Look, we may consider this by a given illustration with a plane mirror, although I still look ahead of the same doubt in case of spherical as well as parabolic mirrors.

Ray Diagram

Ray Diagram of the illustration!

Correction: The unit was meant to be m instead of cm in 5 cm....however it is of no change.

Now, as we are taught in high school, a plane mirror forms a virtual erect image, in this case, a point image A' of a real object, in this case A at the same perpendicular distance from the mirror as same as the perpendicular distance of object from the mirror which in this case is 5m.

As mentioned earlier the question is same that how can a mirror of 1-1.5 cm thickness form images 10-20 meters behind of itself? Moreover, considering the fact that they are polished from behind so as no light can pass through the mirror..(To be honest, this seems of no use but still I am adding it just for a bit of precaution.)

Edit: If you come to a final verdict that it is kind of a optical illusion for our eyes....I would request you to share some light on the fact that how could the length behind the mirror be illusioned at 5m only for a object at 5m, why not 4m? Why not 6m?

Important Edit

Guys, the ques is not about the formation of image but about the distance at which it is formed and the convention that we are using to represent it...

Moreover, below is an explanation which I got from my high school teacher.

Consider an example of looking one's face in a mirror, he said that assume you are standing at a distance of 5 m from the mirror,i.e. your face (object) is at a distance of 5m from the mirror,now the light Ray after striking your face will incident onto the mirror(Till now it has covered 5m) and reflect back to your retina, i.e. where the image will be formed.(Covered another 5m). That is how he argued that this to and fro path of light rays [(5+5)m] is represented as such on paper under the convention.

This is of no doubt that no image is formed behind the mirror it just appears to form there but is actually formed in the retina of one's eye.(A clarification for a answer below)

Now talking of my teacher's reason....this fact seems to work only when the image of eye is being formed...just consider another simple example that is we are standing in front of a mirror at 5m that is wide enough and my friend is standing behind me at a distance of 7m from the mirror in such a way that none of our images overlap each other(Or atleast we don't see them overlapping, however even if they overlap it will not be of any issue but still including this to prevent another answer just concerning all this non-relative to the question stuff). Now the light rays after striking his face are incident onto the glass(covering 7m) and reflecting back into my eyes(covering 5m). This results in image-object distance to be just 12m according to sir's explaination, not which it should be i.e. 14m? I know this view distance is totally observer relative....but this was to include my teacher's reasoning and to further add to the question , that where did he go wrong ?

  • $\begingroup$ The thickness of the mirror is actually irellevant. The light is reflected by the thin layer of aluminum of silver which is just a fraction of millimeter. Even if you have mirror made of a thick slab of polished metal, the thickness of the slab is irellevant. The refelction electromagnetic radion interact just with the superficial layer. $\endgroup$
    – nasu
    Mar 25, 2023 at 13:50
  • $\begingroup$ @nasu Good Sir! This was really of beneficial knowledge for me....Neither were we told like this by our professors nor I visualized the same ever $\endgroup$
    – Darshit ff
    Mar 25, 2023 at 13:53

1 Answer 1


There is no "meeting" of the rays from the object behind the mirror rather a divergent beam of rays, $OA$ and $OB$, which originated from the object, $O$, are reflected off the surface of the mirror $A$ and $B$.
Note that the thickness of the mirror is irrelevant and mirror tend to have the reflecting surface at the back so as to protect the surface from scratching and corrosion.

enter image description here

The divergent reflected rays, $AC$ and $BD$, then enter you eye and the optical system of the eye forms an image on the retina of your eye.

Those divergent rays, $AC$ and $BD$, appear to have come from a region behind the mirror, $I$, which is as far behind the mirror as the object is in front of the mirror.
It is those diverging rays which form an image on the retina and as far as the eye is concerned those rays could have come from an object at position $I$ to form the image on the retina and that is what the eye "sees", a "copy" of the object behind the mirror which is called a virtual image.

  • $\begingroup$ I appreciate all the efforts tbh! But to be a bit more concise in the same matter you may have demonstrated how the distance between the image and object appears to be 10 cm (5+5) apart from each other. Referring to the edit , why it is not 9m or 6m ?? Taking your answer into consideration the total path covered by light must be either more or less than 10 m?? Right? $\endgroup$
    – Darshit ff
    Mar 25, 2023 at 14:20
  • $\begingroup$ I personally do not feel the fact of the image object distance being 10m in this case is validated in the above answer. Thus , saving your reverence, I am completely against the three upvotes for the same. $\endgroup$
    – Darshit ff
    Mar 25, 2023 at 14:55
  • $\begingroup$ @Darshitff Since the angle of incidence is equal to the angle of reflection the geometry of the situation tells one that triangles INA and ONA, and triangles INB and ONB are congruent so IN+ON. $\endgroup$
    – Farcher
    Mar 25, 2023 at 23:30

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