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

Question: (i) the width $\varepsilon$ of the spectral line.

Solution:
The half maximum occurs at:

$$
\delta=2 m \pi \pm \frac{\varepsilon}{2} \text { (b1) }
$$

Given that $T=0.5$, thus:

$$
F \sin ^{2} \frac{\delta}{2}=1 \quad \text { (b2) }
$$

$\varepsilon=\frac{4}{\sqrt{F}}=\frac{2(1-R)}{\sqrt{R}}=\frac{2(1-0.9)}{\sqrt{0.9}}=0.21 \mathrm{rad}$( or 12.03degree) (b3)

Final answer: 0.21 rad