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Ch 17: Superposition
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 17, Problema 65

Engineers are testing a new thin-film coating whose index of refraction is less than that of glass. They deposit a 560-nm-thick layer on glass, then shine lasers on it. A red laser with a wavelength of 640 nm has no reflection at all, but a violet laser with a wavelength of 400 nm has a maximum reflection. How the coating behaves at other wavelengths is unknown. What is the coating’s index of refraction?

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Step 1: Understand the problem. The coating is a thin-film layer on glass, and its behavior is described by interference effects. The red laser (640 nm) experiences destructive interference (no reflection), while the violet laser (400 nm) experiences constructive interference (maximum reflection). The goal is to find the index of refraction of the coating.
Step 2: Recall the condition for destructive interference in thin films. Destructive interference occurs when the optical path difference is an odd multiple of half the wavelength in the film. The formula for the optical path difference is: 2tn, where t is the thickness of the film, n is the index of refraction, and the wavelength is adjusted for the medium.
Step 3: Recall the condition for constructive interference in thin films. Constructive interference occurs when the optical path difference is an integer multiple of the wavelength in the film. The same formula applies: 2tn. For the violet laser, this condition is satisfied.
Step 4: Use the given data to set up equations. For the red laser (destructive interference), the condition is: 2tn=mλ2, where λ is the wavelength in the coating and m is an odd integer. For the violet laser (constructive interference), the condition is: 2tn=mλ, where m is an integer.
Step 5: Solve for the index of refraction n. Substitute the thickness t = 560 nm and the wavelengths of the lasers into the equations. Use the fact that the wavelength in the coating is related to the wavelength in air by λ=λn. Solve the system of equations to find n.

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Index of Refraction

The index of refraction (n) is a dimensionless number that describes how light propagates through a medium. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. A lower index of refraction indicates that light travels faster in that medium compared to glass, which has an index of approximately 1.5.
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Index of Refraction

Thin-Film Interference

Thin-film interference occurs when light waves reflect off the boundaries of a thin layer, such as the coating in this scenario. The reflected waves can interfere constructively or destructively depending on the thickness of the film and the wavelength of the light. This phenomenon explains why certain wavelengths experience maximum reflection while others do not, based on the film's thickness and refractive index.
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Wave Interference & Superposition

Wavelength and Reflection

The wavelength of light plays a crucial role in determining how it interacts with materials. In this case, the red laser (640 nm) experiences no reflection, indicating that it is likely in a condition of destructive interference, while the violet laser (400 nm) reflects maximally, suggesting constructive interference. The relationship between the wavelength, film thickness, and refractive index is essential for predicting the behavior of light in the coating.
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Law of Reflection
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