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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 \(1.20-\mathrm{cm}\) -tall object is \(50.0 \mathrm{~cm}\) to the left of a converging lens of focal length \(40.0 \mathrm{~cm}\). A second converging lens, this one having a focal length of \(60.0 \mathrm{~cm},\) is located \(300.0 \mathrm{~cm}\) to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it \(I_{1}\) ) formed by the lens with a focal length of \(40.0 \mathrm{~cm}\). (b) \(I_{1}\) is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

A candle \(4.85 \mathrm{~cm}\) tall is \(39.2 \mathrm{~cm}\) to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?

The left end of a long glass rod \(6.00 \mathrm{~cm}\) in diameter has a convex hemispherical surface \(3.00 \mathrm{~cm}\) in radius. The refractive index of the glass is \(1.60 .\) Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far; (b) \(12.0 \mathrm{~cm}\) (c) \(2.00 \mathrm{~cm}\).

When a camera is focused, the lens is moved away from or toward the digital image sensor. If you take a picture of your friend, who is standing \(3.90 \mathrm{~m}\) from the lens, using a camera with a lens with an \(85 \mathrm{~mm}\) focal length, how far from the sensor is the lens? Will the whole image of your friend, who is \(175 \mathrm{~cm}\) tall, fit on a sensor that is \(24 \mathrm{~mm} \times 36 \mathrm{~mm} ?\)

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{~cm}\) and \(R_{2}=+28.0 \mathrm{~cm} .\) The index of refraction is \(1.60 .\) (a) Compute the position and size of the image of an object in the form of an arrow \(5.00 \mathrm{~mm}\) tall, perpendicular to the lens axis, \(45.0 \mathrm{~cm}\) to the left of the lens. (b) A second converging lens with the same focal length is placed \(3.15 \mathrm{~m}\) to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens \(45.0 \mathrm{~cm}\) to the right of the first.
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