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Chapter 2: Problem 146

A far-sighted person has a near point of \(100 \mathrm{cm}\). How far in front or behind the retina does the image of an object placed \(25 \mathrm{cm}\) from the eye form? Use the cornea to retina distance of \(2.5 \mathrm{cm}\)

Short Answer

Expert verified
The image forms \(17.5 \mathrm{cm}\) behind the retina.

Step by step solution

01

Use the Lens Formula

First, we will use the lens formula which is formulated as \[1/f = 1/v - 1/u\] where \(f\) is the focal length of the lens, \(v\) is the image distance (distance between the lens and the image) and \(u\) is the object distance (distance between the lens and the object). However, we know the near point (which is the image distance of the farthest point the eye can see clearly) is at \(100 \mathrm{cm}\). Therefore, we can imply that the focal length \(f\) of the eye is at \(100 \mathrm{cm}\). This is because when the object is at the near point, the eye lens focuses the image exactly onto the retina.
02

Deduce the Object Distance

We are given that the object distance \(u\) is \(25 \mathrm{cm}\). It's important to note that in lens formula convention, the object distance is usually considered negative. So \(u = -25 \mathrm{cm}\). Now we substitute the given parameters into the lens formula and solve for \(v\), the image distance.
03

Solve for the Image Distance

Substitute \(f = 100 \mathrm{cm}\) and \(u = -25 \mathrm{cm}\) into the lens formula and solve for \(v\), i.e., \[1/v = 1/100 - 1/(-25) = 0.01 + 0.04 = 0.05\]. Hence, \(v = 1/0.05 = 20 \mathrm{cm}\] from the lens.
04

Determine the Image position relative to the Retina

Since the cornea to retina distance is \(2.5 \mathrm{cm}\) and the image formed is \(20 \mathrm{cm}\) from the cornea-lens, the image forms \(20 - 2.5 = 17.5 \mathrm{cm}\) behind the retina.

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Most popular questions from this chapter

A nearsighted man cannot see objects clearly beyond \(20 \mathrm{cm}\) from his eyes. How close must he stand to a mirror in order to see what he is doing when he shaves?

Will the focal length of a lens change when it is submerged in water? Explain.

A point source of light is \(50 \mathrm{cm}\) in front of a converging lens of focal length \(30 \mathrm{cm} .\) A concave mirror with a focal length of \(20 \mathrm{cm}\) is placed \(25 \mathrm{cm}\) behind the lens. Where does the final image form, and what are its orientation and magnification?

If the lens of a person's eye is removed because of cataracts (as has been done since ancient times), why would you expect an eyeglass lens of about 16 D to be prescribed?

. \(\mathrm{A} 3 \mathrm{cm}\) tall object is placed \(5 \mathrm{cm}\) in front of a convex mirror of radius of curvature \(20 \mathrm{cm} .\) Where is the image formed? How tall is the image? What is the orientation of the image?
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