/*! 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 38 (II) Human vision normally cover... [FREE SOLUTION] | 91影视

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Chapter 33: Problem 38

(II) Human vision normally covers an angle of about \(40^{\circ}\) horizontally. A "normal" camera lens then is defined as follows: When focused on a distant horizontal object which subtends an angle of \(40^{\circ}\), the lens produces an image that extends across the full horizontal extent of the camera's light-recording medium (film or electronic sensor). Determine the focal length \(f\) of the "normal" lens for the following types of cameras: \((a)\) a \(35-\mathrm{mm}\) camera that records images on film \(36 \mathrm{~mm}\) wide; \((b)\) a digital camera that records images on a charge-coupled device \((\mathrm{CCD}) 1.00 \mathrm{~cm}\) wide.

Short Answer

Expert verified
(a) approx. 49.45 mm, (b) approx. 13.74 mm.

Step by step solution

01

Convert Angle to Radians

The angle of the field of view given is \( 40^\circ \). To perform calculations using trigonometric functions, we first convert this angle to radians using the formula: \[\text{Radians} = \frac{\pi}{180} \times \text{Degrees}\]So, the angle in radians is \[40^\circ = \frac{\pi}{180} \times 40 = \frac{2\pi}{9} \text{ radians}\]
02

Understand the Lens Equation

The lens focuses an image such that an object subtending an angle \(\theta\) produces an image of width equal to the sensor/film width. This forms a right triangle with the relationship:\[\tan\left(\frac{\theta}{2}\right) = \frac{\text{half width of sensor}}{f}\]This equation will help us find the focal length \( f \).
03

Calculate Focal Length for 35-mm Camera

The width of the film is \( 36 \text{ mm} \). Thus, half the width is \( 18 \text{ mm} \). Use the formula:\[\tan\left(\frac{20^\circ}{2}\right) = \frac{18}{f}\]Calculate \( f \):\[\tan\left(20^\circ\right) = 0.364\]\[f = \frac{18}{0.364} \approx 49.45 \text{ mm}\]
04

Calculate Focal Length for Digital Camera

The width of the CCD is \( 1.00 \text{ cm} \), which is \( 10 \text{ mm} \). Hence, half the width is \( 5 \text{ mm} \). The calculation is similar:\[\tan\left(20^\circ\right) = \frac{5}{f}\]Calculate \( f \):\[f = \frac{5}{0.364} \approx 13.74 \text{ mm}\]
05

Conclusion

Therefore, the focal length \( f \) for the 35-mm camera is approximately \( 49.45 \text{ mm} \) and for the digital camera is approximately \( 13.74 \text{ mm} \).

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Key Concepts

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

Camera Focal Length
The focal length of a camera lens is a crucial concept in optics. It determines how a camera captures an image and how much of a scene is visible. A lens's focal length is the distance between the lens and the image sensor or film when the subject is in focus.
For a 35-mm camera, the `normal lens` focal length is the one that can capture the horizontal field of view the human eye can naturally see, about 40 degrees. This means that the camera lens must produce an image that fills up the entire width of the film, in this example, 36 mm.
The focal length (\(f\)) is calculated through trigonometric relationships by considering half the sensor's or film's width. The equation used is:
  • \( \tan\left(\frac{\theta}{2}\right) = \frac{\text{half width of sensor}}{f} \)
Using this equation helps determine how far from the lens the sensor or film should be placed to create a clear image.
Trigonometric Functions
Trigonometric functions like tangent, sine, and cosine play vital roles in various calculations in geometry and physics. Specifically, in optics and this exercise, the tangent function helps understand the relationship between an object's visible angle and the dimensions captured by a lens.
To find the focal length, we use the tangent function because it relates the opposite side to the adjacent side in a right triangle. In this case, the `opposite` side is half the width of the camera鈥檚 sensor, and the `adjacent` side is the focal length.
Here's how you use it:
  • Divide the total angle by two to find half of the field of view, converting it if needed, to radians.
  • Apply the formula \( \tan\left(\frac{\theta}{2}\right) = \frac{\text{half width of sensor}}{f} \)
  • Isolate \(f\) to calculate the focal length \(f = \frac{\text{half width of sensor}}{\tan\left(\frac{\theta}{2}\right)} \)
The tangent function is crucial here because it allows us to solve for physical dimensions using angular measurements.
Angle Conversion to Radians
Understanding angles is foundational for solving many geometrical problems, and conversions between degrees and radians are necessary for functions in calculus and trigonometry.
Degrees are most commonly used in everyday contexts, but radians provide a more natural mathematical measurement, especially for scientific calculations. One radian equals the angle created by taking the radius of a circle and wrapping it along the circle's edge, which is approximately 57.3 degrees.
To convert degrees to radians, use the conversion formula:
  • \( \text{Radians} = \frac{\pi}{180} \times \text{Degrees} \)
For example, a 40-degree field of view in degrees translates to radians using this equation:
  • \(40^{\circ} \rightarrow \frac{\pi}{180} \times 40 = \frac{2\pi}{9} \text{ radians} \)
Radian conversion is essential when working with trigonometric functions since many formulas in calculus and physics expect angles in radians.

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Most popular questions from this chapter

(III) In the "magnification" method, the focal length \(f\) of a converging lens is found by placing an object of known size at various locations in front of the lens and measuring the resulting real-image distances \(d_{i}\) and their associated magnifications \(m\) (minus sign indicates that image is inverted). The data taken in such an experiment are given here:\(\begin{array}{rrrrrr}{d_{\mathrm{i}}(\mathrm{cm})} & {20} & {25} & {30} & {35} & {40} \\ {m} & {-0.43} & {-0.79} & {-1.14} & {-1.50} & {-1.89}\end{array}\) (a) Show analytically that a graph of \(m\) vs. \(d_{\text { i should }}\) produce a straight line. What are the theoretically expected values for the slope and the \(y\) -intercept of this line? [Hint: \(d_{\mathrm{o}}\) is not constant.] \((b)\) Using the data above, graph \(m\) vs. \(d_{\mathrm{i}}\) and show that a straight line does indeed result. Use the slope of this line to determine the focal length of the lens. Does the \(y\) -intercept of your plot have the expected value? (c) In performing such an experiment, one has the practical problem of locating the exact center of the lens since \(d_{\mathrm{i}}\) must be measured from this point. Imagine, instead, that one measures the image distance \(d\) from the back surface of the lens, which is a distance \(\ell\) from the lens's center. Then, \(d_{i}=d_{1}^{\prime}+\ell .\) Show that, when implementing the magnification method in this fashion, a plot of \(m\) vs.di will still result in a straight line. How can \(f\) be determined from this straight line?

(II) What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is \(6.4 \mathrm{~m}\) and an eyepiece whose focal length is \(2.8 \mathrm{~cm} ?\)

(II) A stamp collector uses a converging lens with focal length \(28 \mathrm{~cm}\) to view a stamp \(18 \mathrm{~cm}\) in front of the lens. (a) Where is the image located? (b) What is the magnification?

(II) An eye is corrected by a \(-4.50-\mathrm{D}\) lens, \(2.0 \mathrm{~cm}\) from the eye. \((a)\) Is this eye near- or farsighted? \((b)\) What is this eye's far point without glasses?

(II) A 105 -mm-focal-length lens is used to focus an image on the sensor of a camera. The maximum distance allowed between the lens and the sensor plane is \(132 \mathrm{~mm} .\) (a) How far ahead of the sensor should the lens be if the object to be photographed is \(10.0 \mathrm{~m}\) away? \((b) 3.0 \mathrm{~m}\) away? \((c) 1.0 \mathrm{~m}\) away? (d) What is the closest object this lens could photograph sharply?
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