Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 24: Problem 12

Kim says that she was less than 10 ft away from the president when she took a picture of him with her \(50-\mathrm{mm}\) focal length camera lens. The picture shows the upper half of the president's body (or \(3.0 \mathrm{ft}\) of his total height). On the negative of the film, this part of his body is $18 \mathrm{mm}$ high. How close was Kim to the president when she took the picture?

Short Answer

Expert verified
Answer: Kim was approximately 1.006 feet (or approximately 1 foot) away from the president when she took the picture.

Step by step solution

01

Write down the known values

We know the following: 1. Focal length of the camera lens, f = \(50 mm\) 2. Object height (upper half of the president's body), h = \(3 ft\) 3. Image height on the film, h' = \(18 mm\) Note: We need to convert the object height to the same unit as the other measurements (millimeters). There are 304.8 millimeters in a foot, so \(3 ft = 914.4 mm\).
02

Write down the lens equation

The lens equation relating the object distance (d), image distance (d'), and focal length (f) is: $$\frac{1}{d}+\frac{1}{d'}=\frac{1}{f}$$
03

Write down the magnification equation

The magnification (M) is the ratio of the image height (h') to the object height (h). It is also equal to the ratio of the image distance (d') to the object distance (d). $$M=\frac{h'}{h}=\frac{d'}{d}$$
04

Solve for the image distance (d')

Now, we can find the image distance (d') using the magnification equation. Rearranging the equation and plugging in the values, we get: $$d' = d \times \frac{h'}{h} = d \times \frac{18\,\text{mm}}{914.4\,\text{mm}}$$ Substitute this expression for d' in the lens equation: $$\frac{1}{d}+\frac{1}{d \times \frac{18}{914.4}}=\frac{1}{50}$$
05

Solve for the object distance (d)

Now, solve the equation for the object distance (d): $$\frac{1}{d}+\frac{1}{d} \times \frac{914.4}{18}=\frac{1}{50}$$ Multiply both sides of the equation by \(50d\) to get rid of the fractions: $$50 + 50 \times \frac{914.4}{18}=d$$ Now, solve for d: $$d=50 + 50 \times \frac{914.4}{18}\approx 306.8\,\text{mm}$$
06

Convert the object distance to feet

Convert the object distance from millimeters to feet: $$d \approx 306.8\,\text{mm} \times \frac{1\,\text{ft}}{304.8\,\text{mm}}\approx 1.006\,\text{ft}$$
07

Conclusion

Kim was approximately 1.006 feet (or approximately 1 foot) away from the president when she took the picture.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 statue is \(6.6 \mathrm{m}\) from the opening of a pinhole camera, and the screen is \(2.8 \mathrm{m}\) from the pinhole. (a) Is the image erect or inverted? (b) What is the magnification of the image? (c) To get a brighter image, we enlarge the pinhole to let more light through, but then the image looks blurry. Why? (d) To admit more light and still have a sharp image, we replace the pinhole with a lens. Should it be a converging or diverging lens? Why? (e) What should the focal length of the lens be?
A man requires reading glasses with \(+2.0 \mathrm{D}\) power to read a book held \(40.0 \mathrm{cm}\) away with a relaxed eye. Assume the glasses are $2.0 \mathrm{cm}$ from his eyes. (a) What is his uncorrected far point? (b) What refractive power lenses should he use for distance vision? (c) His uncorrected near point is \(1.0 \mathrm{m} .\) What should the refractive powers of the two lenses in his bifocals be to give him clear vision from \(25 \mathrm{cm}\) to infinity?
A converging lens with a focal length of \(3.00 \mathrm{cm}\) is placed $24.00 \mathrm{cm}\( to the right of a concave mirror with a focal length of \)4.00 \mathrm{cm} .\( An object is placed between the mirror and the lens, \)6.00 \mathrm{cm}\( to the right of the mirror and \)18.00 \mathrm{cm}$ to the left of the lens. Name three places where you could find an image of this object. For each image tell whether it is inverted or upright and give the total magnification.
Comprehensive Problems Good lenses used in cameras and other optical devices are actually compound lenses, made of five or more lenses put together to minimize distortions, including chromatic aberration. Suppose a converging lens with a focal length of \(4.00 \mathrm{cm}\) is placed right next to a diverging lens with focal length of \(-20.0 \mathrm{cm} .\) An object is placed \(2.50 \mathrm{m}\) to the left of this combination. (a) Where will the image be located? (b) Is the image real or virtual?
Suppose that the lens system (cornea + lens) in a particular eye has a focal length that can vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm},\) but the distance from the lens system to the retina is only \(1.90 \mathrm{cm} .\) (a) Is this eye nearsighted or farsighted? Explain. (b) What range of distances can the eye see clearly without corrective lenses?
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Electricity and Magnetism

Read Explanation

Electromagnetism

Read Explanation

Circular Motion and Gravitation

Read Explanation

Waves 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.

Sign-up for free