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Chapter 34: Problem 2

The image of a tree just covers the length of a plane mirror \(4.00 \mathrm{~cm}\) tall when the mirror is held \(35.0 \mathrm{~cm}\) from the eye. The tree is \(28.0 \mathrm{~m}\) from the mirror. What is its height?

Short Answer

Expert verified
The height of the tree is \( 32.0 \) meters

Step by step solution

01

Identify the Given Variables

From the exercise, we understand that the height of the mirror, \(h_m\), is 4.00 cm or 0.04 m (converted to meters), the distance between the eye and the mirror, \(d_e\), is 35.0 cm or 0.35 m, and the distance from the tree to the mirror, \(d_t\), is 28.0 m.
02

Blending Similar Triangles

By using the fact that the triangles are similar, because they preserve the ratios of corresponding sides, we will use the following relationship: \( \frac{h_t}{h_m} = \frac{d_t + d_e}{d_e} \) Here, \(h_t\) is height of tree which we're looking for.
03

Calculate the Height of Tree

Rearrange the above equation to find the height of the tree: \(h_t = (d_t + d_e) \cdot \frac{h_m}{d_e}\). Then substitute the data into the equation to find \(h_t = (28.0 m + 0.35 m) \cdot \frac{0.04 m}{0.35 m}\).

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Key Concepts

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

Similar Triangles
In optics, similar triangles often help us make sense of light, images, and their proportions. Let's explore how similar triangles work in the context of this exercise. When two triangles are similar, they have the same shape but different sizes.
This means they have the same angles and their sides are proportional.
  • If one triangle has sides that are double the size of another, the angles will still match, which makes them similar triangles.
  • This concept is useful for calculating unknown distances or sizes.
In our problem, similar triangles are formed by the lines of sight from the eye to the top and bottom of the mirror compared to those from the eye to the tree. Since the triangles share a common base (the eye to the mirror) and the same angles, their sides are proportional.
  • This proportional relationship allows us to find the height of the tree using the known measurements of the mirror and distances.
Plane Mirrors
A plane mirror reflects light uniformly, and it has some cool tricks up its reflective sleeve. When you look into a plane mirror, you see an image that might not be where it seems. In our exercise, the mirror is flat and reflects light directly back, creating a virtual image.
  • This image forms as if it's the same distance behind the mirror as the actual object is in front of it.
  • The height of the image will be the same as the actual object鈥攁 property of plane mirrors.
So, if a tree stands in front of a plane mirror, its image appears behind the mirror at a corresponding location and at the same size.
This property of plane mirrors is essential to solving problems involving reflections and image distances.
Image Formation
Image formation in the context of plane mirrors revolves around how and where we perceive images. The virtual image created by a plane mirror isn't actually at a physical location. Instead,
  • it's a visual perception from the reflected light that seems to originate from behind the mirror.
  • The image size keeps the same ratio as the object, making image formation predictable using similar triangles.
In our exercise, the tree's image fully covers the mirror, so the image has been perfectly formed based on the light traveling from the tree to your eyes via the mirror.
By understanding the distance between the viewer, mirror, and tree, we use the concept of image formation to calculate the tree's height through geometric principles.

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Most popular questions from this chapter

A tank whose bottom is a mirror is filled with water to a depth of \(20.0 \mathrm{~cm}\). A small fish floats motionless \(7.0 \mathrm{~cm}\) under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns \((\mu \mathrm{m})\) is typical near the center of the eye. We shall model the eye as a sphere \(2.50 \mathrm{~cm}\) in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells \(5.0 \mu \mathrm{m}\) in diameter. (a) What is the smallest object you can resolve at a near point of \(25 \mathrm{~cm} ?\) (b) What angle is subtended by this object at the eye? Express your answer in units of minutes \(\left(1^{\circ}=60 \mathrm{~min}\right),\) and compare it with the typical experimental value of about \(1.0 \mathrm{~min} .\) (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

A compound microscope has an objective lens with focal length \(14.0 \mathrm{~mm}\) and an eyepiece with focal length \(20.0 \mathrm{~mm}\). The final image is at infinity. The object to be viewed is placed \(2.0 \mathrm{~mm}\) beyond the focal point of the objective lens. (a) What is the distance between the two lenses? (b) Without making the approximation \(s_{1} \approx f_{1},\) use \(M=m_{1} M_{2}\) with \(m_{1}=-s_{1}^{\prime} / s_{1}\) to find the overall angular magnification of the microscope. (c) What is the percentage difference between your result and the result obtained if the approximation \(s_{1} \approx f_{1}\) is used to find \(M ?\)

A converging lens with a focal length of \(12.0 \mathrm{~cm}\) forms a virtual image \(8.00 \mathrm{~mm}\) tall, \(17.0 \mathrm{~cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is \(45.0 \mathrm{~cm}\) from his eyes instead of the usual \(25.0 \mathrm{~cm}\). (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?
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