will Nikon release professional "digial" lenses ?

Discussion in 'Nikon' started by Michael Schnell, May 25, 2006.

  1. Michael Schnell

    Jeremy Nixon Guest

    If there's a lens designer in the house (and a couple have been around
    here from time to time) they could probably give a better answer, but I'll
    give it a try.

    The whole theory you state above is correct, but it's based on the *theory*
    of the lens. The focal length of the lens, the optical length, is the
    distance from which a pinhole would project an image of the given size.
    The size of that hole is the aperture -- again, in theory, since a large
    hole won't actually project a useful image at all. That's where all the
    glass comes in.

    So, to create a given aperture ratio (f-stop) at a given focal length, you
    don't need an iris of that size, you need an iris that's going to create
    the effect of the hole that size in the theoretical simple "lens". But
    you need a front element of that size in order to be able to accomplish it.
    So the whole system is part of it -- it's not solely the front element's
    size that determines the max aperture -- but the simple part of it is that
    the front element has to be (at least) the size you need, because that's
    the only way to get that amount of light collected. We know the light
    needed is what falls on that physical area, so it just can't be any
    smaller no matter what optical tricks you play further into the lens.
    The iris itself certainly needs to be *some* size, I'm sure (and I don't
    know what size that is, myself), but it doesn't have to be *that* size.

    In fact, we know that can't be required, because a 300mm f/2 lens exists,
    and the lens mount isn't 150mm wide. Nikon sells a 200mm f/2 lens right
    now, and it's not 100mm wide, either.

    And now, someone who actually knows lens design is welcome to fill in the
    holes in my attempt. :)
    Jeremy Nixon, May 26, 2006
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  2. Michael Schnell

    Roy Smith Guest

    What happens when you reverse a lens using a macro ring? The focal length
    stays the same, but the "front" of the lens is now the small end. Does
    that mean that the f-stop becomes smaller when you flip the lens around
    without changing any settings?
    Roy Smith, May 26, 2006
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  3. Michael Schnell

    J. Clarke Guest

    Actually it is. The outside diameter is 124mm. Perhaps the 52mm filters
    are throwing you--that's because the filter mount is behind the front lens.

    The outside diameter of the 300/2 is 183mm, leaving ample room for a 150mm
    diameter front element--this also takes 52mm filters in a drawer.

    Don't confuse the iris with the aperture--the full open aperture is limited
    by the diameter, the iris however may be located at a point in the optical
    path where the beam is narrower and may very well have a significantly
    smaller diameter than the front element. As an example, the iris on an
    F/1.5 Summarit is about half the diameter of the front element.

    The trick that _can_ be done is to design a teleconverter into the lens so
    that for a given effective focal length it can be _shorter_ than otherwise.
    J. Clarke, May 26, 2006
  4. Michael Schnell

    Paul Furman Guest

    Yes, this is my understanding.
    Paul Furman, May 26, 2006
  5. I see that this assumption was wrong. To get the same effect as a
    450/2.8 on a 35mm camera I supposedly need a "DX" 450/2.8 lens.

    Michael Schnell, May 26, 2006
  6. Michael Schnell

    Jeremy Nixon Guest

    I meant the lens mount isn't that wide. The lens obviously is.
    Jeremy Nixon, May 26, 2006
  7. Michael Schnell

    Jeremy Nixon Guest

    I would imagine it must; you can only gather as much light as will fall on
    the area of the front element. But I've never done that or really looked
    into it.
    Jeremy Nixon, May 26, 2006
  8. Michael Schnell

    bjw Guest

    No, it's the focal length divided by the diameter of the entrance
    pupil. The entrance pupil is the image of the iris as viewed
    from the front of the lens. This is the area that can actually
    collect light from the source. Generally with a long-normal or
    tele lens the front element diameter is about the diameter
    of the entrance pupil, and about matches focal length/f-number.
    You can verify this just by looking from the front.

    In many lenses, especially normals and telephotos, the elements
    in front of the iris are positive and magnify the iris, so the
    entrance pupil can be significantly larger than the physical
    iris. This is really obvious if you take a lens apart. This is
    hard to do with 35mm SLR lenses, but easy with large format
    lenses, where the front and rear groups are designed to
    unscrew from the shutter/iris assembly.

    The exit pupil is the image of the iris as viewed from the rear.
    In an asymmetrical lens, the entrance and exit pupils can be quite
    different in diameter. Thus if you turn the lens around,
    you reverse entrance and exit. So a 300/2.8 reversed (if anyone
    were crazy enough to use it that way) is no longer an f/2.8 lens,
    because the formerly-exit pupil is a lot smaller.

    The lens mount diameter does restrict the degrees of freedom
    a lens designer has, because it restricts the size of the rear
    element and exit pupil. I don't know for sure, but suspect this
    is not very important except for ultra-fast lenses. It's mostly
    a marketing point since most major manufacturers have been
    able to make lenses which are so fast that they are too expensive
    for all but a few.

    Many or most users own a lens whose _entrance_ pupil
    is bigger than the mount diameter (every 135/2.8 or 200/4
    for a 35mm camera ...) Tele lenses are slower than
    wide angles in part because of the _entrance_ pupil requirements,
    which mean ridiculous amounts of glass are needed to make
    a 300/2.8, let alone a 200/2, 300/2 etc. (And other issues
    having to do with controlling aberrations as a function of lens
    speed and so on.)

    bjw, May 26, 2006
  9. Michael Schnell

    JPS Guest

    In message <>,
    Does this mean that for any given focal length, you really can't say how
    much diffracted light is in an image, as a percentage of total light,
    based on the f-stop, since the closer the iris is to the entrance pupil,
    the larger its area:diameter ratio?
    JPS, May 27, 2006
  10. Michael Schnell

    JPS Guest

    In message <>,
    Here's a though experiment; you mount the lens in a dark corner of a
    room on a tripod, lens facing the outdoors through the window. Make a
    piece of paper with rectangles on it the size of various sensor and film
    formats. Hold the paper so that the image focuses on each rectangle.
    Same brightness! The lens doesn't know or care what's capturing its
    light; it always acts the same.
    JPS, May 27, 2006
  11. Michael Schnell

    JPS Guest

    In message <e5754q$q0q$02$-online.com>,
    You don't need a DX lens at all.

    As others have already said, 300mm lenses for 35mm cameras are already
    as small as they can be to have their apertures and focal lengths. With
    wide-angle lenses, the case is different and the bulk of the lenses is
    necessary to achieve 35mm full-frame coverage.
    JPS, May 27, 2006
  12. Michael Schnell

    Roy Smith Guest

    That can't be right. The same amount of light (I've forgotten some of my
    physics; maybe it's flux?) will fall on a larger area, so it won't be as

    It's exactly the same thing as back in the neanderthal days when you
    changed the enlarger from printing a 5x7 to printing an 8x10. Same total
    amount of light spread out over a larger area means less brightness and a
    longer exposure.
    Roy Smith, May 27, 2006
  13. Michael Schnell

    Bill Guest

    Actually it is right...you're not thinking about it right.

    The TOTAL amount of light gathered is less on a smaller sensor, but the
    PER PIXEL amount is the same regardless of sensor size.

    Think of a film frame cut into 9 squares (3x3 like a Rubiks cube). Each
    frame will have the same amount of exposure, but each will only have a
    small section of the total image. The center square will be similar to
    what a small digital sensor captures because it's not as large as the
    regular film frame.

    The intensity of light hitting that smaller section will be the same as
    the intensity that hits all of the other sections.

    Try this page to visualize the smaller image circle:

    That's not the same thing - you're using the same amount of light but
    spreading it across a greater surface.

    With digital sensors you're simply capturing a smaller section of the
    total image circle.
    Bill, May 27, 2006
  14. Michael Schnell

    DoN. Nichols Guest

    But it *is* right. Nothing has been done to change the focal
    length of the lens, so it is providing a constant size *image* of what
    is outside the window. The only thing that you are changing is what
    percentage of that element you are actually *using*. The rest simply
    shoots past the paper rectangles and is lost in the room.

    Remember -- to capture an image in focus, each of those
    rectangles of paper must be the same distance behind the lens. Try to
    move the paper to get the light to *fully* illuminate a larger or
    smaller rectangle, and the image will be out of focus.
    For a given focal length lens, and a given reproduction ratio, a
    given area of the subject will produce a constant area of image, so
    you cannot change that without changing either the lens focal length, or
    the distance to the subject.

    What you are talking about is light from a point source, not
    light focused by a lens. Any lens of any focal length of a given
    f-stop, imaging a constant intensity subject (say a gray card) will
    produce the same illumination per square millimeter at *any* distance as
    long as you are within the image of that gray card. This is why f-stops
    are so useful in controlling exposure. With some lenses of very short
    focal lengths, the area of the image of the gray card may be too small
    to be easily measured (though the film or the sensor can be used to
    measure the level with a given shutter speed).

    (Actually -- at extreme distances, in atmosphere, some light
    will be absorbed or diffused by the atmosphere, so let's move this to a
    vacuum, where there is no atmosphere involved in the equation. :)
    Remember -- when you moved the height of the enlarger, you
    *also* had to adjust the lens to negative distance to bring things back
    into focus. In the thought experiment described above, the lens to
    subject distance is held constant (distance from lens to whatever is
    outside that window), and the lens focal length is also constant (did
    anyone *ever* put zoom lenses on enlargers? I hope not.), so the
    distance from the lens to an in-focus image, and the image size remain
    constant. Again, you are simply not using all of that image area with
    smaller rectangles of paper (smaller sensors).

    DoN. Nichols, May 27, 2006
  15. Michael Schnell

    JPS Guest

    In message <>,
    I meant to say, "area:circumference", but what I wrote works, too, as
    diameter and circumference are always in the same proportion for the
    same shape.
    JPS, May 27, 2006
  16. Michael Schnell

    JPS Guest

    In message <>,
    That analogy is inapplicable here; the enlarger can vary the focal
    length, which is what you do for different-sized projections.
    JPS, May 27, 2006
  17. Michael Schnell

    Stacey Guest

    Michael Schnell wrote:

    It wouldn't be any lighter than the non-digital version. The weight is in
    the front piece of glass to get a F2.8 300mm, you can't reduce the size of
    the front element and still get F2.8 300mm in a lens no matter what the
    format. Where weight and size reduction comes from is in the wide angle
    lenses. Go look at say a 150mm F4 35mm format lens vs 6X6 medium format
    (they'll be ~ the same size) then look at 1 50mm F4 (The medium format one
    is HUGE) and you'll see what I'm talking about.
    Stacey, May 27, 2006
  18. Michael Schnell

    Stacey Guest

    And people wonder why the olympus designed for 4/3 version is so expensive..
    Stacey, May 27, 2006
  19. Michael Schnell

    Stacey Guest

    You'd be wrong. I'd be willing to bet that 300 F2.8 would easily cover 6X6
    if not larger if the lens barrel didn't block the light path.
    You'd be wrong again. The F2.8 determines the diameter of the front element,
    not the format in this case.
    Stacey, May 27, 2006
  20. Michael Schnell

    Stacey Guest

    Again in a long lens, it will NOT be smaller and lighter. Go look at the
    4/3 300mm F2.8, it's the same size as the 35mm versions, just higher
    Stacey, May 27, 2006
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