Answer:
The distance between the object is [tex]l=0.0056\ cm[/tex]
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 = [tex]2 \mu m[/tex]
The distance of separation by one cone(not excited ) is [tex]d = 4\mu m = 4*10^{-4}cm[/tex]
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
[tex]\frac{l}{A} =\frac{d}{D}[/tex]
[tex]\frac{l}{28} = \frac{4*10^{-4}}{2}[/tex]
making l the subject we have
[tex]l = \frac{28 *4*10^{-4}}{2}[/tex]
[tex]l=0.0056\ cm[/tex]
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
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, [tex]\frac{l}{0.28} = \frac{4 * 10^{-6} }{D}[/tex]
[tex]l = \frac{0.28 * 4 * 10^{-6} }{0.02}[/tex]
l = 0.000056 m
