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.

Answer 1

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.

Answer:

The objects are 0.000056 m apart

Step-by-step 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)/(0.28) = (4 * 10^(-6) )/(D)`

`l = (0.28 * 4 * 10^(-6) )/(0.02)`

l = 0.000056 m

Answer 2

The distance between the object is
`l=0.0056\ cm`

Step-by-step explanation:

The free body diagram of this setup is on the first uploaded image

From the question

The diameter of closely packed cones in the fovea of the eye is =
`2 \mu m`

The distance of separation by one cone(not excited ) is
`d = 4\mu m = 4*10^(-4)cm`

The distance between the two point-like object is l

The diameter of the eye is D = 2 cm

The distance of the two point-like object from the near point of the eye is A = 28 cm

From the diagram we see that the light from the two point-like object form a triangle of similar base l and d and height D and A

So for a triangle with similar base we have that

`(l)/(A) =(d)/(D)`

`(l)/(28) = (4*10^(-4))/(2)`

making l the subject we have

`l = (28 *4*10^(-4))/(2)`

`l=0.0056\ cm`