Ch 36: Diffraction

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 36: Diffraction

Problem 40

Chapter 36, Problem 40

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.

Verified step by step guidance

Step 1: Understand the problem. The resolving power of the eye is given as 1 arcminute (1/60 degree), and we are tasked with finding the effective diameter of the eye's optical system using Rayleigh's criterion. The wavelength of light is given as λ = 550 nm.

Step 2: Recall Rayleigh's criterion for diffraction-limited resolution. The angular resolution θ (in radians) is given by: θ = 1.22 * (λ / D), where λ is the wavelength of light and D is the diameter of the aperture (in this case, the effective diameter of the eye's optical system).

Step 3: Convert the angular resolution from arcminutes to radians. Since 1 degree = π/180 radians and 1 arcminute = 1/60 degree, the angular resolution θ in radians is: θ = (1/60) * (π/180).

Step 4: Rearrange Rayleigh's criterion formula to solve for D (the effective diameter): D = 1.22 * (λ / θ). Substitute the values for λ (550 nm, converted to meters as 550 * 10^-9 m) and θ (calculated in Step 3).

Step 5: Perform the substitution and simplify the expression for D. This will give the effective diameter of the eye's optical system in meters. Ensure units are consistent throughout the calculation.

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Key Concepts

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

Resolving Power

Resolving power refers to the ability of an optical system to distinguish between two closely spaced objects. It is often measured in angular units, such as arcminutes or degrees. In this context, a resolving power of 1 arcminute means that the eye can differentiate between two points that are separated by an angle of 1/60th of a degree.

Rayleigh's Criterion

Rayleigh's criterion is a formula used to determine the minimum resolvable detail in an optical system due to diffraction. It states that two point sources are considered resolvable when the central maximum of one diffraction pattern coincides with the first minimum of another. This criterion is essential for calculating the effective diameter of an optical system based on its resolving power.

Diffraction Limit

The diffraction limit is the fundamental limit to the resolution of an optical system caused by the wave nature of light. It occurs when light waves spread out as they pass through an aperture, leading to a loss of detail. In the context of the eye, if the resolving power is diffraction limited, it implies that the eye's ability to resolve detail is constrained by the physical properties of light rather than by the eye's optical components.

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Textbook Question

If the planes of a crystal are 3.50 Å (1 Å = 10^-10 m = 1 Ångstrom unit) apart, what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0°, and in what part of the electromagnetic spectrum do these waves lie?

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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

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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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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A laser beam of wavelength λ = 632.8 nm shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are 1.60 μm apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only 0.740 μm apart. Repeat the calculation of part (a) for the DVD.

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