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Chapter 33: Problem 76

A classmate claims that by using a \(40.0-\mathrm{cm}\) focal length mirror, he can project onto a screen a \(10.0-\mathrm{cm}\) tall bird locat ed 100 . \(\mathrm{m}\) away. He claims that the image will be no less than \(1.00 \mathrm{~cm}\) tall and inverted. Will he make good on his claim?

Short Answer

Expert verified
Answer: No, the student's claim is partially correct. Though the projected image is indeed inverted, it is smaller than 1 cm tall (approximately 0.421 cm).

Step by step solution

01

Identify the given information

We are given: 1. Focal length of the mirror (f) = 40 cm 2. Object distance from the mirror (d_o) = 100 m (convert it to cm) 3. Object height (h_o) = 10 cm We need to find the height of the image (h_i) and verify if it is inverted.
02

Convert the distance to a matching unit

Convert the object distance from meters to centimeters: d_o = 100 m * 100 cm/m = 10,000 cm
03

Use the mirror equation to find the image distance (d_i)

The mirror equation is: 1/f = 1/d_o + 1/d_i Rearrange to solve for d_i: 1/d_i = 1/f - 1/d_o Substitute the values: 1/d_i = 1/(40 cm) - 1/(10,000 cm) Now, calculate the value of d_i: d_i = 1/((1/40 - 1/10000)) ≈ 42.11 cm
04

Use the magnification equation to find the image height (h_i)

The magnification equation is: M = h_i / h_o = -d_i / d_o We want to find h_i, so rearrange the equation: h_i = h_o * (-d_i / d_o) Substitute the values: h_i = (10 cm) * (-42.11 cm / 10,000 cm) Calculate the value of h_i: h_i ≈ -0.421 cm
05

Determine if the claim is correct

From our calculations, we find that the height of the image (h_i) is approximately -0.421 cm, which means the image is indeed inverted (negative sign). However, the image is smaller than the claimed 1 cm (0.421 cm < 1 cm). Therefore, the claim is partially correct - the image is inverted, but it is smaller than 1 cm.

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Most popular questions from this chapter

A microscope has a \(2.0-\mathrm{cm}\) focal length eyepiece and a \(0.80-\mathrm{cm}\) objective lens. For a relaxed normal eye, calculate the position of the object if the distance between the lenses is \(16.2 \mathrm{~cm}\)

The objective lens in a laboratory microscope has a focal length of \(3.00 \mathrm{~cm}\) and provides an overall magnification of \(1.0 \cdot 10^{2} .\) What is the focal length of the eyepiece if the distance between the two lenses is \(30.0 \mathrm{~cm}\) ?

An object is placed on the left of a converging lens at a distance that is less than the focal length of the lens. The image produced will be a) real and inverted. c) virtual and inverted. b) virtual and erect. d) real and erect.

Is it possible to design a system that will form an image without lenses or mirrors? If so, how? and what drawbacks, if any, would it have?

Will the magnification produced by a simple magnifier increase, decrease, or stay the same when the object and the lens are both moved from air into water?
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