/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 A student has built a 15 -cm-lon... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 5

A student has built a 15 -cm-long pinhole camera for a science fair project. She wants to photograph her 180 -cm-tall friend and have the image on the film be \(5.0 \mathrm{cm}\) high. How far should the front of the camera be from her friend?

Short Answer

Expert verified
The friend should be 540 cm away from the front of the camera.

Step by step solution

01

Understand the Problem

The problem represents two similar triangles. The first triangle's heights are the friend's height and the distance from the friend to the pinhole. The second triangle's heights are the height of the image and the distance from the image to the pinhole. Due to similarities of triangles, the ratios of corresponding lengths of both triangles are equal.
02

Setting Up an Equation

Now, let's denote the distance from the friend to the pinhole as \(d\). Then the equation from the similarity of triangles will be: \(\frac{180}{d} = \frac{5.0}{15}\) . This equation represents the ratio of the heights of the friend and the image that equals the ratio of the distances from the friend and image to the pinhole. Here, the height of the friend and image are in centimeters, and \(d\) and the distance from the image to the pinhole (15 cm) also must be in the same unit for a correct comparison.
03

Solve the Equation

Next step is to solve the equation from Step 2 to find \(d\). Cross multiply the equation to solve for \(d\): \(180 * 15 = 5.0 * d\). Dividing both sides by 5, we get: \(d = \frac{180 * 15}{5.0}\).
04

Calculate the Value of d

By calculating the numeric value of \(d\), we get: \(d = 540.0 \mathrm{cm}\). This value is the distance from the friend to the camera point (pinhole).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Similar Triangles
In geometry, similar triangles are those that have the same shape but possibly different sizes. However, the angles in similar triangles are always the same, and their corresponding sides are in proportion. In the case of the pinhole camera project, you can imagine two triangles. The first triangle is formed by the friend's height and the distance from her to the camera. The second triangle is formed by the height of the image on the film and the distance from the film to the pinhole. These two triangles are similar because the angles are equal.

This similarity means:
  • The ratio of the friend's height to the height of her image is the same as the ratio of the distance from the friend to the pinhole, and the distance from the film to the pinhole.
  • We use this property to set up equations that help us solve for unknown distances or lengths, like how far the front of the camera should be from the friend.
Ratios in Geometry
Ratios are comparisons of two quantities that are often expressed as fractions. In geometry, they are used to describe relationships between sides of triangles, especially in similar figures. When we say that all corresponding lengths in similar triangles have the same ratio, it means for the two similar triangles with sides of length: length of side 1/length of side 2. The ratio remains constant.

For our exercise, the ratio of the friend's height to the image height is:
  • \[\frac{180 \text{ cm (friend's height)}}{5.0 \text{ cm (image height)}}\]
  • This is set equal to the ratio of the distances from the friend to the pinhole and from the image to the pinhole: \[\frac{d}{15 \text{ cm}}\]

This equality allows you to solve for the unknown distance. Cross multiplying gives you an equation that can be simplified to find the solution.
Exploring Optics in Physics
Optics is a branch of physics that deals with light and its interactions with different materials. In the context of the pinhole camera, the concept of light traveling in straight lines is crucial. Through the tiny hole (or pinhole), light rays pass from the object (like the friend) to form an image on the opposite side of the box (camera) on a film or sensor.

Pinhole cameras exemplify the basic principles of how cameras work: they capture light and project it to form images.
  • The distance between the pinhole and the screen (film) inside the camera box determines the size and clarity of the image.
  • The pinhole essentially acts as a lens, where the light coming from the top and bottom of an object meets at a point to create an inverted image.

With this setup, students can visually understand how angles and distances affect the image produced, linking the principles of geometry and physics together.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sports photographer has a \(150-\) mm-focal-length lens on his camera. The photographer wants to photograph a sprinter running straight away from him at \(5.0 \mathrm{m} / \mathrm{s} .\) What is the speed (in \(\mathrm{mm} / \mathrm{s}\) ) of the sprinter's image at the instant the sprinter is \(10 \mathrm{m}\) in front of the lens?

a. Estimate the diameter of your eycball. b. Bring this page up to the closest distance at which the text is sharp - not the closest at which you can still read it, but the closest at which the letters remain sharp. If you wear glasses or contact lenses, leave them on. This distance is called the nearpoint of your (possibly corrected) eye. Measure it. c. Bstimate the effective focal length of your eye. The effective focal length includes the focusing due to the lens, the curvature of the cornea, and any corrections you wear. Ignore the effects of the fluid in your eye.

You're helping with an experiment in which a vertical cylinder will rotate about its axis by a very small angle. You need to devise a way to measure this angle. You decide to use what is called an optical lever. You begin by mounting a small mirror on top of the cylinder. A laser \(5.0 \mathrm{m}\) away shoots a laser beam at the mirror. Before the experiment starts, the mirror is adjusted to reflect the laser beam directly back to the laser. Later, you measure that the reflected laser beam, when it returns to the laser, has becn deflected sideways by 2.0 mm. Through how many degrees has the cylinder rotated?

There's one angle of incidence \(\beta\) onto a prism for which the light inside an isosceles prism travels parallel to the base and emerges at angle \(\beta\) a. Find an expression for \(\beta\) in terms of the prism's apex angle \(\alpha\) and index of refraction \(n\) b. A laboratory measurement finds that \(\beta=52.2^{\circ}\) for a prism shaped like an equilateral triangle. What is the prism's index of refraction?

A slide projector needs to create a 98 -cm-high image of a \(2.0-\mathrm{cm}\) -tall slide. The screen is \(300 \mathrm{cm}\) from the slide. a. What focal length does the lens need? Assume that it is a thin lens. b. How far should you place the lens from the slide?
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Oscillations

Read Explanation

Medical Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Translational Dynamics

Read Explanation

Astrophysics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.