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Chapter 33: Problem 11

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.

Short Answer

Expert verified
Short answer: The object is located 6 cm from the lens on the other side of the lens.

Step by step solution

01

Write down the lens equation and the magnification equation.

The lens equation is given by: \begin{align} \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \end{align} where \(f\) is the focal length of the lens, \(d_o\) is the object distance, and \(d_i\) is the image distance. The magnification equation is given by: \begin{align} M = -\frac{d_i}{d_o} \end{align} where \(M\) is the magnification. In this problem, we have \(d_i = 24\mathrm{~cm}\) and \(M = 4\).
02

Solve for the object distance using the magnification equation.

Since the image is inverted and 4 times larger, we have \(M = -4\). Using the magnification equation: \begin{align} -4 = -\frac{d_i}{d_o} \end{align} Cross-multiplying the equation, we get: \begin{align} 4d_o = d_i \end{align} We know \(d_i = 24\mathrm{~cm}\), so we can substitute to find \(d_o\): \begin{align} 4d_o = 24 \Rightarrow d_o = 6\mathrm{~cm} \end{align}
03

Determine the correct answer regarding the object's location.

We found that the object is \(6\mathrm{~cm}\) away from the lens. Since the image is inverted and magnified, it must be a real image formed on the other side of the lens by a converging lens. Therefore, the correct answer is: b) \(6 \mathrm{~cm}\) from the lens on the other side of the lens.

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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.

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.

Three converging lenses of focal length \(5.0 \mathrm{~cm}\) are arranged with a spacing of \(2.0 \cdot 10^{1} \mathrm{~cm}\) between them, and are used to image an insect \(2.0 \cdot 10^{1} \mathrm{~cm}\) away. a) Where is the image? b) Is it real or virtual? c) Is it upright or inverted?

The focal length of the lens of a camera is \(38.0 \mathrm{~mm}\). How far must the lens be moved to change focus from a person \(30.0 \mathrm{~m}\) away to one that is \(5.00 \mathrm{~m}\) away?

What is the focal length of a magnifying glass if a \(1.0-\mathrm{mm}\) object appears to be \(10 . \mathrm{mm} ?\)
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