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Chapter 33: Problem 3

A converging lens will be used as a magnifying glass. In order for this to work, the object must be placed at a distance a) \(d_{\mathrm{o}}>f\). c) \(d_{\mathrm{o}}

Short Answer

Expert verified
a) \(d_o > f\) b) \(d_o = f\) c) \(d_o < f\) Answer: c) \(d_o < f\)

Step by step solution

01

Conditions for magnification

A lens works as a magnifying glass when it produces a virtual, upright, and magnified image of the object. For a converging lens to produce such an image, the object must be placed within its focal length, meaning that it should be closer to the lens than the focal length. In other words, the object distance \(d_o < f\). Now let's check each of the given options to see which one matches the condition we found.
02

Evaluate option a

Option a suggests that the object should be placed at a distance \(d_o > f\). If the object is placed at a position farther than the focal length, the converging lens will produce a real and inverted image instead of a virtual and upright one. Hence, the lens will not work as a magnifying glass in this case. So, option a is an incorrect choice.
03

Evaluate option b

Option b suggests that the object should be placed at a distance \(d_o = f\), which means the object will be placed exactly at the focal point on the object side. In this case, the converging lens will form an image at infinity, which is not useful for magnification either. Therefore, option b is also an incorrect choice.
04

Evaluate option c

Option c suggests that the object should be placed at a distance \(d_o < f\). As we mentioned in Step 1, placing the object at a distance less than the lens' focal length produces a virtual, upright, and magnified image, which is the desired condition for a magnifying glass. Thus, option c is the correct choice. Hence, the right distance to place the object for a converging lens to work as a magnifying glass is \(d_o < f\), making option (c) the correct answer.

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Most popular questions from this chapter

An object is placed on the left of a converging lens at a distance that is less than the focal length of the lens. The image produced will be a) real and inverted. c) virtual and inverted. b) virtual and erect. d) real and erect.

Determine the position and size of the final image formed by a system of elements consisting of an object \(2.0 \mathrm{~cm}\) high located at \(x=0 \mathrm{~m},\) a converging lens with focal length \(5.0 \cdot 10^{1} \mathrm{~cm}\) located at \(x=3.0 \cdot 10^{1} \mathrm{~cm}\) and a plane mirror located at \(x=7.0 \cdot 10^{1} \mathrm{~cm}\)

You are experimenting with a magnifying glass (consisting of a single converging lens) at a table. You discover that by holding the magnifying glass \(92.0 \mathrm{~mm}\) above your desk, you can form a real image of a light that is directly overhead. If the distance between the light and the table is \(2.35 \mathrm{~m},\) what is the focal length of the lens?

An unknown lens forms an image of an object that is \(24 \mathrm{~cm}\) away from the lens, inverted, and a factor of 4 larger in size than the object. Where is the object located? a) \(6 \mathrm{~cm}\) from the lens on the same side of the lens b) \(6 \mathrm{~cm}\) from the lens on the other side of the lens c) \(96 \mathrm{~cm}\) from the lens on the same side of the lens d) \(96 \mathrm{~cm}\) from the lens on the other side of the lens e) No object could have formed this image.

Two identical thin convex lenses, each of focal length \(f\), are separated by a distance \(d=2.5 f\). An object is placed in front of the first lens at a distance \(d_{\mathrm{a}, 1}=2 f .\) a) Calculate the position of the final image of an object through the system of lenses. b) Calculate the total transverse magnification of the system. c) Draw the ray diagram for this system and show the final image. d) Describe the final image (real or virtual, erect or inverted, larger or smaller) in relation to the initial object.
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