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Chapter 34: Problem 2

The image of a tree just covers the length of a plane mirror 4.00 cm tall when the mirror is held 35.0 cm from the eye. The tree is 28.0 m from the mirror. What is its height?

Short Answer

Expert verified
The tree's height is approximately 12.96 meters.

Step by step solution

01

Identify Relevant Parameters

We have a mirror that is 4.00 cm tall and placed 35.0 cm from the eye. The object, which is a tree, is 28.0 m away from the mirror. We need to determine the height of the tree.
02

Understanding Similar Triangles

The problem involves similar triangles formed by the line of sight to the mirror and tree. The small triangle consists of the mirror and eye, while the larger triangle is formed by the tree and eye. The ratio of their heights is equal to the ratio of their distances from the eye.
03

Set Up the Proportion

Using the property of similar triangles, we create a proportion between the height of the mirror and the height of the tree. Thus, we have \[ \frac{{\text{height of mirror}}}{{\text{distance from mirror to eye}}} = \frac{{\text{height of tree}}}{{\text{distance from tree to eye}}} \]Substitute in the known values: \[ \frac{4.00 \, \text{cm}}{35.0 \, \text{cm}} = \frac{h}{(28.0 \, \text{m} + 0.35 \, \text{m})} \] (convert cm to m for consistency).
04

Solve for the Tree's Height

Simplify the equation to solve for \( h \), the height of the tree:\[ h = \left( \frac{4.00}{35.0} \right) \times 28.35 \, \text{m} \]Calculate the value of \( h \): \[ h = \left( \frac{1}{8.75} \right) \times 28.35 \, \text{m} \approx 3.24 \, \text{m} \times 28.35 \, \text{m} = 12.96 \, \text{m} \]
05

Conclusion

Given the distances and the measurement of the plane mirror, the height of the tree is approximately 12.96 meters.

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Key Concepts

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

Similar Triangles
When learning about geometric optics and the concept of similar triangles, it's important to remember that these triangles are similar because their corresponding angles are equal, and their sides are in proportion. In this particular problem, two triangles are formed in our mind's eye:
  • The first triangle is small and involves the eye and the plane mirror.
  • The second triangle is large, extending from the eye to the top of the tree.
The similarity in these triangles allows us to use their proportional sides to calculate unknown distances or heights, like the height of the tree in this exercise.
By writing the ratio of the height of the mirror to the distance from the eyes to the mirror and setting it equal to the height of the tree over the complete distance from the eye to the tree (which includes the distance from the eye to the mirror too), we can derive the unknown height accurately.
Mirror Formula
In geometric optics, the mirror formula is a fundamental concept that relates the distances involved when reflections occur. Although not directly applied here, understanding it helps clarify how images are perceived.
  • In general, the mirror formula states: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
  • Here, \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance.
For flat mirrors like the one in the exercise, the image distance is equal to the object distance because the focal length is infinite. Even though we are dealing with flat mirrors here, this exercise uses proportions derived from similar triangles rather than the mirror formula directly. Still, understanding the foundational principles helps bolster our knowledge of how images are formed and measured using geometric optics.
Proportion in Physics
The exercise powerfully illustrates the concept of proportion, which frequently appears in physics to solve problems through relationships between different quantities. Proportions allow us to compare two ratios that are equivalent. In this activity:
  • We compare the ratio of the height of the mirror to the height of the tree.
  • We also compare their respective distances from the observer.
By understanding that the height of the mirror divided by its distance from the observer relates in the same way as the height of the tree to its distance from the observer, we apply a proportion to solve for the tree's height. This reveals the inherent beauty and logic of physics, where simple comparisons can yield solutions to seemingly complex problems. Ensuring unit consistency, like converting centimeters to meters, is essential for maintaining the integrity of these ratios.

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Most popular questions from this chapter

A microscope is focused on the upper surface of a glass plate. A second plate is then placed over the first. To focus on the bottom surface of the second plate, the microscope must be raised 0.780 mm. To focus on the upper surface, it must be raised another 2.10 mm. Find the index of refraction of the second plate.

In one form of cataract surgery the person's natural lens, which has become cloudy, is replaced by an artificial lens. The refracting properties of the replacement lens can be chosen so that the person's eye focuses on distant objects. But there is no accommodation, and glasses or contact lenses are needed for close vision. What is the power, in diopters, of the corrective contact lenses that will enable a person who has had such surgery to focus on the page of a book at a distance of 24 cm?

What is the size of the smallest vertical plane mirror in which a woman of height \(h\) can see her full-length image?

To determine whether a frog can judge distance by means of the amount its lens must move to focus on an object, researchers covered one eye with an opaque material. An insect was placed in front of the frog, and the distance that the frog snapped its tongue out to catch the insect was measured with high-speed video. The experiment was repeated with a contact lens over the eye to determine whether the frog could correctly judge the distance under these conditions. If such an experiment is performed twice, once with a lens of power -9-D and once with a lens of power -15-D, in which case does the frog have to focus at a shorter distance, and why? (a) With the -9-D lens; because the lenses are diverging, the lens with the longer focal length creates an image that is closer to the frog. (b) With the -15-D lens; because the lenses are diverging, the lens with the shorter focal length creates an image that is closer to the frog. (c) With the -9-D lens; because the lenses are converging, the lens with the longer focal length creates a larger real image. (d) With the -15-D lens; because the lenses are converging, the lens with the shorter focal length creates a larger real image.

A screen is placed a distance \(d\) to the right of an object. A converging lens with focal length \(f\) is placed between the object and the screen. In terms of \(f\), what is the smallest value \(d\) can have for an image to be in focus on the screen?
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