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Chapter 36: Problem 46

Photography. A wildlife photographer uses a moderate telephoto lens of focal length 135 \(\mathrm{mm}\) and maximum aperture \(f | 4.00\) to photograph a bear that is 11.5 \(\mathrm{m}\) away. Assume the wavelength is 550 \(\mathrm{nm}\) . (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
At f/4, the smallest feature is ~0.0268 mm; at f/22, it's ~0.147 mm.

Step by step solution

01

Diffraction Limit Formula

In optics, the smallest resolvable feature is determined by the diffraction limit, which can be calculated using the formula: \( \text{Width of smallest feature} = 1.22 \times \frac{\lambda \cdot f}{d} \), where \(\lambda\) is the wavelength of light, \(f\) is the focal length of the lens, and \(d\) is the diameter of the aperture.
02

Calculate the Aperture Diameter for f/4

An f-number (f/N) relates to the lens's focal length and the diameter of the aperture, given by \( N = \frac{f}{d} \). For \( f/4.0 \), the diameter \( d \) is \( \frac{135 \text{ mm}}{4} = 33.75 \text{ mm} \).
03

Solve for Smallest Feature at f/4

Using the calculated diameter \( d = 33.75 \text{ mm} \) and the given wavelength \( \lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m} \), substitute these into the diffraction limit formula: \[\text{Width} = 1.22 \times \frac{550 \times 10^{-9} \text{ m} \cdot 135 \times 10^{-3} \text{ m}}{33.75 \times 10^{-3} \text{ m}}\]This evaluates to approximately \( \text{Width} \approx 2.68 \times 10^{-5} \text{ m} \).
04

Calculate the Aperture Diameter for f/22

For \( f/22.0 \), the diameter \( d \) is \( \frac{135 \text{ mm}}{22} = 6.136 \text{ mm} \).
05

Solve for Smallest Feature at f/22

Using the new diameter \( d = 6.136 \text{ mm} \), substitute into the formula: \[\text{Width} = 1.22 \times \frac{550 \times 10^{-9} \text{ m} \cdot 135 \times 10^{-3} \text{ m}}{6.136 \times 10^{-3} \text{ m}}\]This evaluates to approximately \( \text{Width} \approx 1.47 \times 10^{-4} \text{ m} \).

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Key Concepts

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

Telephoto Lens
A telephoto lens is a type of camera lens that is designed to provide a longer focal length. This allows photographers to zoom in on distant subjects and make them appear larger in the frame. Typically used in wildlife photography, telephoto lenses enable you to capture close-up images of subjects even if you are physically far away.

These lenses have a unique optical design that reduces the physical length of the lens, which is why they are crucial for outdoor and action photography.
  • They have a narrower field of view compared to standard lenses.
  • Are often used to blur the background and isolate the subject.
  • Can vary in focal length, with some starting at 70mm and others going beyond 600mm.
Understanding how telephoto lenses work can help you achieve vibrant and clear images, particularly when photographing intricate details from afar.
Aperture
In photography, the aperture is the opening in the lens through which light enters the camera. It plays a vital role in controlling the amount of light that reaches the camera sensor and is measured using f-numbers, such as f/4 or f/22.

The aperture size affects both the exposure and the depth of field in an image. A smaller f-number (like f/4) indicates a larger aperture that lets in more light, whereas a larger f-number (like f/22) represents a smaller aperture allowing less light.
  • Larger apertures (small f-number) are great for low-light conditions and creating a shallow depth of field to blur the background.
  • Smaller apertures (large f-number) are used to achieve a larger depth of field where more of the scene is in focus.
Using the right aperture is crucial in achieving the desired artistic effect and technical quality in your photographs.
Wavelength
The wavelength in the context of photography usually refers to the wavelength of light, typically measured in nanometers (nm). Different wavelengths correspond to different colors of light, which can impact the appearance and resolution of an image.

In the diffraction limit formula, wavelength plays a key role in determining the smallest resolvable feature that a lens can capture. For example, light at 550 nm generally corresponds to green light, which is in the middle of the visible spectrum.
  • Shorter wavelengths (blue/violet light) can resolve finer details than longer wavelengths (red light).
  • The average wavelength used in many photography calculations is around 550 nm, as it represents the peak sensitivity of human vision.
Understanding how wavelength affects resolution can help photographers manage factors like diffraction and sharpness in their images.
Focal Length
Focal length is a critical characteristic of a camera lens, indicating the distance from the lens to the image sensor when the subject is in focus. It is usually measured in millimeters, as seen in the 135 mm telephoto lens used in the exercise.

Different focal lengths impact how much of a scene is captured and are labeled based on their view perspective: wide-angle, standard, or telephoto.
  • Shorter focal lengths (like 24-35 mm) provide a wider field of view, suitable for landscapes.
  • Longer focal lengths (like the 135 mm telephoto lens) offer a narrow field of view, ideal for distant subjects.
The focal length thus determines the magnification level and the angle of view of a photograph. For photographers, choosing the right focal length is crucial for capturing images with the desired composition and perspective.

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Most popular questions from this chapter

A slit 0.240 \(\mathrm{mm}\) wide is illuminated by parallel light rays of wavelength 540 \(\mathrm{nm}\) . The diffraction pattern is observed on a screen that is 3.00 \(\mathrm{m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(6.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2} .\) (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

Light of wavelength 585 \(\mathrm{nm}\) falls on a slit 0.0666 \(\mathrm{mm}\) wide. (a) On a very large distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calclating all the anglest (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest that \(m\) can be 7 (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

Tsunamit On December \(26,2004,\) a violent magnitude 9.1 earthquake 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 \(\mathrm{h}\) and the speed of the wave to be 800 \(\mathrm{km} / \mathrm{h}\) . 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 \(\mathrm{km}\) , while the distance between the southern end of Australia and Antarctica is about 3700 \(\mathrm{km}\) . 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.

Monochromatic light of wavelength 580 \(\mathrm{nm}\) passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at \(\pm 90.0^{\circ},\) so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta=45.0^{\circ}\) to the intensity at \(\theta=0 ?\)

X rays of wavelength 0.0850 \(\mathrm{nm}\) are scattered from the atoms of a crystal. The second-order maximum in the Bragg reflection occurs when the angle \(\theta\) in Fig. 36.23 is \(21.5^{\circ} .\) What is the spacing between adjacent atomic planes in the crystal?
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