The difference in the number of wavelengths, | vedclass

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(B) Let $L$ be the thickness of the air and vacuum columns.The number of wavelengths in vacuum is $N_v = \frac{L}{\lambda_0}$.The number of wavelength

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$2 	ext{ m}$

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(B) Let $L$ be the thickness of the air and vacuum columns.The number of wavelengths in vacuum is $N_v = \frac{L}{\lambda_0}$.The number of wavelengths in air is $N_a = \frac{L}{\lambda_a} = \frac{L}{\lambda_0 / \mu_a} = \frac{L \mu_a}{\lambda_0}$.The difference in the number of wavelengths is given as $N_a - N_v = 1$.$\frac{L \mu_a}{\lambda_0} - \frac{L}{\lambda_0} = 1$.$L \left( \frac{\mu_a - 1}{\lambda_0} ight) = 1$.$L = \frac{\lambda_0}{\mu_a - 1}$.Given $\lambda_0 = 6000 	imes 10^{-10} 	ext{ m}$ and $\mu_a = 1.0003$.$L = \frac{6000 	imes 10^{-10}}{1.0003 - 1} = \frac{6 	imes 10^{-7}}{0.0003} = \frac{6 	imes 10^{-7}}{3 	imes 10^{-4}} = 2 	imes 10^{-3} 	ext{ m} = 2 	ext{ mm}$.

The difference in the number of wavelengths, when yellow light propagates through air and vacuum columns of the same thickness, is $1$. The thickness of the air column is (Refractive index of air $\mu_a = 1.0003$, Wavelength of yellow light in vacuum $\lambda_0 = 6000 \text{ Å}$)

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