What happens when an object is placed at the focal point of a convex lens?

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When an object is placed at the focal point of a convex lens, the refracted rays do not intersect, resulting in an image distance of infinity, meaning no image is formed on a screen. In astronomical telescopes, the first image created by the objective lens is positioned at the eyepiece's focal point, allowing the eye to perceive the image as it focuses parallel rays on the retina. This principle applies similarly to magnifying glasses, where the final image is also at infinity, requiring the eye to convert these parallel rays into a real image. The distinction between real and virtual images is clarified, with virtual images appearing to diverge from a point on the object side of the lens. Ultimately, the eye acts as an additional lens, enabling the perception of images formed by the convex lens despite the initial lack of intersection of the light rays.
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Concerning the Convex Lens and my diagram attached, when an object is placed exactly at the principle focus (focal point) of the convex lens, if we apply the concept of Light Ray Diagram, the Principle Ray and Centre Ray after refraction does not intersect. From here, with 2 refracted ray extending until infinity, image distance is infinity .Notes on Physicsclassroom said that no image will be formed as there is no intersection of the light rays.

However in our astronomical telescope, the first image formed by the Objective lens is adjusted until it is located exactly at the focal point of the eyepiece , causing the distance between the objective lens and the eyepiece to be the summation of the focal length of objective lens and the eyepiece. Does this means during this arrangement the image also can be formed ??

Than what will we see when the object is placed exactly at the focal length of a magnifying glass(convex lens) ?
#Is this moment is actually the momentarily blurring of the image due to the Transformation of the Largest Inverted Real Image to the Largest Upright Virtual Image ??
 
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Hey guys i really need your help! Explain to me please.
 
I have no idea what you are trying to say. "Momentary blurring"? If you place an object at a distance equal to the absolute value of the focal length of a convex lens, away from that convex lens, what is the image distance and what does the sign mean? Is it really infinity? And in an astronomical telescope, which really doesn't have anything to do with the convex lens problem, what is the final image distance?
 
That's the problem i got. Sorry for unclear statement. Because if we use 2 rays to illustrate the location of the image formed from an object placed exactly at the focal point of the convex lens, the 2 rays refracted do not intersect. Image distance is infinity. Could i say in this way? And at this moment, what will we see on the other side of the lens?
 
The eye-piece may form the image at infinity. This means that if you try to catch the image on a screen it won't work.
But you are not trying to capture the image but rather you put your eye in the path of these parallel rays. The lenses in the eye will bend the parallel rays from the eyepiece into an image on the retina.
So the bottom line is that you have another lens after the eyepiece (the lens in the eye)..
 
Yh Hoo said:
...if we use 2 rays to illustrate the location of the image formed from an object placed exactly at the focal point of the convex lens, the 2 rays refracted do not intersect. Image distance is infinity.

No, that is not correct. The rays do intersect, just not in the space beyond the lens. What kind of image is this? Do you understand what real and virtual images are?
 
i think i do. Virtual image is formed when light rays seems to intersect in the side of the lens where the object is placed but actually the refracted rays that enter our eyes does not intersect right ? Real image is formed when refracted rays intersect on image side right ?
However in this case because of they are parallel rays.. they will not actually intersect right ? but they too do no appear to diverge from a point at object side.. Should it be a virtual image ?
 
1/o + 1/i = 1/f; f is negative, o is positive, so i is negative and not infinite. The rays exiting the lens are not parallel, they converge somewhere - where do they converge, and what does that mean?
 
Okay. By using the equation 1/o + 1/i = i/f. In this case the object distance,o is exactly equal to the focal length,f.(As in my attachment) so, 1/i = 0. Image distance,i = infinity . Is it ?
And according to the picture, the 2 light rays on the left hand side of the lens is parallel right ?
 
  • #10
I give up. BTW, there is no "attachment" visible.
 
  • #11
come on professor. I really do not understand what you are trying to imply and i really need explanation from adorable people like you . I have attached the photo. My problem is will the 2 refracted rays intersect?
 

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  • #12
nasu said:
The eye-piece may form the image at infinity. This means that if you try to catch the image on a screen it won't work.
But you are not trying to capture the image but rather you put your eye in the path of these parallel rays. The lenses in the eye will bend the parallel rays from the eyepiece into an image on the retina.
So the bottom line is that you have another lens after the eyepiece (the lens in the eye)..

that means we will be able to see the image when the object is placed exactly at the focal point of the convex lens right ?
 
  • #13
Ah, my eyes saw "convex lens" and thought of a convex mirror, which has a negative focal length - my mistake. Yes, the image is at infinity, meaning the lens by itself does not form an image. In order to form an image, you need another lens, which is what your eye does - it takes the parallel light rays and focuses them down to make a real image on your retina.

Same deal with a telescope - final image is at infinity, and your eye focuses the parallel rays down on your retina.

Same deal with the magnifying glass. Final images is at infinity, and your eye focuses the parallel rays down on your retina.

If you replace your eyeball with a piece of film, then in each case you do not see an image.
 
  • #14
Yea. Thanks a lot professor! I have got what i demand for.
 
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