• raleigh Criterea

    From whburling@outlook.com@21:1/5 to All on Mon Oct 23 12:15:03 2017
    Text books often define the Raleigh Criterea as Sin(theta) = 1.22* Lambda/D

    Should not the criterea be, TAN(theta) = 1.22 * Lambda/D

    I am suggesting the criterea should be with tan from looking at its derivation from Bessel function

    r = 3.832/pi * L*Lambda/D to first point
    r = 1.22 * L*Lambda/D
    r/L = 1.22 * lambda/D
    tan (theta) = r/L = 1.22 * Lambda/D

    if i could draw a picture of r and L, that would be great, but I will have to draw it with words. I am assuming r is the distance between the edge of the aires disc to the center (its radius) and the L is the focal length which is the triangle side
    which forms a right angle with the radius. the hypotenuse is not directly defined; hence the application of tan and not sin which requires knowing the hypotenuse value.

    I appreciate that if the angle is small, then sin(theta) = tan(theta)
    some people say sin(theta) can equal theta but I can not see that.

    sin (1E-5) = 1.745E-7 degrees
    tan (1E-5) = 1.745E-7 degrees
    but 1E-5 does not equal 1.245E-7

    Please help me understand. I truly am in the space of trying to understand.

    Thank you
    bil

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  • From Phil Hobbs@21:1/5 to whburling@outlook.com on Mon Oct 23 16:08:28 2017
    On 10/23/2017 03:15 PM, whburling@outlook.com wrote:
    Text books often define the Raleigh Criterea as Sin(theta) = 1.22* Lambda/D

    Should not the criterea be, TAN(theta) = 1.22 * Lambda/D

    I am suggesting the criterea should be with tan from looking at its derivation from Bessel function

    r = 3.832/pi * L*Lambda/D to first point
    r = 1.22 * L*Lambda/D
    r/L = 1.22 * lambda/D
    tan (theta) = r/L = 1.22 * Lambda/D

    if i could draw a picture of r and L, that would be great, but I will have to draw it with words. I am assuming r is the distance between the edge of the aires disc to the center (its radius) and the L is the focal length which is the triangle side
    which forms a right angle with the radius. the hypotenuse is not directly defined; hence the application of tan and not sin which requires knowing the hypotenuse value.

    I appreciate that if the angle is small, then sin(theta) = tan(theta)
    some people say sin(theta) can equal theta but I can not see that.

    sin (1E-5) = 1.745E-7 degrees
    tan (1E-5) = 1.745E-7 degrees
    but 1E-5 does not equal 1.245E-7

    Please help me understand. I truly am in the space of trying to understand.

    Thank you
    bil

    Sure, no worries. We all start out there.

    First of all, the small angle approximation. sin(theta) ~= theta for
    small theta, but it has to be in radians, not degrees. Mathematically,
    that's because

    sin x = x - x**3/3! + x**5/5! -x**7/7! + ....

    Thus the relative error in approximationg sin x ~= x is about
    (x**3/3!)/x = x**2/6.

    Thus for x < 0.1 radian (~5.7 degrees), the relative error is less than
    0.17%, which is pretty good for most things.

    Next, the Fourier transform relation.

    The conjugate variables in 1-D Fourier optics are x/lambda and u =
    sin(theta), so all the angular dependences come out in terms of
    sin(theta). That is, Fourier optics connects the spatial and angular
    domains. The natural place to measure the spatial pattern is on a
    plane, and the natural place to measure the angular pattern is on the
    inside of a large sphere. If you cut a ping-pong ball in half, it makes
    a useful scatterometer for visual observations. (I have several in my
    drawer.)

    If the Rayleigh pattern were a function of tan(theta), the fringe
    spacing would go to zero as you go to +-90 degrees, whereas in real life
    the fringe period can't be less than than lambda. What actually
    happens, if you use a very narrow pinhole and project the fringes on the
    inside of your ping-pong ball is that you'll see the fringes get a bit
    broader as you go towards theta = pi/2 rather than narrower, and they
    cut off at some lowish order like 5 or 10 or 100 depending on the
    diameter of the pinhole.

    If you project the pattern on a plane perpendicular to the beam axis,
    the fringes get much wider as you go out in angle.

    Cheers

    Phil Hobbs



    --
    Dr Philip C D Hobbs
    Principal Consultant
    ElectroOptical Innovations LLC
    Optics, Electro-optics, Photonics, Analog Electronics

    160 North State Road #203
    Briarcliff Manor NY 10510

    hobbs at electrooptical dot net
    http://electrooptical.net

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