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A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture \(f/\)4.00 to photograph a bear that is 11.5 m away. Assume the wavelength is 550 nm. (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f/\)22.0, what would be the width of the smallest resolvable feature on the bear?

Short Answer

Expert verified
(a) 0.23 mm; (b) 1.27 mm.

Step by step solution

01

Understanding the diffraction limit

To find the smallest feature size that can be resolved, we use the Rayleigh criterion for diffraction-limited resolution, which states that the minimum resolvable feature size, \( w \), is given by \( w = \frac{1.22 \times \lambda \times D}{d} \), where \( \lambda \) is the wavelength, \( D \) is the distance between the lens and the object, and \( d \) is the diameter of the aperture.
02

Calculating the aperture diameter for f/4.0

The focal length \( f = 135 \text{ mm} \), and the aperture number \( N = 4.0 \), so the diameter of the aperture \( d \) can be calculated using \( d = \frac{f}{N} = \frac{135}{4} \). Hence, \( d = 33.75 \text{ mm} \).
03

Calculating the smallest resolvable feature using f/4.0

Plugging in the values, we have \( \lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m} \), \( D = 11.5 \text{ m} \), and \( d = 33.75 \times 10^{-3} \text{ m} \). So, \( w = \frac{1.22 \times 550 \times 10^{-9} \times 11.5}{33.75 \times 10^{-3}} \approx 0.23 \text{ mm} \).
04

Calculating the aperture diameter for f/22.0

For \( N = 22.0 \), the aperture diameter \( d = \frac{135}{22} \). This gives \( d = 6.136 \text{ mm} \).
05

Calculating the smallest resolvable feature using f/22.0

Using the same formula as before with \( d = 6.136 \times 10^{-3} \text{ m} \), we get \( w = \frac{1.22 \times 550 \times 10^{-9} \times 11.5}{6.136 \times 10^{-3}} \approx 1.27 \text{ mm} \).
06

Conclusion

The smallest feature on the bear that can be resolved when the lens is opened to its maximum aperture \( f/4.0 \) is approximately \( 0.23 \text{ mm} \), and when stopped down to \( f/22.0 \), it is approximately \( 1.27 \text{ mm} \).

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Key Concepts

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

Diffraction-limited resolution
The ability of optical systems, like cameras and telescopes, to distinguish two closely spaced objects is influenced by the phenomenon of diffraction. This limit is known as the diffraction-limited resolution, a concept arising from the wave nature of light. When light passes through an aperture, it bends around the edges, creating an interference pattern that limits the detail that can be perceived.
Rayleigh's criterion offers a practical way to estimate this limit. According to the Rayleigh criterion, two objects are considered resolvable when the central maximum of the diffraction pattern of one image falls on the first minimum of the other. For circular apertures, this can be simplified mathematically to:
  • Minimum resolvable angle: \[ \theta = 1.22 \frac{\lambda}{d} \]
  • Minimum resolvable feature size: \[ w = \frac{1.22 \times \lambda \times D}{d} \]
In this context, \( \lambda \) is the wavelength of light, \( D \) is the distance to the object, and \( d \) is the aperture diameter. Understanding this concept helps photographers and scientists design systems that maximize clarity.
Aperture diameter
The aperture of a lens plays a critical role in determining the resolution of an image. It is the size of the lens opening that allows light to enter the camera or telescope. Measured in millimeters, the aperture diameter, \( d \), is vital for the Rayleigh criterion and is calculated as:
\[ d = \frac{f}{N} \]
  • \( f \): focal length of the lens
  • \( N \): f-number (or f-stop) that indicates the size of the aperture relative to the focal length
For example, in the exercise, at \( f/4.0 \), the aperture diameter is calculated as 33.75 mm, while at \( f/22.0 \), it shrinks to approximately 6.136 mm. A larger aperture diameter allows more light through, increasing the potential for better resolution but reducing the depth of field. Conversely, stopping down increases depth but decreases resolution.
Wavelength
Wavelength, denoted as \( \lambda \), is a fundamental characteristic of light that defines the distance between successive peaks of a wave. It is typically measured in nanometers (nm) for visible light, with 550 nm being a common reference for calculations involving typical daylight conditions.
The wavelength of light significantly impacts the diffraction-limited resolution. Longer wavelengths result in larger diffraction patterns, which can limit resolution, while shorter wavelengths enhance it. In practical terms, this is why ultraviolet or blue light, with shorter wavelengths, can achieve higher resolutions compared with red light.
  • In the scenario, the wavelength is used in the formula for the minimum resolvable feature size: \[ w = \frac{1.22 \times 550 \times 10^{-9} \times 11.5}{d} \]
This formula highlights how the wavelength interacts with the aperture diameter and distance to determine resolution limits.
Focal length
Focal length, represented as \( f \), is a crucial lens specification that affects both the field of view and the size of objects captured within an image. It is the distance from the lens to the sensor (or film) where light converges to a point.
  • A shorter focal length results in a wider field of view, suitable for landscape photography.
  • A longer focal length, like the 135 mm used in this exercise, narrows the view but magnifies distant subjects, ideal for wildlife photography.
When considering diffraction, the focal length impacts the total light path and, consequently, the potential resolution as it interacts with the aperture setting. The combination of focal length and aperture diameter defines the f-number, influencing depth of field and clarity:
\[ N = \frac{f}{d} \]Understanding focal length aids in achieving the desired balance between subject magnification and sharpness, making it a key element in photographic and observational setups.

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Most popular questions from this chapter

On December 26, 2004, a violent earthquake of magnitude 9.1 occurred off the coast of Sumatra. This quake triggered a huge tsunami (similar to a tidal wave) that killed more than 150,000 people. Scientists observing the wave on the open ocean measured the time between crests to be 1.0 h and the speed of the wave to be 800 km/h. Computer models of the evolution of this enormous wave showed that it bent around the continents and spread to all the oceans of the earth. When the wave reached the gaps between continents, it diffracted between them as through a slit. (a) What was the wavelength of this tsunami? (b) The distance between the southern tip of Africa and northern Antarctica is about 4500 km, while the distance between the southern end of Australia and Antarctica is about 3700 km. As an approximation, we can model this wave's behavior by using Fraunhofer diffraction. Find the smallest angle away from the central maximum for which the waves would cancel after going through each of these continental gaps.

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