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Chapter 33: Problem 8

Where is the image formed if an object is placed \(25 \mathrm{~cm}\) from the eye of a nearsighted person. What kind of a corrective lens should the person wear? a) Behind the retina. Converging lenses. b) Behind the retina. Diverging lenses. c) In front of the retina. Converging lenses. d) In front of the retina. Diverging lenses.

Short Answer

Expert verified
Short Answer: The image is formed in front of the retina and the person should wear diverging lenses to correct their vision.

Step by step solution

01

Understand the characteristics of a nearsighted eye

A nearsighted person has difficulty seeing objects that are far away but can see nearby objects clearly. This happens because the light coming from distant objects converges in front of the retina, causing a blurry image. The retina is the light-sensitive layer at the back of the eye that processes the light entering the eye.
02

Analyze the given information

The object is placed 25 cm away from the eye, and we need to determine where the image forms in a nearsighted person's eye.
03

Determine the location of the image formed

Since a nearsighted person has trouble seeing distant objects, the light from the object placed 25 cm away from the eye will converge in front of the retina. Therefore, the image is formed in front of the retina.
04

Determine the type of corrective lens needed

To correct nearsightedness, the corrective lens should cause the light to converge on the retina instead of in front of it. Diverging lenses can accomplish this by spreading the light out slightly, allowing it to converge at the correct position on the retina. Therefore, the nearsighted person should wear diverging lenses.
05

Choose the correct option

Based on our analysis, the image is formed in front of the retina, and the person should wear diverging lenses to correct their vision. Thus the correct option is: d) In front of the retina. Diverging lenses.

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Most popular questions from this chapter

An object is moved from a distance of \(30 \mathrm{~cm}\) to a distance of \(10 \mathrm{~cm}\) in front of a converging lens of focal length \(20 \mathrm{~cm}\). What happens to the image? a) Image goes from real and upright to real and inverted. b) Image goes from virtual and upright to real and inverted. c) Image goes from virtual and inverted to real and upright. d) Image goes from real and inverted to virtual and upright. e) None of the above.

What kind of lens is used in eyeglasses to correct the vision of someone who is a) nearsighted? b) farsighted?

You have found in the lab an old microscope, which has lost its eyepiece. It still has its objective lens, and markings indicate that its focal length is \(7.00 \mathrm{~mm}\). You can put in a new eyepiece, which goes in \(20.0 \mathrm{~cm}\) from the objective. You need a magnification of about 200. Assume you want the comfortable viewing distance for the final image to be \(25.0 \mathrm{~cm}\). You find in a drawer eyepieces marked \(2.00-, 4.00-,\) and \(8.00-\mathrm{cm}\) focal length. Which is your best choice?

A telescope is advertised as providing a magnification of magnitude 41 using an eyepiece of focal length \(4.0 \cdot 10^{1} \mathrm{~mm}\). What is the focal length of the objective?

When a diver with \(20 / 20\) vision removes her mask underwater, her vision becomes blurry. Why is this the case? Does the diver become nearsighted (eye lens focuses in front of retina) or farsighted (eye lens focuses behind retina)? As the index of refraction of the medium approaches that of the lens, where does the object get imaged? Typically, the index of refraction for water is 1.33 , while the index of refraction for the lens in a human eye is \(1.40 .\)
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