/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A jeweler is wearing a \(20 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A jeweler is wearing a \(20 \mathrm{D}\) magnifying lens directly in front of his eye. If his near point is a typical \(25 \mathrm{cm},\) how close can he hold a gem that he is inspecting?

Short Answer

Expert verified
The jeweler can hold the gem as close as \(5cm\) from his eye.

Step by step solution

01

Understanding the given parameters

First, let's understand what the given data means. Here, \(20D\) refers to the power of lens in Diopters and \(25cm\) is the near point, which is the nearest distance at which an object can be viewed distinctly without any strain on eyes.
02

Calculating the focal length of the lens

The power \(P\) of a lens is defined as the reciprocal of the focal length \(f\) (in meters). That means, \(P = 1/f\). Here, \(P = 20D\). We can convert this to meters and find the focal length, \(f = 1/P = 1/20 = 0.05m\) or \(5cm\).
03

Finding out the minimum distance

Now, when the magnifying glass is used, it's typically held close to the eye. In such a scenario, the closest an object can be held from the eye, while still appearing distinct, is at the focal point of the magnifying glass. Therefore, the minimum distance the gem can be held from the jeweler's eye is equal to the focal length of the lens, which is \(5cm\).

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Key Concepts

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

Magnifying Lens
A magnifying lens is a simple optical device designed to make objects appear larger than they are. This happens because the lens bends the light waves, enlarging the image that reaches the eye. Magnifying lenses are typically used for tasks requiring detailed examination, such as inspecting gemstones, reading small text, or observing intricate details in a small object.

These lenses have a convex shape, meaning they curve outward on both sides. This design allows them to converge light rays at a focal point, creating a magnified image of the object. By bringing the object to the correct distance relative to the lens, the viewer can observe a larger, clearer image. A jeweler, for example, uses a magnifying lens close to the eye to inspect details in a gem, as it helps in visualizing tiny facets and inclusions.
Diopter
Diopter is the unit of measurement used to express the optical power of a lens. It measures how much the lens can bend light, which directly affects how much it will magnify objects.

For instance:
  • A lens with a power of 1 diopter will focus parallel rays of light at one meter.
  • A 20 diopter lens, as in the provided example, focuses light much closer, at 0.05 meters (or 5 cm), indicating higher magnification power.
Understanding diopters is crucial for selecting the right lens for specific tasks, ensuring that the object is visible clearly and comfortably to the observer.
Focal Length
The focal length of a lens is the distance from the lens to the point where it converges light rays to form a sharp image. This is a critical concept in optics, as it determines how magnified the viewed object will be.

More specifically:
  • Short focal lengths offer higher magnification and are perfect for tasks such as inspecting jewelry.
  • Long focal lengths provide less magnification, suitable for viewing larger, further-away objects.
Calculating the focal length is simple. It's the reciprocal of the lens's power in diopters when expressed in meters. So, for a 20D lens, the focal length is 1/20 = 0.05 meters, or 5 cm. This tells us how close objects should be held to appear clearly magnified.
Near Point
The near point of the eye is the closest distance at which an object can be clearly focused on, without causing eye strain or discomfort. For a young adult with normal vision, this is typically around 25 cm.

In optics and lens applications:
  • Tools like magnifying glasses effectively reduce the near point, allowing smaller, nearby objects to be viewed clearly.
  • Anyone using a magnifying lens can position the object at distances shorter than the natural near point, given the magnification provided by the lens.
In the context of the exercise, the jeweler can hold the gem significantly closer than the typical 25 cm near point, thanks to the lens's magnifying effect, down to the focal length of the lens, which is 5 cm in this case.
Power of Lens
The power of a lens is an essential factor that influences how effective a magnifying lens will be in practical use. It is calculated as the reciprocal of the focal length in meters, and the unit of measurement is diopters.

This guides:
  • The strength of the lens: Higher power lenses have shorter focal lengths and provide greater magnification.
  • Applications: A jeweler's lens with 20 diopters, for example, is powerful enough to allow inspecting fine details in gemstones.
Understanding the power of a lens helps in determining how specific lenses can be best utilized, especially in fields requiring precision, such as jewelry making or watch repair.

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Most popular questions from this chapter

A narrow beam of light with wavelengths from 450 nm to \(700 \mathrm{nm}\) is incident perpendicular to one face of a prism made of crown glass, for which the index of refraction ranges from \(n=1.533\) to \(n=1.517\) for those wavelengths. The light strikes the opposite side of the prism at an angle of \(40^{\circ} .\) What is the angular spread of the beam as it leaves the prism?

Two lightbulbs are \(1.0 \mathrm{m}\) apart. From what distance can these lightbulbs be marginally resolved by a small telescope with a 4.0-cm-diameter objective lens? Assume that the lens is limited only by diffraction and \(\lambda=600 \mathrm{nm}\).

The objective lens of a microscope has a focal length of \(5.0 \mathrm{mm} .\) What eyepiece focal length will give the microscope an overall angular magnification of \(350 ?\) Assume a length \(L=160 \mathrm{mm}\).

Two converging lenses with focal lengths of \(40 \mathrm{cm}\) and \(20 \mathrm{cm}\) are \(10 \mathrm{cm}\) apart. A \(2.0-\mathrm{cm}\) -tall object is \(15 \mathrm{cm}\) in front of the \(40-\mathrm{cm}-\) focal-length lens. a. Use ray tracing to find the position and height of the image. To do this accurately use a ruler or paper with a grid. Determine the image distance and image height by making measurements on your diagram. b. Calculate the image height and position relative to the second lens. Compare with your ray-tracing answers in part a.

Rachel has good distant vision but has a touch of presbyopia. Her near point is \(0.60 \mathrm{m}\). When she wears \(+2.0 \mathrm{D}\) reading glasses, what is her near point? Her far point?

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