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Chapter 17: Problem 124

An arrow \(2.00 \mathrm{~cm}\) long points toward a lens with the focal length \(f=30.0 \mathrm{~cm}\). The arrow lies along the lens's axis, extending from \(74.0 \mathrm{~cm}\) to \(76.0 \mathrm{~cm}\) from the lens, as indicated in Figure 17.53. What is the length of the arrow's image? (Hint: Use the thin- lens equation to locate the image of each end of the arrow.)

Short Answer

Expert verified
The length of the arrow's image is \(3.0\, \text{cm}\).

Step by step solution

01

Determine the position of the object

The problem states that the arrow extends from \(74.0\, \text{cm}\) to \(76.0\, \text{cm}\) from the lens. The arrow is \(2.00\, \text{cm}\) long, and points towards the lens along its axis. So, the front end of the arrow is at a distance \(d_1 = 74.0\, \text{cm}\), and the back end is at \(d_2 = 76.0\, \text{cm}\) from the lens.
02

Use the thin-lens equation to find the image of the arrow's front

The thin-lens equation is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). First, let's calculate the image position for the front of the arrow with \(d_o = d_1 = 74.0\, \text{cm}\) and \(f = 30.0\, \text{cm}\). Thus: \[ \frac{1}{30} = \frac{1}{74} + \frac{1}{d_{i1}} \]Solving for \(d_{i1}\), we find:\[ \frac{1}{d_{i1}} = \frac{1}{30} - \frac{1}{74} \approx 0.01757 \]\[ d_{i1} \approx \frac{1}{0.01757} \approx 56.9\, \text{cm} \]
03

Find the image of the arrow's back

Now calculate the image position for the back of the arrow with \(d_o = d_2 = 76.0\, \text{cm}\). Using the thin-lens equation:\[ \frac{1}{30} = \frac{1}{76} + \frac{1}{d_{i2}} \]Solving for \(d_{i2}\):\[ \frac{1}{d_{i2}} = \frac{1}{30} - \frac{1}{76} \approx 0.01668 \]\[ d_{i2} \approx \frac{1}{0.01668} \approx 59.9\, \text{cm} \]
04

Calculate the length of the image

The length of the image can be calculated by the difference between the image positions of the front and back of the arrow. Thus, the length is:\[ \text{Image Length} = |d_{i2} - d_{i1}| = |59.9 - 56.9| = 3.0\, \text{cm} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Image Formation
Image formation is a fascinating concept in physics, where lenses are used to create images by bending light rays. When light rays pass through a lens, they are refracted or bent to form images on the other side. In the case of the thin lens equation, it allows us to determine where these images are formed. The position and size of the image depend on characteristics like the focal length of the lens and the object's position.

In our exercise, we have an arrow positioned in front of a lens. By using the thin lens equation, we can calculate the images of the arrow's front and back ends. This allows us to deduce not only where the image is formed but also the length of the image. Such calculations are essential in optics and have practical applications in cameras, microscopes, and eyeglasses.
Focal Length
The focal length is one of the fundamental properties of a lens. It is the distance from the lens to the focal point, where parallel light rays converge after passing through the lens. A positive focal length indicates a converging lens, which focuses light to a point.In the problem, the lens has a focal length of 30.0 cm. This means that parallel rays of light would converge 30.0 cm from the lens. The focal length helps us use the thin lens equation

\( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
to find the image distance \(d_i\) where the image is formed, given the object distance \(d_o\) and the known focal length \(f\). Understanding focal length is critical for designing systems that focus light appropriately, from photographic lenses to scientific instruments.
Physics Problem Solving
Physics problem solving often involves breaking down a problem into manageable steps with a clear application of physical laws. In solving the given exercise, we start by understanding the problem's setup and identifying relevant data, such as the object's distance from the lens and the lens's focal length.We then apply the thin lens equation, step by step, to determine key values such as the positions of the images for the arrow's ends.
  • This involves substituting known values into the equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) and solving for unknowns.
  • Iterating through these calculations methodically ensures accuracy.
This structured approach not only simplifies complex problems but helps in understanding the core concepts better. By following such techniques, you improve your problem-solving skills and gain a deeper grasp of physics principles in varied contexts.

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Most popular questions from this chapter

What focal length must an IOL have if it is to focus an object at a distance of \(45 \mathrm{~cm}\) on the retina of the eye, which is \(2.9 \mathrm{~cm}\) from the IOL? A. \(0.021 \mathrm{~cm}\) C. \(2.7 \mathrm{~cm}\) B. \(0.37 \mathrm{~cm}\) D. \(3.1 \mathrm{~cm}\)

An object is a distance \(2 f\) from a convex lens. (a) Use a ray diagram to find the approximate location of the image. (b) Is the image upright or inverted? (c) Is the image real or virtual? Explain.

Calculate A lens is \(6.2 \mathrm{~cm}\) from the CCD sensor in a camera. If the lens is focused on an object \(3.8 \mathrm{~m}\) away, what is the focal length of the lens?

An intracorneal ring is a small plastic device implanted in a person's cornea to change its curvature. By changing the shape of the cornea, the intracorneal ring can correct a person's vision. (a) If a person is nearsighted, should the ring increase or decrease the cornea's curvature? (b) Choose the best explanation from among the following: A. The intracorneal ring should increase the curvature of the cornea so that it bends light more. This will allow it to focus light coming from far away. B. The intracorneal ring should decrease the curvature of the cornea so that it is flatter and bends light less. This will allow parallel rays from far away to be focused.

One of the many works published by the Greek astronomer Ptolemy (A.D. circa 90-170) was Optics. In this book Ptolemy reported the results of refraction experiments he conducted by observing light passing from air into water. Two of his results are as follows: (1) angle of incidence \(=10.0^{\circ}\), angle of refraction \(=8.00^{\circ} ;(2)\) angle of incidence \(=20.0^{\circ}\), angle of refraction \(=15.5^{\circ}\). Find the percent error for each of Ptolemy's measurements, assuming that the index of refraction of water is \(1.33\).
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