Take the following optical arrange for a concave mirror:

enter image description here

Where the letters stand for Object, Image, Focus, Center

For it, we can use the equation below to calculate image distance formation:

$$\large x_o^{-1}+x_i^{-1}=f^{-1}$$

where $x$ stands for distance to the mirror, $o$ for object $i$ for image.


Do we need to be exactly on the beams intersection ($x_i$) to see the image?

I'm not very familiar with curve mirrors, but it doesn't make sense in my head, as objects can be seen with us standing on different positions.

  • $\begingroup$ Geometric optics is not my strong suit but isn't the statement that the image is formed at x_i just a statement that if you look there appears to be an object at the point in space? $\endgroup$ – jacob1729 May 17 '18 at 17:48
  • $\begingroup$ @jacob1729 nice, im not sure. I think is the point where, if you put the eye, you see an image, but not if you are apart from it... $\endgroup$ – user153036 May 17 '18 at 17:52
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    $\begingroup$ The image point is where all rays come to a single focus. If you are away from it, you can't get a clear image... except our eyes are better than that and given two diverging rays will trace them back to work out the source. I think the image should act like a point source in this case. Will be good if someone comes along with a nice answer though. $\endgroup$ – jacob1729 May 17 '18 at 17:55

There is not an image at I and nothing around it. It does not suddenly appear and disappear as you move through the position I. What happens is that the image moves in and out of focus.

The image is most in focus (most clear) at the position I. This means that each point in the object becomes a point in the image, because (as in your ray diagram) all rays from a single point source intersect at a single point.

At other positions around I the image is blurred, ie out of focus. The further away from I the more blurred the image is. Each point source of light from the object becomes a disk of light in the image. The further from I the larger these disks are, because the rays converge/diverge. Also, the more they overlap with neighbouring disks. This is what makes the image blurred.

  • $\begingroup$ What @sammygerbil said is correct, if a diffuser (like a piece of paper) is moved along the axis: the image comes to a focus on the diffuser when the diffuser is at I. However, if you look toward the mirror from the left side, you will see an image floating at position I, and your eye will relay the image to focus on your retina. No diffuser required in that case. $\endgroup$ – S. McGrew May 18 '18 at 13:25
  • $\begingroup$ Real images and their formation might be useful. $\endgroup$ – sammy gerbil May 18 '18 at 13:50
  • $\begingroup$ Nice. What I can't understand yet is, on that point of image formation, there is no need of an eye, it means that if we put a diffuser the image is clear isn't it? But what if we put the eye? $\endgroup$ – user153036 May 18 '18 at 15:20
  • $\begingroup$ Putting your eye there would be like putting your eye on the object. You would not be able to focus your eye on it. $\endgroup$ – S. McGrew May 18 '18 at 16:41

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