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Ch 36: Diffraction
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
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Capitolo 36, Problema 17a

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 mm wide. At the point in the pattern 3.25° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. What is the wavelength of the radiation?

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Understand the relationship between the phase difference and the path difference in a single-slit diffraction pattern. The phase difference \( \Delta \phi \) is related to the path difference \( \Delta x \) by the formula \( \Delta \phi = \frac{2 \pi \Delta x}{\lambda} \), where \( \lambda \) is the wavelength of the radiation.
Determine the path difference \( \Delta x \) for the wavelets from the top and bottom of the slit. The path difference is given by \( \Delta x = a \sin \theta \), where \( a \) is the width of the slit (0.105 mm) and \( \theta \) is the angle from the central maximum (3.25°).
Substitute the expression for \( \Delta x \) into the phase difference formula: \( \Delta \phi = \frac{2 \pi (a \sin \theta)}{\lambda} \). Rearrange this equation to solve for the wavelength \( \lambda \): \( \lambda = \frac{2 \pi a \sin \theta}{\Delta \phi} \).
Convert all quantities to consistent units. Ensure the slit width \( a \) is in meters (0.105 mm = 0.105 × 10^{-3} m), the angle \( \theta \) is in radians (3.25° = 3.25 × \(\frac{\pi}{180}\) radians), and the phase difference \( \Delta \phi \) is in radians (already given as 56.0 rad).
Substitute the numerical values for \( a \), \( \sin \theta \), and \( \Delta \phi \) into the formula for \( \lambda \) to calculate the wavelength. The result will be in meters, which can be converted to nanometers if needed (1 m = 10^9 nm).

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Diffraction

Diffraction is the bending of waves around obstacles and the spreading of waves when they pass through narrow openings. In the context of light, diffraction patterns arise when waves encounter a slit, leading to interference effects that create a series of bright and dark fringes. The extent of diffraction depends on the wavelength of the light and the size of the slit.
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Diffraction

Phase Difference

Phase difference refers to the difference in the phase of two waves at a given point in time, often measured in radians. In diffraction, the phase difference between wavelets emanating from different parts of a slit affects the interference pattern observed. A phase difference of 2π radians corresponds to a full cycle of the wave, while other values lead to constructive or destructive interference.
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Phase Constant of a Wave Function

Wavelength

Wavelength is the distance between successive peaks (or troughs) of a wave, typically denoted by the symbol λ. It is a fundamental property of waves, including electromagnetic radiation, and is inversely related to frequency. In diffraction problems, the wavelength can be determined using the geometry of the setup and the observed phase differences, as it influences the pattern's spacing and intensity.
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