I have been confused about this question for a while. Like how can the placement of object exactly on focal point of concave produce no image? What is it about placing the object on the focal point itself that reflected rays are caused to not converge and thus not produce image? Also since object is located right at point of focal point, why can't image be superimposed on where the object is, then?
[Physics] How to the placement of object on focal point produce no image
opticsreflection
Related Solutions
The focal length tell us how much the light rays will be bent. Think of the light rays as a paper cone, just like the one you get when you buy a snow cone. The mouth of the cone is the size of the lens, and the point of the cone is the focal point. The length of the cone is the focal length. Now picture a cone where the point is very close to the mouth. It would be a very steep short cone. The light rays would come in at very steep angles, and after they crossed the focal point they would spread out quickly. In fact, they would make another cone leaving the focal point that matches the cone that entered the focal point. The rays would continue on spreading out wider and wider. Now picture a cone that is very long, say three feet, a novelty snow cone, notice how these rays come into focus at a much smaller angle, and after they pass the focal point they will form another long cone on there way out. Well, there you have it, the focal length tells us how long the snow cone is. I hope your not offended by my simple examples and language in this answer. It's how I think of focal length. With a little imagination you can see how different cone sizes would be applicable to different applications.
The image could be real or virtual. We'll start with a real image. Also, we'll consider a point object and an ideal lens.
For a real image of a point to be formed, the rays emitted by or reflected from that point have to converge at some other point in space.
If a point (blue dot on the diagrams below) is placed in a focal plane of a convex lens and its rays, collected by the lens, are coming out parallel to each other, they, obviously, are not going to to converge and, therefore, are not going to form an image.
If a point is placed in front of the focal plane, the rays are going to converge and form a real image.
If a point is placed behind the focal plane (i.e. between the focal plane and the lens), the rays are going to diverge and, therefore are not going to form a real image. If the diverging rays are extended backwards, they will meet at some point (of the apparent divergence) behind the lens, forming a virtual image.
Hopefully, this clarifies the picture.
Best Answer
When the object is at the focus the image will be formed at infinity.
As the object gets closer and closer to the focal point, the image gets further and further from the mirror. When the object is between the mirror and the focal point the image is behind the mirror (virtual); when the object further from the mirror than the focal point the image is in front of the mirror (real).
At the focal point the image switches from being at $-\infty$ to $+\infty$.This is the nature of parallel lines. A slight deviation from parallel makes the intersection switch from one side to the other.
If you want the image and object to be at the same place on the optical axis, the object should be located at the centre of curvature. The image will be real and inverted and the same size as the object.
Both cases are illustrated in diagrams at The Physics Classroom.