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Chapter 23: Problem 57

(a) For a converging lens with a focal length of \(3.50 \mathrm{cm}\) find the object distance that will result in an inverted image with an image distance of \(5.00 \mathrm{cm} .\) Use a ray diagram to verify your calculations. (b) Is the image real or virtual? (c) What is the magnification?

Short Answer

Expert verified
Based on the solution provided, what is the object distance, image type, and magnification for a converging lens with a focal length of 3.50 cm and an image distance of 5.00 cm. Object distance: 8.75 cm Image type: Real and inverted Magnification: -0.571

Step by step solution

01

Use the lens formula to find the object distance

To find the object distance, we can use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) Where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. We are given the values of \(f = 3.50 \mathrm{cm}\) and \(d_i = 5.00 \mathrm{cm}\). Plugging the values into the formula, we get: \( \frac{1}{3.50} = \frac{1}{d_o} + \frac{1}{5.00} \) Now, let's solve for \(d_o\).
02

Solve for the object distance (do)

Let's isolate \(d_o\) in our equation: \( \frac{1}{d_o} = \frac{1}{3.50} - \frac{1}{5.00} \) Taking the inverse of the equation on both sides to get the value of \(d_o\): \( d_o = \frac{1}{\frac{1}{3.50} - \frac{1}{5.00}} \) Using a calculator to solve this expression, we find: \( d_o = 8.75 \mathrm{cm} \) So, the object distance that results in an inverted image with an image distance of 5.00 cm for a converging lens of focal length 3.50 cm is 8.75 cm.
03

Verify the result with a ray diagram

To verify our result, we would draw a ray diagram of the converging lens, the object, and the image. The object would be placed at 8.75 cm from the lens on the principal axis, and the lens would have a focal length of 3.50 cm. After drawing the parallel and focus rays and the rays through the center of the lens, we should see that the image is indeed inverted and 5.00 cm away on the other side of the lens.
04

Determine if the image is real or virtual

A real image is formed when light rays converge, while a virtual image is formed when light rays appear to diverge from a point. In this case, since the image is formed on the other side of the lens where light rays converge, we can conclude that the image is real.
05

Calculate magnification

The magnification of the lens is the ratio of the image height to the object height, which is given by the ratio of the image distance to the object distance: \( M = \frac{h_i}{h_o} = \frac{d_i}{d_o} \) We have already calculated the object distance (\(d_o = 8.75 \mathrm{cm}\)) and the image distance (\(d_i = 5.00 \mathrm{cm}\)). Plugging those values into the formula, we get: \( M = \frac{5.00}{8.75} \) Using a calculator to solve this expression, we get: \( M = 0.571 \) Since the image is inverted, the magnification is -0.571. This means the image is reduced in size and inverted.

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Most popular questions from this chapter

Xi Yang is practicing for his driver's license test. He notices in the rearview mirror that a tree, located directly behind the automobile, is approaching his car as he is backing up. If the car is moving at $8.0 \mathrm{km} / \mathrm{h}$ in reverse, how fast relative to the car does the image of the tree appear to be approaching?
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