Subfield: Optics

Figure 1 shows a Fabry-Perot (F-P) etalon, in which air pressure is tunable. The F-P etalon consists of two glass plates with high-reflectivity inner surfaces. The two plates form a cavity in which light can be reflected back and forth. The outer surfaces of the plates are generally not parallel to the inner ones and do not affect the back-and-forth reflection. The air density in the etalon can be controlled. Light from a Sodium lamp is collimated by the lens L1 and then passes through the F-P etalon. The transmitivity of the etalon is given by $T=\frac{1}{1+F \sin ^{2}(\delta / 2)}$, where $F=\frac{4 R}{(1-R)^{2}}, \mathrm{R}$ is the reflectivity of the inner surfaces, $\delta=\frac{4 \pi n t \cos \theta}{\lambda}$ is the phase shift of two neighboring rays, $\mathrm{n}$ is the refractive index of the gas, $\mathrm{t}$ is the spacing of inner surfaces, $\theta$ is the incident angle, and $\lambda$ is the light wavelength.

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Figure 1

The Sodium lamp emits D1 $(\lambda=589.6 \mathrm{~nm})$ and D2 $(589 \mathrm{~nm})$ spectral lines and is located in a tunable uniform magnetic field. For simplicity, an optical filter F1 is assumed to only allow the D1 line to pass through. The D1 line is then collimated to the F-P etalon by the lens L1. Circular interference fringes will be present on the focal plane of the lens L2 with a focal length $\mathrm{f}=30 \mathrm{~cm}$. Different fringes have the different incident angle $\theta$. A microscope is used to observe the fringes. We take the reflectivity $\mathrm{R}=90 \%$ and the inner-surface spacing $\mathrm{t}=1 \mathrm{~cm}$.

Some physical constants: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}, c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$.
Context question:
(a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane.
Context answer:
$m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) }
$\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3)  }$
$D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5)  }$


Extra Supplementary Reading Materials:

(b)  As shown in Fig. 2, the width $\varepsilon$ of the spectral line is defined as the full width of half maximum (FWHM) of light transmitivity T regarding the phase shift $\delta$. The resolution of the F-P etalon is defined as follows: for two wavelengths $\lambda$ and $\lambda+\Delta \lambda$, when the central phase difference $\Delta \delta$ of both spectral lines is larger than $\varepsilon$, they are thought to be resolvable; then the etalon resolution is $\lambda / \Delta \lambda$ when $\Delta \delta=\varepsilon$. For the vacuum case, the $\mathrm{D} 1$ line ( $\lambda=589.6 \mathrm{~nm}$ ), and because of the incident angle $\theta \approx 0$, take $\cos \theta \approx 1.0$, please calculate:

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Figure 2
Context question:
(i) the width $\varepsilon$ of the spectral line.
Context answer:
\boxed{0.21}


Context question:
(ii) the resolution $\lambda / \Delta \lambda$ of the etalon.
Context answer:
\boxed{$1.01 \times 10^{6}$}


Question: (c)  As shown in Fig. 1, the initial air pressure is zero. By slowly tuning the pin valve, air is gradually injected into the F-P etalon and finally the air pressure reaches the standard atmospheric pressure. On the same time, ten new fringes are observed to produce from the center of the ring patterns on the focal plane. Based on this phenomenon, calculate the refractive index of air $n_{\text {air }}$ at the standard atmospheric pressure.

Solution:
From Question (a), we know that the order of the 1st fringe near the center of ring patterns is $m=33921$ at the vacuum case $(\mathrm{n}=1.0)$. When the air pressure reaches the standard atmospheric pressure, the order of the 1 st fringe becomes $\mathrm{m}+10$, so we have:

$n_{\text {air }}=\frac{m+10}{2 t / \lambda}=\frac{33931}{33921}=1.00029$. (c1)

Final answer: $1.00029$