- #36
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Perhaps it will be convincing to treat this question as a compound mirror and lens problem and put some math in it. Let ##o_1## = distance of man's face in post #19.A.T. said:You can always make the hole bigger and a put a lens in it. But for starters the OP learn to distinguish between the real image formed by the mirror, and the real image in the eye/camera.
First we find the position of the real image produced by the mirror relative to the mirror: ##\dfrac{1}{o}+\dfrac{1}{i_1}=\dfrac{1}{f_1}.##
The real image produced by the mirror is a virtual object for the lens at a negative distance ##o'=-(i_1-o)## from the lens.
This gives a second equation ##\dfrac{1}{-(i_1-o)}+\dfrac{1}{i_2}=\dfrac{1}{f_2}##
Solving the system of two equations and two unknowns yields ##i_1## and ##i_2##.
Judging from OP's photographs, ##f_1\approx 60~\mathrm{cm};~~o\approx 90~\mathrm{cm}##. With a "commonly accepted value" from the web ##f_2=2.4~\mathrm{cm}##, I got ##i_2=+2.34~\mathrm{cm}##.
If we assume that ##f_2## is the distance from the eye lens to the retina, this value places the image in front of the retina which means that the image would be (slightly?) out of focus. The image will be interpreted as "flipped" because there is no flip in the second equation since the object is virtual. Furthermore, the answer for ##i_2##, which I do not provide because I think it would be instructive for OP to derive it, is multiplied by an overall factor of ##f_2##. This says that even if the eye muscles change the focal length of the lens, the image will still not be in focus. The reason why ##i_2## and ##f_2## are close is because ##o## and ##f_1## are close in value and much larger that ##f_2##.