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Single-slit Diffraction Problem

  1. Aug 12, 2017 #1
    1. The problem statement, all variables and given/known data
    A spy camera is said to be able to read the numbers on a car’s license plate. If the numbers on the plate are 5.0 cm apart, and the spy satellite is at an altitude of 160 km, what must be the diameter of the camera’s aperture? (Assume light with a wavelength of 550 nm.)

    2. Relevant equations
    Single-slit diffraction: wsinθ = (where w is the width of the slit, m = 1, 2, 3, ... for destructive interference)
    Small-angle approximation: sinθ ≈ tanθ = y/D (where y is the vertical height above the central axis, D is the distance between the slit(s) and the screen) for small θ

    3. The attempt at a solution
    Since the goal of the problem is to solve for w, I used the small-angle approximation to set up the following equation: /w = y/D. I then rearranged the variables to isolate w, ending up with w = mλy/D. Now, I know that D = 160 * 103 m and λ = 500 * 10-9 m. I'm having trouble, however, figuring out what to do for m and y. Since the numbers on the plate are 5.0 cm apart, I'm assuming that, every 5.0 cm, we have constructive interference occurring. Since there will definitely be constructive interference along the central axis, I figured that the first-order minimum would occur halfway in-between the central bright fringe and the first-order bright fringe, which would give m = 1 and y = 2.5 cm. However, this does not give me the correct answer; if I try instead to use m = 1 and y = 5.0 cm, this does give me the right answer, but I can't come up with a sketch of the situation that would explain why. Some assistance with the intuition here would really be appreciated.

    Thank you very much in advance for your help!
  2. jcsd
  3. Aug 13, 2017 #2


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  4. Aug 13, 2017 #3

    Charles Link

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    The central maximum for single slit diffraction, (where ## m=0 ## and constructive interference occurs) is the item of interest here. Most of the energy is in this central part of the image which becomes a small blob (instead of a point) of width ## \Delta x=f \Delta \theta ## when focused by the lens with focal length ## f ##. The width of this blob is often taken to be ## \Delta \theta=\lambda/w ##. (Sometimes with a factor like 1.22 in front of ## \lambda/w ##. To actually read the numbers, it will need to have a resolution even better than the ## \Delta \theta=5 \, cm/160 \, km ## that they give you, but that is apparently the ## \Delta \theta ## they want you to use. ## \\ ## Additional info/question that might be helpful: What is the spacing ## \Delta x ## in the focal plane of the lens of two numbers on the license plate that are separated by ## 5 \, cm ##? ## \\ ## And additional item: The far-field diffraction pattern from light incident on the lens (such as perfectly parallel rays which are going through the aperture of width ## w ## (=the width ## w ## becomes essentially a slit width)) gets imaged by the lens in the focal plane of the lens. An angular spread ## \Delta \theta ## in the far-field diffraction pattern gets converted to a distance ## \Delta x ## in the focal plane given by ## \Delta x=f \Delta \theta ##. The light (all parallel rays on the lens) that would go to angle ## \theta ## in the far field winds up being focused at position ## x=f \, \theta ## in the focal plane of the lens. Thereby, the far-field diffraction pattern that would be created by a set of incident parallel rays on the lens winds up as the image that results in the focal plane, rather than getting the perfect focused point that the ray trace theory would give. (Instead of a perfectly focused point, you get a very small focused spot or blob).
    Last edited: Aug 13, 2017
  5. Aug 13, 2017 #4
    @scottdave and @Charles Link Thanks very much to both of you for the help; I think I've gotten the right answer now.
  6. Dec 30, 2017 #5
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