The difference in the number of waves when ye | vedclass

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(A) Let the thickness of the columns be $t$. The number of waves in vacuum is $N_v = t / \lambda$ and the number of waves in air is $N_a = t / \lambda

Vedclass · Accepted Answer

$2$

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(A) Let the thickness of the columns be $t$. The number of waves in vacuum is $N_v = t / \lambda$ and the number of waves in air is $N_a = t / \lambda_a$,where $\lambda_a = \lambda / \mu_{air}$.Given that the difference in the number of waves is $1$,we have $N_v - N_a = 1$.Substituting the expressions: $t / \lambda - t / (\lambda / \mu_{air}) = 1$.$t / \lambda - (t \cdot \mu_{air}) / \lambda = 1$.$(t / \lambda) (1 - \mu_{air}) = 1$.Since $\mu_{air} = 1.0003$,the magnitude of the difference is $(t / \lambda) (\mu_{air} - 1) = 1$.$t = \lambda / (\mu_{air} - 1) = 6000 	imes 10^{-10} \, m / (1.0003 - 1) = 6000 	imes 10^{-10} / 0.0003$.$t = 6000 	imes 10^{-10} / (3 	imes 10^{-4}) = 2000 	imes 10^{-6} \, m = 2 	imes 10^{-3} \, m = 2 \, mm$.

The difference in the number of waves when yellow light propagates through air and vacuum columns of the same thickness is $1$. The thickness of the air column is $....\,mm$. [Refractive index of air $= 1.0003$,wavelength of yellow light in vacuum $= 6000\,\mathring{A}$]

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