/*! 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 67 The area occupied by one frame o... [FREE SOLUTION] | 91影视

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Chapter 24: Problem 67

The area occupied by one frame on 35 -mm film is \(24 \mathrm{mm}\) by $36 \mathrm{mm}-\( see the figure with Problem \)16 .$ The focal length of the camera lens is \(50.0 \mathrm{mm}\). A picture is taken of a person $182 \mathrm{cm}$ tall. What is the minimum distance from the camera for the person to stand so that the image fits on the film? Give two answers; one for each orientation of the camera.

Short Answer

Expert verified
Answer: The minimum distance for the person to stand from the camera so that their image fits on the film is 1949.5mm in portrait orientation and 2924.2mm in landscape orientation.

Step by step solution

01

List the known values

In this problem, we know the following: - Frame dimensions: 24mm x 36mm - Focal length (f): 50mm - Height of person (h): 182cm = 1820mm (converted to mm)
02

Find the minimum object distance in portrait orientation

In this case, the height of the frame is 36mm and we need to fit the person in that height. We can use the magnification formula: \(M = \frac{h'}{h} = \frac{q}{p}\) We know that height of person (h) is 1820mm and the maximum height in the frame (h') is 36mm. Then, \(M = \frac{36}{1820} = \frac{q}{p}\) Now, we will use the lens formula: \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\) Substitute the magnification formula into the lens formula and solve for p: \(\frac{1}{50} = \frac{1}{p} + \frac{p}{36p}\) First we will calculate the value of q: \(q = Mp = \frac{36}{1820}p\) Now substitute this value in the lens formula: \(\frac{1}{50} = \frac{1}{p} + \frac{1}{\frac{36}{1820}p}\) Solving for p, we get: \(p_{portrait} = 1949.5mm\)
03

Find the minimum object distance in landscape orientation

In this case, the height of the frame is 24mm, and we need to fit the person in that height. We can use the magnification formula again: \(M = \frac{h'}{h} = \frac{q}{p}\) We know that height of person (h) is 1820mm and the maximum height in the frame (h') is 24mm. Then, \(M = \frac{24}{1820} = \frac{q}{p}\) Using the same method as in Step 2, calculate the value of q: \(q = Mp = \frac{24}{1820}p\) Now substitute this value in the lens formula: \(\frac{1}{50} = \frac{1}{p} + \frac{1}{\frac{24}{1820}p}\) Solving for p, we get: \(p_{landscape} = 2924.2mm\)
04

Report the minimum object distances in both orientations

The minimum distance for the person to stand from the camera so that their image fits on the film in portrait orientation is 1949.5mm, and in landscape orientation is 2924.2mm.

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Most popular questions from this chapter

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?
Repeat Problem \(40(\mathrm{c})\) using a different eyepiece that gives an angular magnification of 5.00 for a final image at the viewer's near point \((25.0 \mathrm{cm})\) instead of at infinity.
A person on a safari wants to take a photograph of a hippopotamus from a distance of \(75.0 \mathrm{m} .\) The animal is \(4.00 \mathrm{m}\) long and its image is to be \(1.20 \mathrm{cm}\) long on the film. (a) What focal length lens should be used? (b) What would be the size of the image if a lens of \(50.0-\mathrm{mm}\) focal length were used? (c) How close to the hippo would the person have to be to capture a \(1.20-\mathrm{cm}-\) long image using a 50.0 -mm lens?
You have a set of converging lenses with focal lengths $1.00 \mathrm{cm}, 10.0 \mathrm{cm}, 50.0 \mathrm{cm},\( and \)80.0 \mathrm{cm} .$ (a) Which two lenses would you select to make a telescope with the largest magnifying power? What is the angular magnification of the telescope when viewing a distant object? (b) Which lens is used as objective and which as eyepiece? (c) What should be the distance between the objective and eyepiece?
An object is located \(10.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens is a second converging lens, \(30.0 \mathrm{cm}\) from the first lens, of focal length \(10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify by using the lens equations.
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