Question

The difference in the number of wavelengths when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is:
(Refractive index of air ${{\mu }_{a}}=1.0003$; wavelength of yellow light in vacuum = $6000\overset{\text{o}}{\mathop{\text{A}}}\,$)
(A). $1.8mm$
(B). $2mm$
(C). $2cm$
(D). $2.2cm$

Answer 1

Light propagates between air and vacuum columns. The thickness of columns is the same. The number of wavelengths that can propagate through each medium depends on the thickness of that medium and the wavelength of light. Using the above relation, we find the difference in the number of wavelengths and use that equation to calculate thickness.

Formula used:
$\mu =\dfrac{{{\lambda }_{0}}}{\lambda }$
$n=\dfrac{t}{\lambda }$
$\dfrac{x}{{{\lambda }_{air}}}-\dfrac{x}{{{\lambda }_{vacuum}}}=1$

Complete step-by-step answer:
The refractive index is the ratio of the speed of light in vacuum to the speed of light in a medium. It does not have any unit. Therefore,
$\mu =\dfrac{c}{v}$
Here, $\mu$ is the refractive index of a medium
$c$ is the speed of light in vacuum
$v$ is the speed of light in medium
The refractive index is related to the wavelength as-
$\mu =\dfrac{{{\lambda }_{0}}}{\lambda }$ - (1)
Here, $\lambda$ is the wavelength in medium
${{\lambda }_{0}}$ is the wavelength in vacuum
Given that yellow light is used, it propagates through columns of air and vacuum, the thickness of air and vacuum columns is same. Let the thickness of the columns be $x$.
The number of wavelengths is given by-
$n=\dfrac{t}{\lambda }$ - (2)
Here, $n$ is the number of wavelengths
$t$ is the thickness
The difference in the number of wavelengths is one, therefore, using eq (1) we get,
$\dfrac{x}{{{\lambda }_{air}}}-\dfrac{x}{{{\lambda }_{vacuum}}}=1$
Here, ${{\lambda }_{air}}$ is the wavelength of yellow light in air column
${{\lambda }_{vacuum}}$ is the wavelength of yellow light in the vacuum column
Solving the above equation, we get,
$\begin{aligned} & \dfrac{x}{{{\lambda }_{air}}}-\dfrac{x}{{{\lambda }_{vacuum}}}=1 \\\ & \Rightarrow x\left( \dfrac{1}{{{\lambda }_{air}}}-\dfrac{1}{{{\lambda }_{vacuum}}} \right)=1 \\\ \end{aligned}$
Given, ${{\mu }_{a}}=1.0003$, $6000\overset{\text{o}}{\mathop{\text{A}}}\,$. Given values are substituted in the above equation to get,
$\dfrac{x}{\lambda }\left( {{\mu }_{air}}-{{\mu }_{vacuum}} \right)=1$ [Substituting from eq (1)]
$\begin{aligned} & \Rightarrow \dfrac{x}{\lambda }\left( 1.0003-1 \right)=1 \\\ & \Rightarrow x=\dfrac{6000\times {{10}^{-10}}}{0.0003} \\\ & \therefore x=2mm \\\ \end{aligned}$
Therefore, the thickness of the air and vacuum columns is $2mm$.

So, the correct answer is “Option B”.

Note: Refractive index is unitless because it is a ratio of similar quantities. It also depends on the density of a material. As the density increases, the refractive index also increases and vice versa. Light is an electromagnetic wave and can even propagate without a medium or in vacuum.