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Chapter 23: Problem 70

An object that is \(6.00 \mathrm{cm}\) tall is placed \(40.0 \mathrm{cm}\) in front of a diverging lens. The magnitude of the focal length of the lens is \(20.0 \mathrm{cm} .\) Find the image position and size. Is the image real or virtual? Upright or inverted?

Short Answer

Expert verified
Question: Determine the position, type (real or virtual), orientation (upright or inverted), and size of the image formed by a diverging lens with a focal length of 20 cm when an object with a height of 6 cm is placed 40 cm from the lens. Answer: The image is formed at a distance of 13.3 cm on the same side of the lens as the object. The image is virtual, upright, and has a size of 2 cm.

Step by step solution

01

Identify given information

We are given the object's height (\(h_o = 6.00 \mathrm{cm}\)), the object's distance from the lens (\(d_o = 40.0 \mathrm{cm}\)), and the magnitude of the lens's focal length (\(|f| = 20.0 \mathrm{cm}\)). As it's a diverging lens, the focal length is negative, so \(f = -20.0 \mathrm{cm}\).
02

Apply the lens formula

The lens formula relates the object's distance (\(d_o\)), the image's distance (\(d_i\)), and the focal length (\(f\)): \( \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} \) Plug in the given values: \( \frac{1}{40.0} + \frac{1}{d_i} = \frac{1}{-20.0} \) Now, find the image's distance (\(d_i\)).
03

Calculate the image's distance

Solve for \(d_i\): \( \frac{1}{d_i} = \frac{1}{-20.0} - \frac{1}{40.0}\) \( \frac{1}{d_i} = -\frac{3}{40}\) \(d_i = -\frac{40}{3} \approx -13.3 \mathrm{cm}\) The negative sign indicates that the image is formed on the same side as the object, which means it's a virtual image.
04

Apply the magnification equation

Now, we will use the magnification equation to find the image's height (\(h_i\)): \(M = -\frac{d_i}{d_o} = \frac{h_i}{h_o}\) \(h_i = M \cdot h_o\) We will first calculate the magnification (\(M\)) and then find the image's height.
05

Calculate the magnification and image's height

Calculate the magnification (\(M\)) and the image's height (\(h_i\)): \(M = -\frac{-13.3}{40.0} = \frac{13.3}{40.0} = \frac{1}{3}\) \(h_i = \frac{1}{3} \times 6.00 = 2.00 \mathrm{cm}\) Since the magnification factor is positive, the image is upright.
06

Describe the image

The image is found at a distance of \(13.3 \mathrm{cm}\) on the same side of the lens as the object. The image is virtual and upright, with a size of \(2.00 \mathrm{cm}\).

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Most popular questions from this chapter

A converging lens has a focal length of \(8.00 \mathrm{cm} .\) (a) What are the image distances for objects placed at these distances from the thin lens: \(5.00 \mathrm{cm}, 14.0 \mathrm{cm}\) $16.0 \mathrm{cm}, 20.0 \mathrm{cm} ?$ In each case, describe the image as real or virtual, upright or inverted, and enlarged or diminished in size. (b) If the object is $4.00 \mathrm{cm}$ high, what is the height of the image for the object distances of \(5.00 \mathrm{cm}\) and \(20.0 \mathrm{cm} ?\)
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