Question

In Fig. 35-48, an airtight chamber of length $\mathit{d}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ is placed in one of the arms of a Michelson interferometer. (The glass window on each end of the chamber has negligible thickness.) Light of wavelength l $\mathbf{\lambda }\mathbf{=}\mathbf{500}\mathbf{}\mathbf{nm}$ is used. Evacuating the air from the chamber causes a shift of 60 bright fringes. From these data and to six significant figures, find the index of refraction of air at atmospheric pressure.

Short answer

The refractive index of the air at atmospheric pressure is $1.0003$.

Step-by-step solution

Given data

Wavelength of light $\lambda =500 \text{nm}$

Chamber of length$d=5.0\mathrm{cm}$

Principal of Michelson interferometer

Fiber optic Michelson interferometer employs the same principle of splitting a laser beam and inserting the optical path difference between the arms.

Concept used

The light source emerges light that strikes the encounter beam splitter. Encounter beam splitter is nothing but a plane mirror which is inclined at $45°$ with the horizontal this encounter beam splitter transmits half of the light through horizontal mirror and reflects the other through the vertical mirror.

Finally, these two light rays completely reflect from the vertical and horizontal mirrors and enters the telescope.

The path difference between the reflected ray and transmission ray is,

$\Delta \lambda ={\lambda }_v-{\lambda }_h$

Here,$\Delta \lambda$ is the path difference between the reflected ray, ${\lambda }_v$ is the path length of the reflected ray, and ${\lambda }_h$is the path length of the transmitted ray.

The expression for the path length of the reflected ray is,

${\lambda }_v=2Ln$

Here, L is thickness of material or chamber length and n is the refractive index of the medium.

The expression for the path length of the transmitted ray is,

${\lambda }_h=2L$

Substitute 2L for ${\lambda }_h$ and $2Ln$for ${\lambda }_v$ in the equation $\Delta \lambda ={\lambda }_v-{\lambda }_h$.

$\begin{array}{rcl}\Delta \lambda & =& 2Ln-2L\\ & =& 2L\left(n-1\right)\end{array}$

The relation between the path difference and phase difference of the light rays is,

$\Delta \varphi =\frac{2\pi }{\lambda }\left(\Delta \lambda \right)$

Here, $\Delta \varphi$ is the phase difference of the light rays and $\lambda$ is the wavelength of the light rays.

The fringe pattern of the light rays can shift by one fringe for one each phase change of light rays.

Then the expression for the phase change of the light rays is,

$\Delta \varphi =2\pi N_B$

Here, $N_B$ is the number of bright fringes.

Determine the index of refraction of air at atmospheric pressure

Substitute $2\pi N_B$ for $∆\varphi$ and $2L\left(N-1\right)$ for $∆\lambda$ and rewrite it for n.

$\begin{array}{rcl}2\pi N_B& =& \frac{2\pi }{\lambda }\left(2L\left(N-1\right)\right)\\ N_B& =& \frac{2Ln}{\lambda }-\frac{2L}{\lambda }\\ n& =& \frac{\lambda }{2L}\left(N_B+\frac{2L}{\lambda }\right)\\ & =& \frac{\lambda N_B}{2L}+1\end{array}$

Substitute 500 nm for $\lambda$, 60 for and for $N_B$, and 5.0 cm for L in the equation $n=\frac{\lambda N_B}{2L}+1$, solve for n.

$\begin{array}{rcl}n& =& \left(\frac{\left(500\text{nm}\right)\left(60\right)}{\left(2\right)\left(5.0\text{cm}\right)}\right)+1\\ & =& \left(\frac{\left(500\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{cm}}\right)\right)}{\left(2\right)\left(5.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}\right)+1\\ & =& 1.0003\end{array}$

Therefore, the refractive index of the air at atmospheric pressure is $1.0003$.