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Chapter 19: Problem 17

A farsighted person has a near point of \(50 \mathrm{cm}\). What strength lens, in diopters, is needed to bring his near point to \(25 \mathrm{cm} ?\)

Short Answer

Expert verified
The additional strength required for the lens to bring the near point from 50 cm to 25 cm is 2 diopters.

Step by step solution

01

Determine the initial and final near points

Initially, the near point is given as 50 cm or \(0.5 \mathrm{m}\). We need to reduce this to 25 cm or \(0.25 \mathrm{m}.\) Here remember that the near point corresponds to the focal length of the lens required for vision correction.
02

Calculate the existing and required lens strength

Using the formula for lens strength (in diopters), \(Power = 1/focal length\), calculate the existing lens strength as \(Power_{initial} = 1/0.5 = 2 \mathrm{D}\). For the required near point of 0.25 m, the needed lens strength is \(Power_{required} = 1/0.25 = 4 \mathrm{D}\)
03

Compute the strength of the required additional lens

To compute the strength of the additional lens required to bring the near point from 50 cm to 25 cm - subtract the initial lens power from the required power. So, \(Power_{additional} = Power_{required} - Power_{initial} = 4 \mathrm{D} - 2 \mathrm{D} = 2 \mathrm{D}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diopters
Diopters are a unit of measurement used in optics to indicate the refractive power of lenses. They are an essential concept in understanding how lenses work, especially when it comes to vision correction. Let's break this down further:
  • The diopter (D) expresses the strength of a lens. It is calculated as the inverse of the focal length measured in meters. The formula is: \[ \text{Diopter} = \frac{1}{\text{focal length (m)}} \]
  • Positive diopters (+) are used for converging lenses, which bring light rays together. This is the type commonly used in glasses or contacts for farsighted individuals.
  • Negative diopters (-) apply to diverging lenses, which spread light rays apart and are used to correct nearsightedness.
So, when changing the near point from 50cm to 25cm, calculating the diopter power involved is key to finding the correct strength for vision correction. A value in diopters helps optometrists recommend lenses that compensate for vision issues effectively.
Focal Length
Focal length is a critical aspect of lens evaluation in optics. It defines the distance between the lens and the point where it converges or diverges light rays. This is known as the focal point. Understanding focal length means knowing how lenses can sharpen or correct our vision.
  • Focal length is expressed in meters and is inversely related to diopters. A shorter focal length means a higher diopter value, and hence, stronger power of the lens.
  • The formula to find the focal length if you know the diopter is simple: \[ \text{Focal length (m)} = \frac{1}{\text{Diopter}} \]
  • In the context of vision correction, adjusting the focal length directly relates to shifting the near or far points of a person's vision, thereby improving clarity at different distances.
For our exercise, adjusting the focal length from 50cm to 25cm effectively shifts the near point closer, aligning with the typical reading distance for better visual comfort.
Vision Correction
Vision correction involves adjusting the focal point where light converges on the retina. Glasses or contact lenses are used to change this focal point so that images are clearer.
  • For farsightedness (hyperopia), objects close up are blurry because the image focuses behind the retina. Converging lenses (with positive diopters) help to push the focal point forward.
  • Nearsightedness (myopia), on the other hand, involves distant objects being blurred as focus falls in front of the retina. Diverging lenses (negative diopters) spread the light out more to move the focus back.
  • Each person's degree of correction is personalized, based on tests that indicate how lenses need to change the path of light to achieve perfect focus on the retina.
In our scenario, by shifting the near point from 50 cm to 25 cm using lenses of appropriate strength, the near point matches a more natural range for a farsighted person, offering improved comfort and reduced eye strain when performing tasks like reading.

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Most popular questions from this chapter

Your telescope has an objective lens with a focal length of \(1.0 \mathrm{m} .\) You point the telescope at the moon, then realize that the eyepiece is missing. You can still see the real image of the moon formed by the objective lens if you place your eye a little past the image so as to view the rays diverging from the image plane, just as rays would diverge from an object at that location. What is the angular magnification of the moon if you view its real image from \(25 \mathrm{cm}\) away, your near-point distance?

Your task in physics lab is to make a microscope from two lenses. One lens has a focal length of \(10 \mathrm{cm},\) the other a focal length of \(3.0 \mathrm{cm} .\) You plan to use the more powerful lens as the objective, and you want its image to be \(16 \mathrm{cm}\) from the lens, as in a standard biological microscope. a. How far should the objective lens be from the object to produce a real image \(16 \mathrm{cm}\) from the objective? b. What will be the magnification of your microscope?

A student has built a 20 -cm-long pinhole camera for a science fair project. She wants to photograph the Washington Monument, which is \(167 \mathrm{m}(550 \mathrm{ft})\) tall, and to have the image on the detector be \(5.0 \mathrm{cm}\) high. How far should she stand from the Washington Monument?

The Hubble Space Telescope has a mirror diameter of \(2.4 \mathrm{m}\). Suppose the telescope was used to photograph the surface of the moon from a distance of \(3.8 \times 10^{8} \mathrm{m} .\) What is the distance between two objects that the telescope can barely resolve? Assume the wavelength of light is \(600 \mathrm{nm}\).

At a distance of 6 meters, a person with average vision is able to clearly read letters \(1.0 \mathrm{cm}\) high. Approximately how large do the letters appear on the retina? (Assume that the retina is \(1.7 \mathrm{cm}\) from the lens.)
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