The difference in the number of wavelengths,w | 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

Vedclass · Accepted Answer

$2 	ext{ mm}$

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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 $1$:$N_a - N_v = 1$$\frac{L \mu_a}{\lambda_0} - \frac{L}{\lambda_0} = 1$$\frac{L}{\lambda_0} (\mu_a - 1) = 1$$L = \frac{\lambda_0}{\mu_a - 1}$Given $\lambda_0 = 6000 	ext{ Å} = 6000 	imes 10^{-10} 	ext{ m} = 6 	imes 10^{-7} 	ext{ m}$ and $\mu_a = 1.0003$.$L = \frac{6 	imes 10^{-7}}{1.0003 - 1} = \frac{6 	imes 10^{-7}}{0.0003} = \frac{6 	imes 10^{-7}}{3 	imes 10^{-4}} = 2 	imes 10^{-3} 	ext{ m}$.$L = 2 	ext{ mm}$.

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

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