/*! 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 36 When a camera is focused, the le... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(85-\mathrm{mm}\) 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
Lens-to-film distance: 86.91 mm. Image too tall for 36 mm film.

Step by step solution

01

Understanding the Problem

We need to calculate the distance from the lens to the film when focusing on an object 3.90 m away with a camera lens of 85 mm focal length. Additionally, we must determine if the person's entire image can fit on a film with dimensions 24 mm by 36 mm.
02

Applying the Lens Equation

Use the lens formula to find the distance from the lens to the film. The lens formula 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. Here, \( f = 85 \ \mathrm{mm} \) and \( d_o = 3.90 \ \mathrm{m} = 3900 \ \mathrm{mm} \).
03

Calculating the Image Distance

Rearrange the lens formula to solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \). Substitute \( f = 85 \ \mathrm{mm} \) and \( d_o = 3900 \ \mathrm{mm} \) into the equation: \( \frac{1}{d_i} = \frac{1}{85} - \frac{1}{3900} \). Calculate \( \frac{1}{d_i} \), then find \( d_i \).
04

Completing Image Distance Calculation

Calculate \( \frac{1}{85} = 0.01176 \) and \( \frac{1}{3900} = 0.000256 \). Thus, \( \frac{1}{d_i} = 0.01176 - 0.000256 = 0.011504 \). Invert \( 0.011504 \) to find \( d_i = 86.91 \ \mathrm{mm} \).
05

Checking if the Image Fits on the Film

Determine the magnification using \( M = \frac{d_i}{d_o} \). Compute \( M = \frac{86.91}{3900} \approx 0.02228 \). The image height is \( M \times 1750 \ \mathrm{mm} = 38.99 \ \mathrm{mm} \), which exceeds the film's 36 mm height.
06

Conclusion

The lens needs to be positioned 86.91 mm from the film to focus on the friend. However, the image will be 38.99 mm tall, which is too tall for a 36 mm film.

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Key Concepts

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

Focal Length
The focal length of a lens is a crucial concept in camera optics. It is the distance from the lens to the point where light rays converge to form a clear image. In our example, the focal length is given as 85 mm. This measurement is vital because it dictates how much of the scene will be captured and how large the objects in the image will appear. Shorter focal lengths capture wider scenes, while longer focal lengths provide greater magnification and a narrower field of view.
The focal length also influences the depth of field, affecting how much of the image is in focus. Understanding the relationship between focal length, field of view, and magnification helps in creating desired photographic effects.
Image Distance
The image distance refers to the distance from the lens to where the image is formed, in this case, the film. This is determined using the lens equation: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]
Here, \(f\) is the focal length, \(d_o\) is the object distance (distance from the object to the lens), and \(d_i\) is the image distance. In our scenario, to focus an image of the friend standing 3.90 meters away, the calculated image distance is approximately 86.91 mm.
This formula is pivotal as it helps photographers and filmmakers adjust the lens position effectively to capture sharp images at various distances.
Magnification
Magnification in camera optics represents how much larger or smaller the image appears compared to the object. It is calculated using the formula: \[M = \frac{d_i}{d_o}\]
For our example, with \(d_i = 86.91 \ \mathrm{mm}\) and \(d_o = 3900 \ \mathrm{mm}\), the magnification is approximately 0.02228.
This means the image on the film is about 2.2% the size of the actual object.
Understanding magnification is essential, particularly in determining whether the entire subject will fit within the film's dimensions. In our scenario, the magnified image height was calculated as 38.99 mm, which is larger than the 36 mm the film can accommodate.
Camera Optics
Camera optics encompasses the way lenses in cameras focus light to produce an image. The lens system allows cameras to adjust focus, thereby changing image clarity and composition. By moving the lens closer or further from the film, photographers can focus on subjects at varying distances, as evidenced by the need to set the lens 86.91 mm from the film for the photo.
Camera lenses are designed to manipulate light based on principles of refraction, resulting in clear and sharp images. Different lens types and settings impact various aspects of the captured image, including focus, depth of field, and magnification, which all interconnect to create a desired photographic outcome.

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Most popular questions from this chapter

What should be the index of refraction of a transparent sphere in order for paraxial rays from an infinitely distant object to be brought to a focus at the vertex of the surface opposite the point of incidence?

Three-Dimensional Image. The longitudinal magnification is defined as \(m^{\prime}=d s^{\prime} / d s .\) It relates the longitudinal dimension of a small object to the longitudinal dimension of its image. (a) Show that for a spherical mirror, \(m^{\prime}=-m^{2}\) . What is the significance of the fact that \(m^{\prime}\) is always negative? (b) A wire frame in the form of a small cube 1.00 \(\mathrm{mm}\) on a side is placed with its center on the axis of a concave mirror with radius of curvature 150.0 \(\mathrm{cm}\) . The sides of the cube are all either parallel or perpendicular to the axis. The cube face toward the mirror is 200.0 \(\mathrm{cm}\) to the left of the mirror vertex. Find (i) the location of the image of this face and of the opposite face of the cube; (ii) the lateral and longitudinal magnifications; (iii) the shape and dimensions of each of the six faces of the image.

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