/*! 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 44 When a camera is focused, the le... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 44

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

Short Answer

Expert verified
The lens is 86.89 mm from the film. The whole image won't fit on the film.

Step by step solution

01

Understand the Problem

The object (friend) is 3.90 m = 3900 mm away from the lens. The lens has a focal length of 85 mm. We need to find the image distance (distance from lens to film). Additionally, we need to check if the image of a 1750 mm tall friend will fit on film measuring 24 mm by 36 mm.
02

Apply the Lens Formula

Use the lens formula, which is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Plug in the known values: \( \frac{1}{85} = \frac{1}{3900} + \frac{1}{d_i} \).
03

Solve for the Image Distance

Rearrange the formula to solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{85} - \frac{1}{3900} \). Calculate \( \frac{1}{85} \approx 0.011765 \) and \( \frac{1}{3900} \approx 0.000256 \). Subtract these results to get \( \frac{1}{d_i} \approx 0.011509 \). Then, solve for \( d_i \) by taking the reciprocal: \( d_i \approx \frac{1}{0.011509} \approx 86.89 \) mm.
04

Check the Image Size

Use magnification formula \( m = \frac{h_i}{h_o} = \frac{d_i}{d_o} \). Let \( h_o = 1750 \) mm, we already have \( d_o = 3900 \) mm, and found \( d_i = 86.89 \) mm. Find \( h_i \): \( h_i = h_o \cdot \frac{d_i}{d_o} = 1750 \times \frac{86.89}{3900} \approx 39.0 \) mm.
05

Conclusion about Fitting the Image

The film size is 24 mm by 36 mm. Our calculated image height \( h_i \) is approximately 39.0 mm. Since 39.0 mm is greater than the height of the film (24 mm), the entire image won't fit on the film.

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.

Lens Formula
In the world of optics, lenses play a pivotal role in focusing images. The lens formula connects the object distance, the image distance, and the focal length. It is given by the equation: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \).This formula helps in determining the distance at which a sharp image is formed on the other side of the lens.
  • \(f\) represents the focal length.
  • \(d_o\) is the object distance (distance from the object to the lens).
  • \(d_i\) is the image distance (distance from the lens to the image on the film).
The equation implies that if you know two of these parameters, you can find the third. It is an essential part of optics, especially in photographic terms where adjusting these parameters can bring an image into focus.
Focal Length
Focal length is crucial in determining how lenses focus light. It is the distance between the lens and the point where incoming parallel rays converge. Think of it as the lens's power to bend light. In cameras, lenses with short focal lengths give a wider field of view, while those with long focal lengths zoom in.
  • In our problem, the lens has a focal length of 85 mm. This means it is relatively 'zoomed in' compared to lenses with shorter focal lengths.
  • Focal length affects not only focus but also the scale and perspective of the image formed.
Understanding focal length allows photographers and optical engineers to predict how lenses will project images onto surfaces like film or digital sensors.
Image Distance
Image distance is the parameter that describes where the image will be formed after the light passes through the lens. It can be calculated using the lens formula once the object distance and focal length are known.
For instance, in our scenario:
  • Using an 85 mm lens and a subject 3900 mm away, the image is projected 86.89 mm from the lens.
  • This distance (image distance) is crucial, as it tells you where to place the film to capture the sharpest image.
Understanding image distance helps in setting up cameras correctly to ensure that the resulting pictures are focused and sharp.
Magnification
Magnification describes how much larger or smaller the image of the object appears when formed on the film or sensor. It involves the size relationship between the object and its image.
The magnification formula is given by: \( m = \frac{h_i}{h_o} = \frac{d_i}{d_o} \).
  • \( h_i \) is the height of the image (how tall the image appears on the film).
  • \( h_o \) is the height of the object (how tall the object really is).
In the scenario, a friend's image of 175 cm is magnified to about 39 mm on the film. If magnification results in an image larger than the film format, the whole scene won't fit.
Properly managing magnification is critical for photographers ensuring their subjects fit well within the frame.
Film Format
Film format denotes the size of the photographic film, which ultimately influences how much of an image can be captured. It defines the limit for both image height and width.
In our example, the film format is 24 mm by 36 mm. This means:
  • An image of 39 mm height won't completely fit into the film's vertical constraints.
  • The aspect ratio and size govern framing decisions and how elements within an image are positioned.
Selecting the appropriate film format is essential when planning compositions, ensuring that crucial elements aren't cropped. It's especially important when working with limited film size in traditional photography and sensor size in digital formats.

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 certain microscope is provided with objectives that have focal lengths of \(16 \mathrm{mm}, 4 \mathrm{mm},\) and 1.9 \(\mathrm{mm}\) and with eye-pieces that have angular magnifications of \(5 \times\) and \(10 \times .\) Each objective forms an image 120 \(\mathrm{mm}\) beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable.

A converging lens with a focal length of 7.00 \(\mathrm{cm}\) forms an image of a \(4.00-\mathrm{mm}\)-tall real object that is to the left of the lens. The image is 1.30 \(\mathrm{cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

You wish to project the image of a slide on a screen 9.00 \(\mathrm{m}\) from the lens of a slide projector. (a) If the slide is placed 15.0 \(\mathrm{cm}\) from the lens, what focal length lens is required? (b) If the dimensions of the picture on a \(35-\mathrm{mm}\) color slide are 24 \(\mathrm{mm} \times 36 \mathrm{mm}\) , what is the minimum size of the projector screen required to accommodate the image?

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?

Combination of Lenses II. Two thin lenses with a focal length of magnitude \(12.0 \mathrm{cm},\) the first diverging and the second converging, are located 9.00 \(\mathrm{cm}\) apart. An object 2.50 \(\mathrm{mm}\) tall is placed 20.0 \(\mathrm{cm}\) to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it erect or inverted? (Hint: See the preceding two problems.)
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Physics of Motion

Read Explanation

Quantum Physics

Read Explanation

Turning Points in Physics

Read Explanation

Medical Physics

Read Explanation

Fundamentals of Physics

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.