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How to understand the image of lens at infinity

  1. Aug 15, 2009 #1
    An object at the focal length distance from the lens is imaged at infinity,Do this mean that under this situation, our eyes could not see the image? but as our eyes could see the stars from infinity, do this mean that the image of lens which is discussed above is just viewed through screen ,and is different from what is viewed by eyes directly? Then,what could we see in this situation ?
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  3. Aug 15, 2009 #2


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    Hi navigator! :smile:
    Nope … it means that we have to focus our eyes "at infinity" to see it, just as for anything in the distance. :wink:
  4. Aug 15, 2009 #3
    tiny-tim, is that right? If the object is at the focus, the rays will go out parallel and they won't intersect. I would think that the formation of an image is impossible.

    Put the object very near the focus but not right at it, and then you can form a clearly-focused image on a screen a million miles away, but then the screen will be too far away to see it. Or -- bring the screen close enough to see it, and accept the fact that the image will be blurry.

    I will appreciate being corrected if I said that wrong. Thanks.
    Last edited: Aug 15, 2009
  5. Aug 15, 2009 #4


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    It depends how much blur you are willing to accept and how wide the lens is (i.e. the lens aperture), but for most purposes a screen that is only, say, 10 m away will be sharp enough, and for small lenses even closer than that.
  6. Aug 16, 2009 #5


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    Hi mikelepore! :smile:
    ah, I was only answering about the effect on the eyes. :wink:

    But, as DrGreg :smile: says, although the effect on a screen will be slightly blurred, you'll only have to go a short distance for it not to be really noticeable.

    Try a pair of binoculars, and see how far away you have to look before focussing ceases to make any difference … that'll be "infinity"! :smile:
  7. Aug 16, 2009 #6
    Thanks for your replies. I still wonder whether we can see the image, because our eyes also act as lens during our observation and would focus the collimited light to form an image, so we would alway see an image for our eyes can focus automatically? But the fact is that when I look forward through a convex lens with different distance, there are alway exist a distance at which the view is totally blurry , how to explain this phenomenon?
  8. Aug 16, 2009 #7

    Andy Resnick

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    What you are describing relates to the core concept of 'imaging'.

    As a quick response, and under the usual simplifying assumptions, it's more correct to discuss the image not as existing on a plane at infinity, but instead by using *angular coordinates* to describe the image. That's why most diffraction patterns are described in terms of angles rather than sizes.

    Think about what your second sentence really says- distant objects are imaged onto our retinas as the fourier transform of the shape of the object. The optical field at our pupil is a plane wave (for all practical purposes), and the image at our retina is the Fourier transform of the optical field at our pupil.

    If a transmissive object, illuminated by a plane wave, is placed at the front focal plane of a lens, then as a screen is moved further and further away from the lens, the optical field approaches the spatial Fourier Transform of the object.

    Des that help?
  9. Aug 16, 2009 #8


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    I am facinated by the motion of the image as the object moves through the focal point.
    consider the http://hyperphysics.phy-astr.gsu.edu/Hbase/geoopt/lenseq.html" [Broken]:

    [tex] \frac 1 f = \frac 1 o + \frac 1 i [/tex]


    [tex] i = \frac {fo} {f-o} [/tex]

    From this you can see that at f= o the image distance is undefined, so really you cannot say that an image is formed at infinity just that the image distance approaches infinity. As the object passes through the focal point the image distance changes sign. The real image moves to positive infinity disappears then reappears as a virtual image at negative infinity.
    This behavior can be observed with a magnifying lens. Just look through it and watch the image as your eye passes through the focal point.

    For the mathematically inclined if you do a coordinate transform consisting of a [itex] \frac \pi 4 [/itex] radian rotation you will see the standard from of a hyperbola. Moving the object through the focal distance forces the image to cross an asymptote. That means the image jumps from one branch of the hyperbola to the other.

    To find the focal distance of a lens simply observe the image formed by a distant object. In general you can treat any distance over 10xf as equivalent to infinity i.e. "distant".
    Last edited by a moderator: May 4, 2017
  10. Aug 16, 2009 #9
    When O is outside f the rays converge to a real image(convex lens) and when O is inside f the rays diverge from a virtual image.At f the rays emerge parallel and I dont think we can classify the image as being real or virtual.The eye should be considered as an integral part of the whole system and this will focus the parallel rays.Anyway,thats the theory but if a point object were placed at f, a tiny change in object distance can result in a huge change in image distance and so,with a real extended object, different parts of it will have large different image distances which,for want of better words,will result in a resultant "blurred image".
    Last edited: Aug 17, 2009
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