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Chapter 33: Problem 99

What is the magnifying power of a +4.0 -D lens used as a magnifier? Assume a relaxed normal eye.

Short Answer

Expert verified
The magnifying power of the lens is 2.

Step by step solution

01

Understanding Magnifying Power

Magnifying power, or magnification, of a lens is given by the formula for a relaxed eye: \( M = 1 + \frac{D}{F} \), where \( D \) is the least distance of distinct vision and \( F \) is the focal length of the lens.
02

Identifying Provided Values and Assumptions

The least distance of distinct vision \( D \) is generally taken as 25 cm for a normal eye. The power of the lens \( P \) is given as +4.0 diopters, which is the reciprocal of the focal length \( F \) when expressed in meters. Hence, \( F = \frac{1}{P} = \frac{1}{4.0} \) meters.
03

Calculating Focal Length

Convert the focal length from meters to centimeters for compatibility with \( D \). Since \( F = 0.25 \) meters, \( F = 25 \) cm.
04

Using the Magnifying Power Formula

Substitute the values for \( D \) and \( F \) into the magnifying power formula: \[ M = 1 + \frac{25}{25} = 1 + 1 = 2. \]
05

Conclusion of Magnifying Power

The magnifying power of the lens is 2, meaning it magnifies objects to twice their size when viewed with a relaxed eye.

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Key Concepts

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

Magnifying Power
When using lenses, the magnifying power is a measure of how much larger a lens can make an object appear. This value is essential for understanding how effective a lens is in magnifying objects. In mathematical terms, for a relaxed eye, it is expressed as:\[ M = 1 + \frac{D}{F} \]where:
  • \( M \) is the magnifying power.
  • \( D \) represents the least distance of distinct vision, commonly 25 cm for normal vision.
  • \( F \) is the focal length of the lens.
A positive magnifying power indicates that the lens is converging and enlarging the object, rather than scattering the light.
Diopters
Diopters measure the optical power of a lens and are significant in dictating how a lens affects vision. A lens with a positive diopter value, like +4.0 D, indicates a converging lens which brings parallel rays of light to a focus. The formula that relates diopters to focal length is:\[ P = \frac{1}{F} \]where:
  • \( P \) is the power in diopters.
  • \( F \) is the focal length in meters.
Therefore, a lens with +4.0 D has a focal length of 0.25 meters (or 25 cm), which is crucial for determining the magnifying power.
Focal Length
The focal length of a lens is the distance over which parallel rays of light are brought to a focus. It is a core property determining a lens's convergence or divergence capability. For our given lens with a +4.0 diopter value:- The focal length is calculated by taking the reciprocal of the diopter value.- Thus, \( F = \frac{1}{P} = \frac{1}{4.0} \), which equals 0.25 meters.This indicates that the lens focuses light 25 cm from itself, making it essential to use this focal length in magnification calculations.
Least Distance of Distinct Vision
The least distance of distinct vision is the minimum distance at which the eye can comfortably focus on an object. For most people with normal vision, this distance is about 25 cm. This standard measurement helps define metrics like magnifying power efficiently. Utilizing a lens changes the effective least distance, improving how well we can see small details at a shorter distance, effectively doubling the size of these details when the magnifying power is 2, as was calculated in our solution. This concept ensures that when we use magnifying lenses, the reduced strain and increased clarity enhance everyday tasks and scientific observations.

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Most popular questions from this chapter

(II) Human vision normally covers an angle of about \(40^{\circ}\) horizontally. A "normal" camera lens then is defined as follows: When focused on a distant horizontal object which subtends an angle of \(40^{\circ}\), the lens produces an image that extends across the full horizontal extent of the camera's light-recording medium (film or electronic sensor). Determine the focal length \(f\) of the "normal" lens for the following types of cameras: \((a)\) a \(35-\mathrm{mm}\) camera that records images on film \(36 \mathrm{~mm}\) wide; \((b)\) a digital camera that records images on a charge-coupled device \((\mathrm{CCD}) 1.00 \mathrm{~cm}\) wide.

(II) A lighted candle is placed 36 \(\mathrm{cm}\) in front of a converging lens of focal length \(f_{1}=13 \mathrm{cm},\) which in turn is 56 \(\mathrm{cm}\) in front of another converging lens of focal length \(f_{2}=16 \mathrm{cm}\) (see Fig. \(47 ) .(a)\) Draw a ray diagram and estimate the location and the relative size of the final image. (b) Calculate the position and relative size of the final image.

(II) The eyepiece of a compound microscope has a total length of 2.80 \(\mathrm{cm}\) and the objective lens has \(f=0.740 \mathrm{cm} .\) I an object is placed 0.790 \(\mathrm{cm}\) from the objective lens, calculate (a) the distance between the lenses when the microscopeiadjusted for a relaxed eve, and \((b)\) the total magnification.

Sam purchases \(+3.50-\mathrm{D}\) eyeglasses which correct his faulty vision to put his near point at 25 \(\mathrm{cm}\) . (Assume he wears the lenses 2.0 \(\mathrm{cm}\) from his eyes.) (a) Calculate the focal length of Sam's glasses. (b) Calculate Sam's near point without glasses. (c) Pam, who has normal eyes with near point at \(25 \mathrm{cm},\) puts on Sam's glasses. Calculate Pam's near point with Sam's glasses on.

(I) Both surfaces of a double convex lens have radii of \(31.4 \mathrm{~cm} .\) If the focal length is \(28.9 \mathrm{~cm},\) what is the index of refraction of the lens material?
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