# Eye Optics/Emmert's law: Where to place scaled copies of an object so they have identical images on the retina?

I read that the center of projection of the human eye is the entrance pupil. So given a light ray which intersects the objects position and the center of the aperture stop, all copies of said object when placed along that line should cover the exact same position/area on the retina (assuming they are scaled according to their distance).

However, doesn't this ignore the effect of accomodation of the human eye: If the original object is quite near, but a scaled copy along the ray is quite distant, the focal length of the lens is changed and thus a different position and area of the retina is covered.

I tried to visualize this by using the ray simulator at https://ricktu288.github.io/ray-optics/simulator/:

Image A:

Lens focussed on distance of Object A which is projected onto A' on the retina. Object copies B and C are placed at fixed distance either along a ray via the aperture stop (B) or via the center of the lens (C).

Image B:

Lens is focussed on distance of Object B/C which are projected onto B* and C* on the retina. Both are in focus (difficult to see in the picture), but B* is further away from A' than C*. Object B would appear larger than Object C.

So I guess this boils down to the question:

Given the position P of an object A and a lens system, what is the point Q which creates a ray PQ so that all scaled copies of A (A', A'', ...) at different distances along that PQ ray appear at the same position and area on the image created via the lens system.

Is there such a point Q? Is it the center of the lens (i.e. is C* == A') or one of its nodal points?

I am thankful for any pointer to help me understand this!

• I suppose that you need to use the thin lens equation (en.wikipedia.org/wiki/Lens#Imaging_properties), but with different focal distances. Now, the focal distance is not the same as the distance to the object - so one needs some data on how the focal distance of an eye adjusts to the distance to the object. You may try asking this question in the physics community. Commented Dec 12, 2020 at 13:13