \cdot 10^3} \right)\\
    & \approx 2.4 \cdot 10^{-6}\ \text{rad}
    \end{align*}
    [/tex]

    The Rayleigh criterion then gives

    [tex]\begin{align*}
    d &= \frac{1.22 \lambda}{\sin \theta_R}\\
    & \approx 0.27\ m
    \end{align*}
    [/tex]

    Does this make sense?

    - Warren
     
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