Chapter 23. Wavelength of light in a medium (23-8)

Question

C4N3hMC7HudvMGr6fbW8ed1ZUyKIfMBaPGtABtFzGwNluCiJ
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Question

KkdbBAMj8I9lsfJU3hPMWu+PraJUhQmJp/925Q==
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Question

Yu1kI3nxS0l1G5MhK1z7LHDC21wJq8IO4FHJgzX2zhMQmdTR7DrEGg==
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Question

ymgI2IiqDwWFJc9R5x4NPguZvokiV8VkaLtTwDDx+DQqcNkkNbmVLQ==
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Question

tQ8gxafg9ivHLsen5BjMXxolnNeOMPDIH6o1gg==
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Review

Equation 23-8 says that the wavelength is shorter in a medium with a higher index of refraction, where the propagation speed is slower. For example, red light that has a wavelength \(\lambda_{\mathrm{vacuum}} = 750\) nm in vacuum has a wavelength in water (\(n\) = 1.33) equal to \(\lambda = \lambda_{\mathrm{vacuum}} / n = (750 \mathrm{nm}) / (1.33) = 564\) nm. The frequency of this light is the same in both media: \(f = c / \lambda_{\mathrm{vacuum}} = (3.00 \times 10^8 \mathrm{m}/\mathrm{s}) / (750 \times 10^{-9} \mathrm{m}) = 4.00 \times 10^{14}\) Hz in vacuum and \(f = v/\lambda = c / (n\lambda) = (3.00 \times 10^8 \mathrm{m}/\mathrm{s}) / ((1.33)(564 \times 10^{-9} \mathrm{m}))= 4.00 \times 10^{14}\) Hz in water.