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Ch 36: Diffraction
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 36: DiffractionProblem 44
Chapter 36, Problem 44

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?

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Step 1: Understand the problem. The question involves resolving the smallest feature on the bear using a lens. This is determined by the diffraction limit, which depends on the wavelength of light, the aperture size, and the focal ratio (f-number). The formula for the angular resolution is θ = 1.22 * (λ / D), where λ is the wavelength of light and D is the diameter of the aperture.
Step 2: Relate the aperture diameter to the f-number. The f-number (f/N) is defined as the ratio of the focal length (f) to the aperture diameter (D). Rearrange this to find D = f / N. For part (a), the maximum aperture corresponds to f/4.00, so D = 135 mm / 4.00.
Step 3: Calculate the angular resolution θ for part (a). Use the formula θ = 1.22 * (λ / D), where λ = 550 nm (convert to meters: 550 × 10^(-9) m) and D is the aperture diameter calculated in Step 2. This gives the angular resolution in radians.
Step 4: Convert the angular resolution to a linear resolution on the bear. The linear resolution (Δx) is related to the angular resolution by Δx = θ * L, where L is the distance to the object (11.5 m in this case). Perform this calculation for part (a).
Step 5: Repeat the process for part (b). For f/22.0, calculate the new aperture diameter D = 135 mm / 22.0. Then, use the same steps to find the angular resolution θ and the corresponding linear resolution Δx for the stopped-down aperture.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Resolution in Optics

Resolution refers to the ability of an optical system to distinguish between closely spaced objects. It is determined by the lens's aperture and the wavelength of light used. The smaller the resolvable feature, the better the resolution. In photography, resolution is crucial for capturing fine details in images, especially when using telephoto lenses.
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Aperture and f-stop

The aperture of a lens controls the amount of light that enters and affects the depth of field in an image. The f-stop number indicates the size of the aperture; a lower f-stop (like f/4) means a larger opening, allowing more light and resulting in a shallower depth of field. Conversely, a higher f-stop (like f/22) reduces the aperture size, increasing depth of field but decreasing light intake.
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Rayleigh Criterion

The Rayleigh Criterion is a formula used to determine the minimum resolvable detail in an optical system. It states that two point sources are resolvable when the central maximum of one diffraction pattern coincides with the first minimum of another. This criterion is influenced by the wavelength of light and the aperture size, making it essential for calculating the smallest resolvable feature in photography.
Related Practice
Textbook Question

The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (380 - 750 nm). The Arecibo radio telescope in Puerto Rico is 305 m (1000 ft) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength 75 cm. Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon?

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Textbook Question

If you can read the bottom row of your doctor’s eye chart, your eye has a resolving power of 1 arcminute, equal to 1/60 degree. If this resolving power is diffraction-limited, to what effective diameter of your eye’s optical system does this correspond? Use Rayleigh’s criterion and assume λ = 550 nm.

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Textbook Question

The VLBA (Very Long Baseline Array) uses a number of individual radio telescopes to make one unit having an equivalent diameter of about 8000 km. When this radio telescope is focusing radio waves of wavelength 2.0 cm, what would have to be the diameter of the mirror of a visible-light telescope focusing light of wavelength 550 nm so that the visible-light telescope has the same resolution as the radio telescope?

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