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Ch. 35 - Diffraction
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 34, Problem 51

A diffraction grating has 15,000 rulings in its 1.9 cm width. Determine (a) its resolving power in first and second orders, and (b) the minimum wavelength resolution (∆λ) it can yield for λ = 410 nm.

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Determine the number of rulings (N) on the diffraction grating by dividing the total number of rulings by the width of the grating. Use the formula: \( N = \text{total rulings} \).
Calculate the resolving power \( R \) for the first and second orders using the formula \( R = mN \), where \( m \) is the order of diffraction and \( N \) is the number of rulings.
Substitute \( m = 1 \) for the first order and \( m = 2 \) for the second order into the resolving power formula to find \( R \) for each case.
To find the minimum wavelength resolution \( \Delta \lambda \), use the formula \( \Delta \lambda = \frac{\lambda}{R} \), where \( \lambda \) is the given wavelength (410 nm) and \( R \) is the resolving power calculated in the previous step.
Substitute the values of \( R \) for the first and second orders into the formula for \( \Delta \lambda \) to determine the minimum wavelength resolution for each case.

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Key Concepts

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

Diffraction Grating

A diffraction grating is an optical component with a periodic structure that disperses light into its constituent wavelengths. The number of rulings per unit length determines its ability to separate different wavelengths. In this case, the grating has 15,000 rulings over a width of 1.9 cm, which is crucial for calculating its resolving power.
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Resolving Power

Resolving power is a measure of an optical system's ability to distinguish between closely spaced wavelengths. For a diffraction grating, it is defined as the ratio of the wavelength to the minimum resolvable wavelength difference (∆λ). The resolving power can be calculated using the formula R = nN, where n is the order of diffraction and N is the total number of rulings.
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Wavelength Resolution (∆λ)

Wavelength resolution (∆λ) refers to the smallest difference in wavelength that can be distinguished by the diffraction grating. It is inversely related to the resolving power; higher resolving power results in smaller ∆λ. For a given wavelength, the minimum wavelength resolution can be calculated using the formula ∆λ = λ/R, where λ is the wavelength of light being analyzed.
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Textbook Question

(II) White light passes through a 640-slit/ mm diffraction grating. First-order and second-order visible spectra (“rainbows”) appear on the wall 32 cm away as shown in Fig. 35–40. Determine the widths ℓ₁ and ℓ₂ of the two “rainbows” (400 nm to 700 nm). In which order is the “rainbow” dispersed over a larger distance?

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Textbook Question

(II) X-rays of wavelength 0.138 nm fall on a crystal whose atoms, lying in planes, are spaced 0.315 nm apart. At what angle Φ (relative to the surface, Fig. 35–28) must the X-rays be directed if the first diffraction maximum is to be observed?

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Textbook Question

Red laser light from a He–Ne laser (λ = 632.8 nm) creates a second-order fringe at 53.2° after passing through a grating. What is the wavelength λ of light that creates a first-order fringe at 21.2°?

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Textbook Question

Suppose the angles measured in Problem 42 were produced when the spectrometer (but not the source) was submerged in water. What then would be the wavelengths (in air)?

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Textbook Question

You want to design a spy satellite to photograph license plate numbers. Assuming it is necessary to resolve points separated by 2 cm with 550-nm light, and that the satellite orbits at a height of 130 km, what minimum lens aperture (diameter) is required?

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Textbook Question

(II) (a) Suppose for a conventional X-ray image that the X-ray beam consists of parallel rays. What would be the magnification of the image? (b) Suppose, instead, that the X-rays come from a point source (as in Fig. 35–31) that is 15 cm in front of a human body which is 25 cm thick, and the film is pressed against the person’s back. Determine and discuss the range of magnifications that result.


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