The closely packed cones in the fovea of the eye have a diameter of about 2 μm. For the eye to discern two images on the fovea as distinct, assume that the images must be separated by at least one cone that is not excited, by a distance of about 4 μm. If these images are of two point-like objects at the eye's 28-cm near point, how far apart are these barely resolvable objects? Assume the eye's diameter (cornea-to-fovea distance) is 2.0 cm.

Correct Question:

The closely packed cones in the fovea of the eye have a diameter of about 2 μm. For the eye to discern two images on the fovea as distinct, assume that the images must be separated by at least one cone that is not excited, by a distance of about 4 μm.If these images are of two point-like objects at the eye's 28-cm near point, how far apart are these barely resolvable objects? Assume the eye's diameter (cornea-to-fovea distance) is 2.0 cm.

The objects are 0.000056 m apart

Explanation:

From the diagram attached to this solution,

let the diameter of the eye = D = 2 cm 0.02 m

the distance between the objects = l

Position of the eye lens = E

the distance between objects and the lens of the eye = 28 cm = 0.28 m

The images are separated by a distance of 4 μm = 4 * 10⁻⁶m

Observing the attached diagram properly, ΔABE and ΔECD are similar

Therefore,

l = 0.000056 m

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach

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