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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
For maximum aperture, the smallest feature is about 0.23 mm. For the stopped down aperture, it's about 1.28 mm.

Step by step solution

01

Understand Resolving Power Formula

To determine the smallest resolvable feature, we will use the formula for resolving power based on diffraction limits: \( \text{Resolution} = 1.22 \frac{\lambda L}{D}\), where \(\lambda\) is the wavelength of light, \(L\) is the distance to the object, and \(D\) is the diameter of the aperture.
02

Calculate Diameter for Maximum Aperture

For the maximum aperture \(f/4.0\), the diameter \(D\) can be calculated from the focal length \(f\). With \(f/\text{stop} = f/D\), we rearrange to find \(D = 135 \, \mathrm{mm}/4\), so \(D = 33.75 \, \mathrm{mm}\).
03

Apply Formula for Maximum Aperture

Substitute the known values into the formula: \(\lambda = 550 \, \mathrm{nm} = 550 \times 10^{-9} \, \mathrm{m}\), \(L = 11.5 \, \mathrm{m}\), and \(D = 33.75 \times 10^{-3} \, \mathrm{m}\). The resolution is: \(\text{Resolution} = 1.22 \frac{550 \times 10^{-9} \, \mathrm{m} \times 11.5 \, \mathrm{m}}{33.75 \times 10^{-3} \, \mathrm{m}}\). Calculate this to find the resolution.
04

Calculate Diameter for Stopped Down Aperture

For the aperture \(f/22.0\), calculate \(D = 135 \, \mathrm{mm}/22\) which yields \(D = 6.14 \, \mathrm{mm} = 6.14 \times 10^{-3} \, \mathrm{m}\).
05

Apply Formula for Stopped Down Aperture

Using the same formula, substitute \(D = 6.14 \times 10^{-3} \, \mathrm{m}\). The resolution becomes: \(\text{Resolution} = 1.22 \frac{550 \times 10^{-9} \, \mathrm{m} \times 11.5 \, \mathrm{m}}{6.14 \times 10^{-3} \, \mathrm{m}}\). Compute this to determine the smallest resolvable feature size.

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Key Concepts

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

Diffraction Limits
Have you ever wondered why no optical device can produce perfectly sharp images? The limitation comes from the concept of diffraction limits. Diffraction refers to the slight bending of light as it passes around the edges of an object or through an aperture, such as a camera lens. This bending causes the light waves to interfere with one another, and beyond a certain point, this interference limits how fine the details can be resolved in an image.

The diffraction limit can be explained through the resolving power formula. This formula allows you to calculate the smallest detail size your lens can distinguish. The formula is: \( \text{Resolution} = 1.22 \frac{\lambda L}{D} \). In this expression:
  • \( \lambda \) represents the wavelength of light,
  • \( L \) is the distance from the lens to the object,
  • \( D \) indicates the diameter of the lens's aperture.
When this limit is reached, features closer together than the resolution limit will seem to blend together. This is particularly important in photography and microscopy, as it sets a fundamental bound to the detail visible in images.
Focal Length
Focal length is a critical aspect of any lens's design and functionality. It is defined as the distance from the lens to the sensor plane when the subject is in focus. The focal length determines the field of view and magnification power of the lens.

In the context of our problem, the focal length of 135 mm signifies a moderate telephoto lens. Telephoto lenses have longer focal lengths and are excellent for capturing distant subjects as they provide a zoomed-in perspective. Nevertheless, with increased focal length, it's crucial to maintain adequate aperture size to gather sufficient light, thereby balancing exposure and depth of field.
  • Shorter focal lengths offer a wider angle of view, suitable for landscapes.
  • Longer focal lengths offer narrow angles, more suitable for distant subjects.
Ultimately, the focal length of your lens will greatly influence the composition and framing of your photographs.
Aperture Diameter
Aperture diameter matters a lot when it comes to capturing quality images. The aperture is the opening in the lens through which light travels before hitting the camera sensor. Its diameter is directly influenced by the f-stop value: the smaller the f-stop, the larger the aperture.

In our scenario:
  • For an aperture of \( f/4.0 \), the diameter \( D \) is 33.75 mm.
  • For an aperture of \( f/22.0 \), the diameter \( D \) reduces to 6.14 mm.
A larger aperture (small f-stop) means more light reaches the sensor, which is beneficial in low light environments and creates a shallow depth of field, blurring the background. Conversely, a smaller aperture allows for greater depth of field, making more of the scene appear in focus, which is ideal for landscapes. This adjustment directly affects the resolution and detail captured in the image due to diffraction limits.
Wavelength
The wavelength of light plays an essential role in determining the resolution capabilities of an optical system. Wavelength is the distance between consecutive peaks of a wave, and light waves vary by color within the spectrum.

In our problem, we use a wavelength of 550 nm, which is typical for green light and often represents average visible light conditions. Different wavelengths (colors) will diffract differently. For instance:
  • Shorter wavelengths (blue/violet) have higher energy and can provide better resolution in microscopes.
  • Longer wavelengths (red/orange) tend to have less resolution.
Therefore, a precise understanding of the wavelength used is crucial in calculating the resolution limit because it directly affects the resolving power of a lens according to the formula earlier discussed.

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Most popular questions from this chapter

A glass sheet is covered by a very thin opaque coating. In the middle of this sheet there is a thin scratch 0.00125 \(\mathrm{mm}\) thick. The sheet is totally immersed beneath the surface of a liquid. Parallel rays of monochromatic coherent light with wavelength 612 \(\mathrm{nm}\) in air strike the sheet perpendicular to its surface and pass through the scratch. A screen is placed in the liquid a distance of 30.0 \(\mathrm{cm}\) away from the sheet and parallel to it. You observe that the first dark fringes on either side of the central bright fringe on the screen are 22.4 \(\mathrm{cm}\) apart. What is the refractive index of the liquid?

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