# 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. – Vadim Dec 12 '20 at 13:13