0
$\begingroup$

enter image description here

The picture shows the image formed between c and the focus when the object is placed beyond the centre of curvature(c).But my question is since all parallel rays falling on a spherical mirror must pass though the focus after reflection,is there also a point sized image at f? wouldn't all the rays emerging parallel from every point on the object meet at the focus after reflection? So does that mean that wherever i keep the object there will always be a point sized image at f?

$\endgroup$
  • $\begingroup$ The ray from A that passes through C does not pass through F, so F is not on the image plane. $\endgroup$ – lionelbrits Jan 11 '15 at 13:07
  • $\begingroup$ @lionelbrits i am not considering that ray from A that passes through C.I am talking about all the parallel rays that come from the object. The parallel rays have not been portrayed in the picture but you get the point. $\endgroup$ – SMcCK Jan 11 '15 at 15:07
  • $\begingroup$ When considering image formation, the parallel rays have measure zero among all the rays, and thus do not form an image by themselves. Physically, the "image" formed by the parallel rays alone would be too faint to be seen. $\endgroup$ – lionelbrits Jan 11 '15 at 15:20
  • $\begingroup$ @lionelbrits what exactly do you mean by 'measure zero'? sorry but i'm new to this. $\endgroup$ – SMcCK Jan 11 '15 at 15:44
  • 1
    $\begingroup$ Probably not very mathematically exact, but an image is formed when light from one spot takes many paths and converges to another spot. You can think of the final spot as being the sum of infinitely many paths, each having almost no intensity. It takes infinitely many rays to give you something visible. If you only counted the parallel rays, being so few, they would add up to virtually nothing. It helps to ask why we don't see light rays when we are far away from the focal point (even though they clearly pass through the viewing plane). $\endgroup$ – lionelbrits Jan 11 '15 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.