/*! 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 11 An unknown lens forms an image o... [FREE SOLUTION] | 91影视

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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

When a diver with \(20 / 20\) vision removes her mask underwater, her vision becomes blurry. Why is this the case? Does the diver become nearsighted (eye lens focuses in front of retina) or farsighted (eye lens focuses behind retina)? As the index of refraction of the medium approaches that of the lens, where does the object get imaged? Typically, the index of refraction for water is 1.33 , while the index of refraction for the lens in a human eye is \(1.40 .\)

The typical length of a human eyeball is \(2.50 \mathrm{~cm} .\) a) What is the effective focal length of the two-lens system made from a normal person's cornea and lens when viewing objects far away? b) What is the effective focal length for viewing objects at the near point?

A beam of parallel light, \(1.00 \mathrm{~mm}\) in diameter passes through a lens with a focal length of \(10.0 \mathrm{~cm}\). Another lens, this one of focal length \(20.0 \mathrm{~cm},\) is located behind the first lens so that the light traveling out from it is again parallel. a) What is the distance between the two lenses? b) How wide is the outgoing beam? 33.41 How large does a \(5.0-\mathrm{mm}\) insect appear when viewed with a system of two identical lenses of focal length \(5.0 \mathrm{~cm}\) separated by a distance \(12 \mathrm{~cm}\) if the insect is \(10.0 \mathrm{~cm}\) from the first lens? Is the image real or virtual? Inverted or upright?

Will the magnification produced by a simple magnifier increase, decrease, or stay the same when the object and the lens are both moved from air into water?

Two refracting telescopes are used to look at craters on the Moon. The objective focal length of both telescopes is \(95.0 \mathrm{~cm}\) and the eyepiece focal length of both telescopes is \(3.80 \mathrm{~cm} .\) The telescopes are identical except for the diameter of the lenses. Telescope A has an objective diameter of \(10.0 \mathrm{~cm}\) while the lenses of telescope \(\mathrm{B}\) are scaled up by a factor of two, so that its objective diameter is \(20.0 \mathrm{~cm}\). a) What are the angular magnifications of telescopes \(A\) and \(B\) ? b) Do the images produced by the telescopes have the same brightness? If not, which is brighter and by how much?
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