di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Electromagnetism


Question: (b) Given $\omega_{0}$, find such $\vec{v}_{0}$ (direction and magnitude), so that the center of mass will travel in a circle. Find the circle radius $R_{c}$ and the coordinates $x_{c}, y_{c}$ of its center. You don't need to prove the uniqueness of the solution.

Solution:
The net force can be calculated as:

$$
\begin{aligned}
\vec{F}= & \vec{F}_{+}+\vec{F}_{-}=2 q(\vec{\omega} \times \vec{r}) \times \vec{B}= \\
& -2 q(\vec{\omega}(\vec{B} \cdot \vec{r})-\vec{r}(\vec{B} \cdot \vec{\omega}))=2 q B \omega \vec{r}=B \omega \vec{p}
\end{aligned}
$$

where $\vec{p}$ is a dipole moment $(|\vec{p}|=q d=2 q r$ and the direction aligns with $\vec{r}$ ).

When $C$ orbits a circle, $\vec{F}$ acts as a centripetal force, i.e. it points to the center of the circle. Since $\vec{F} \| \vec{p}$, the dipole is always in line with the center of the orbit. Therefore, the orbital angular velocity of $C$ is equal to the angular velocity of rotation of the dipole about $C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_dce611b73b4c47125558g-1.jpg?height=342&width=646&top_left_y=183&top_left_x=1182)

The magnitude of the orbital velocity is:

$$
v_{0}=\left|\omega_{0}\right| R_{c}
$$

From Newton's second law, and accounting that the total mass of the dipole is $2 m$ :

$$
\frac{2 m v_{0}^{2}}{R_{c}}=\frac{p B v_{0}}{R_{c}}
$$

i.e. the magnitude of velocity is:

$$
v_{0}=\frac{p B}{2 m}=\frac{q B d}{2 m}
$$

and the radius of the orbit is:

$$
R_{c}=\frac{v_{0}}{\left|\omega_{0}\right|}=\frac{q B d}{2 m\left|\omega_{0}\right|}
$$

The coordinates of the center of the circle are:

$$
\left(x_{c}, y_{c}\right)=\left( \pm R_{c}, 0\right)
$$

where the "+" sign corresponds to $\omega_{0}>0$, i.e. counterclockwise rotation, and the "-" sign —to clockwise rotation. In either case, the initial velocity should point to the negative $Y$ direction:

$$
\vec{v}_{0}=-\frac{q d B}{2 m} \hat{\jmath}
$$

Final answer: $\vec{v}_{0}=-\frac{q d B}{2 m} \hat{\jmath}$ and $R_{c}=\frac{q B d}{2 m|\omega_{0}|}$ and $(x_{c}, y_{c})=( \pm R_{c}, 0)$
Subfield: Thermodynamics

## 1 Ice pellets

An interesting weather phenomenon can occur when the temperature profile in the atmosphere shows an inversion. The solid blue line in figure 1 shows such a temperature profile. The inversion occurs at heights between $1 \mathrm{~km}$ and $2 \mathrm{~km}$.

Under these conditions snow falling through the atmosphere (partially) melts in the warmer layer and (partially) freezes again before reaching the ground in the form of "ice pellets".

<img_4520>

Figure 1: Atmospheric temperature $T$ vs. height $h$ above the ground.

Assume that a small, spherical ice droplet almost completely melts while falling through the atmospheric layer between $h_{\mathrm{A}}$ and $h_{\mathrm{B}}$ where the temperature is above freezing point.

Question: a. Determine the mass fraction of the droplet that freezes before reaching the ground.

Solution:
With the given assumptions of constant droplet size and constant atmospheric density the drag force leads to a constant velocity $v$ for the fall of the droplet through the atmosphere.

We can assume that the droplet temperature $T_{\mathrm{d}}$ above $h_{\mathrm{A}}$ follows the atmosphere temperature profile (also see below) and remains constant and equal to $T_{0}=0^{\circ} \mathrm{C}$ during the melting below that height.

For small temperature differences $\Delta T=T-T_{\mathrm{d}}$ the heat exchange rate is proportional to this difference $\mathrm{d} Q / \mathrm{d} t=\kappa \Delta T$. The factor $\kappa$ depends on the droplet geometry and its velocity, as well as on the air density. Since these are constant $\kappa=$ const.

In the region between $h_{\mathrm{A}}$ and $h_{\mathrm{B}}$ the droplet is therefore heated at a rate

$$
\mathrm{d} Q=\kappa\left(T-T_{0}\right) \mathrm{d} t=-\kappa\left(T-T_{0}\right) \frac{\mathrm{d} h}{v}, \tag{1}
$$

The total heat transfer between $h_{\mathrm{A}}$ and $h_{\mathrm{B}}$ is

$$
Q=\frac{\kappa A}{v}=m L \tag{2}
$$

where $A=5.0 \mathrm{~km}{ }^{\circ} \mathrm{C}$ is the area between the temperature curve and the height-axis in the region between heights $h_{\mathrm{A}}$ and $h_{\mathrm{B}}$. The right hand side equates the heat with the latent heat necessary to completely melt the ice droplet of mass $m$.

In the region below $h_{\mathrm{B}}$ the liquid droplet partially freezes again. During this process the temperature is again constant. The mass fraction $\eta$ of liquid freezing before reaching the ground can again be derived from the area $A^{\prime}=4.0 \mathrm{~km}{ }^{\circ} \mathrm{C}$ between the curve and the height-axis.

$$
Q^{\prime}=-\frac{\kappa A^{\prime}}{v}=-\eta m L \tag{3}
$$

Dividing (3) by (2) gives the mass fraction

$$
\eta=\frac{A^{\prime}}{A}=\frac{4}{5}=0.80 \tag{4}
$$

Final answer: $\frac{4}{5}$
Subfield: Electromagnetism


Question: (i)  Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$.

Solution:
An electron with charge $-e(e>0)$ experiences the Lorentz force due to the perpendicular magnetic field and the electric force

$$
m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E})
$$

where $\vec{v}$ is the velocity of the electron.

Final answer: $$
m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E})
$$
where $\vec{v}$ is the velocity of the electron.
Subfield: Electromagnetism

The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin.


Figure 1 (a)
<img_4407>


Figure 1 (b)
<img_4500>

Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE.




(a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions.
Context question:
(i)  Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$.
Context answer:
$$
m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E})
$$
where $\vec{v}$ is the velocity of the electron.


Question: (ii)  Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case.

Solution:
In the stationary regime, the acceleration vanishes. Hence

$$
\vec{v}_{s} \times \vec{B}+\vec{E}=0
$$

The velocity can be expressed as

$$
\vec{v}_{s}=\frac{\vec{E} \times \vec{B}}{B^{2}}
$$

whose magnitude is simply $v_{s}=E / B$.

Final answer: $E / B$.
Subfield: Electromagnetism


Question: (i)  Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field $\vec{B}$. When the initial electron velocity $\vec{v}$ is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(0,0,0)$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4f93d3ceab09e952853dg-1.jpg?height=283&width=574&top_left_y=818&top_left_x=724)

Figure 6

Solution:
The electron motion equation

$$
m \frac{\mathrm{d} \vec{v}}{\mathrm{~d} t}=-e \vec{v} \times \vec{B}
\tag{1}
$$

(1) 

Let $\vec{\omega}_{c}=-e \vec{B} / m$, we have

$$
\frac{\mathrm{d} \vec{v}}{\mathrm{~d} t}=-\vec{\omega}_{c} \times \vec{v}
\tag{2}
$$

Since $\vec{B}=B \hat{z}$, thus Equation (2) can be written to be

$$
\left\{\begin{array}{c}
\frac{\mathrm{d} v_{x}}{\mathrm{~d} t}=\omega_{c} v_{y} \\
\frac{\mathrm{d} v_{y}}{\mathrm{~d} t}=-\omega_{c} v_{x} \\
\frac{\mathrm{d} v_{z}}{\mathrm{~d} t}=0
\end{array}\right.
\tag{3}
$$

The general solutions of Equation (3) are

$$
\left\{\begin{array}{c}
v_{x}=v_{\perp} \cos \left(\omega_{c} t+\varphi_{y}\right) \\
v_{y}=v_{\perp} \cos \left(\omega_{c} t+\varphi_{y}\right) \\
v_{z}=v_{z}(t=0)
\end{array}\right.
\tag{4}
$$

where $\varphi_{x, y}$ are the initial phases. Due to $v_{z}(t=0)=0$, it is indicated that the motion of an electron remains to be perpendicular to the magnetic field. By further solving Equation (4), we obtain the motion trajectory of the electron,

$$
\left\{\begin{array}{c}
x=r_{c} \sin \omega_{c} t \\
y=r_{c} \cos \omega_{c} t \\
\quad z=0
\end{array}\right.
\tag{5}
$$


or

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=r_{c}^{2} \\
z=0
\end{array}\right.
$$

where $\left|r_{c}\right|=\frac{v_{\perp}}{\left|\omega_{c}\right|}=\frac{m v_{\perp}}{e B}$ is the gyro radius.

Final answer: $$
\left\{\begin{array}{l}
x^{2}+y^{2}=r_{c}^{2} \\
z=0
\end{array}\right.
$$

where $\left|r_{c}\right|=\frac{v_{\perp}}{\left|\omega_{c}\right|}=\frac{m v_{\perp}}{e B}$ is the gyro radius.
Subfield: Electromagnetism


Question: (i)  Please explain the generation mechanism of the electric current by a schematic drawing.

Solution:
Since the magnetic field and the plasma are uniform $z$, the orbits of ions and electrons can project into in the x-y plane. From the results of (a), we know that an ion has a left-hand circular motion and an electron has a right-hand circular motion. Due to the linear increase of the plasma density in $x$, the number of ions with upward motion is less than that with downward motion at a given $x$ position, which leads a net upward ion flow. Similarly, electrons have a net downward flow. Combining the ion and election flows, we have a net upward electric current as illustrated below in schematic drawing.

![](https://cdn.mathpix.com/cropped/2023_12_21_337a5d5ddb5cdc22a0ddg-1.jpg?height=639&width=577&top_left_y=2005&top_left_x=334)

Final answer: 开放性回答
Subfield: Electromagnetism


Question: (ii)  Based on the motion constant of magnetic moment, please determine what the condition should be satisfied for the angle $\theta$ between the initial electron velocity $\vec{v}$ and the magnetic field at the point " $\mathrm{P}_{2}$ "ifan electron will not escape from the magnetic mirror field.

Solution:
From the motion constant of the magnetic moment of an electron, the perpendicular velocity increases with increase of the magnetic field, which means the parallel velocity decreases due to the conservation of the total kinetic energy. When the electron arrives at the point " $\mathrm{P}_{3}$ ", its parallel velocity decreases to zero, then the electron will not escape from the magnetic mirror field. Thus, the initial velocityshould be

$$
\frac{v_{\perp 0}^{2}}{B_{0}} \geq \frac{v^{2}}{B_{m}}
$$

Since $v_{\perp 0}=v \sin \theta$, we obtain

$$
\theta_{c r} \geq \arcsin \left(\sqrt{\frac{B_{0}}{B_{m}}}\right)
$$

i.e., this is the condition for the electron confined in the magnetic mirror field.

Final answer: $\theta_{c r} \geq \arcsin \left(\sqrt{\frac{B_{0}}{B_{m}}}\right)$
Subfield: Electromagnetism


Question: (e)  Earth's dipole magnetic field lines (blue lines) are shown in Figure 8. The spiral trajectory of a charged particle (red curve) is assumed to be confined in the $y=0$ plane since the gradient and the curvature of the magnetic field can be ignored. If a charged particle with the mass $m$, charge $q$, and velocity $\vec{v}$ is initially located at the equatorial position $\left[x=6 \mathrm{R}_{\mathrm{E}}, y=0, z=0\right]$ and the angle between the electron velocity $\vec{v}$ and the magnetic field is $\theta$ initially, please determine what the condition should be satisfied for $\theta$ if the charged particle arrives below $200 \mathrm{~km}$ of its altitude at the latitude $60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_ae61c9b6b1a575bacd50g-1.jpg?height=545&width=908&top_left_y=2103&top_left_x=583)
Figure 8

Solution:
Since the guiding center of the electron motion is always confined in the $\mathrm{y}=0$ plane, the Earth's dipole magnetic field $\vec{B}=\frac{B_{0} R_{E}^{3}}{r^{5}}\left(-3 x z \hat{x}+\left(x^{2}-2 z^{2}\right) \hat{z}\right)$. At the initial position $\left(6 R_{E}, 0,0\right)$ of an electron, the magnetic field strength



$$
B_{i}=\frac{B_{0} R_{E}^{3}}{r^{5}}\left(x^{2}\right)=\frac{B_{0}}{216}
$$



At the altitude $\mathrm{H}=200 \mathrm{~km}$ and the latitude $\theta_{L}=60^{\circ}$, thus

$$
B_{L}=\left(B_{x}^{2}+B_{z}^{2}\right)^{1 / 2}=\frac{\sqrt{13}}{2} B_{0}
$$

With the same argument in (d) (ii), the condition for the electron to arrive in below the altitude $\mathrm{H}=200 \mathrm{~km}$ with the latitude $\theta_{L}=60^{\circ}$ is

$$
\theta<0.05=2.8^{\circ} .
$$

Final answer: $\theta<2.8^{\circ}$
Subfield: Electromagnetism


Question: (ii) Please give the colors of auroras at the altitudes $\mathrm{H}=160 \mathrm{~km}$ and $\mathrm{H}=220 \mathrm{~km}$. (Hint: The dependence of the collision frequency of atmosphericmolecules on the atmospheric density is $v=v_{0} \rho / \rho_{0}$, where $v_{0} \approx 10^{9} / \mathrm{s}$ is the collision frequency of atmosphericmolecules at sea level.The excited oxygen atom will lose a part of its energy when it collides with other neutral molecules. )

Solution:
At the altitude $H=160 \mathrm{~km}$, the collision frequency of atmosphericmolecules is

$$
v=\frac{v_{0} \rho}{\rho_{0}}=v_{0} e^{-\frac{\rho_{0} g H}{P_{0}}} \approx 0.93 / s
\tag{17}
$$



Since the lifetime of the oxygen atom in the first excited state lasts about 110s, the oxygen atom will collide with other molecules over one hundred times. Thus, the high-frequent collision between particles can de-excite oxygen atoms before they have a chance to radiate. But, the oxygen atoms in the second excited state will emit photons since their lifetime is so short. Thus, aurora at $\mathrm{H}=160 \mathrm{~km}$ is resulted from emission of the oxygen atoms in the second excited state. The wavelength is

$$
\lambda=\frac{h c}{e \Delta V}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{\left(1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)(2.21 \mathrm{eV})}=562 \mathrm{~nm} .
$$

The color for this wavelength is green.



At the altitude $H=220 \mathrm{~km}$, the collision frequency of atmosphericmolecules is

$$
v=\frac{v_{0} \rho}{\rho_{0}}=v_{0} e^{-\frac{\rho_{0} g H}{P_{0}}} \approx 3.8 \times 10^{-4} / \mathrm{s}
\tag{18}
$$

All oxygen atoms in the first and second excited states have a chance to radiate since the collision frequency is so low. Because the number of the oxygen atoms in the first excited state is much larger than in the second excited state, we observe the aurora color is from the emission of the oxygen atoms in the first excited state. The wavelength is

$$
\lambda=\frac{h c}{e \Delta V}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{\left(1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)(1.96 \mathrm{eV})}=633 \mathrm{~nm}
$$



The color for this wavelength is red.

Final answer: when H=160km, the color is green
when H=220km, the color is red
Subfield: Optics


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.

Solution:
The transmittivity of the F-P etalon is given by:

$$
T=\frac{1}{1+F \sin ^{2} \frac{\delta}{2}}
$$

For bright fringes, we have

$$
\begin{aligned}
& T=1 \text { i.e. } \sin ^{2} \frac{\delta}{2}=0 \\
& \frac{\delta}{2}=m \pi \\
& 2 n t \cos \theta=m \lambda
\end{aligned}
$$

For $\mathrm{n}=1.0, \mathrm{t}=1 \mathrm{~cm}, \lambda=589.6 \mathrm{~nm}$, thus:

$$
\cos \theta_{i}=\frac{m_{i}}{2 n t / \lambda}=\frac{m_{i}}{33921.3} \quad \text { (a1)}
$$

Because of $\cos \theta \leq 1$, so the orders of the first three fringes are:

$$
m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) }
$$

The incident angles of the first three fringes are:

$$
\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3)  }
$$

The fringe diameter is given by:

$$
D_{i}=2 f \tan \theta_{i} \approx 2 f \theta_{i} \quad \text { (a4) }
$$

For the focal length $\mathrm{f}=30 \mathrm{~cm}$, thus:

$$
D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5)  }
$$

Final answer: $m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) }
$\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3)  }$
$D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5)  }$
Subfield: Optics


Question: (d)  Energy levels splitting of Sodium atoms occurs when they are placed in a magnetic field. This is called as the Zeeman effect. The energy shift given by $\Delta E=m_{j} g_{k} \mu_{B} B$, where the quantum number $\mathrm{m}_{\mathrm{j}}$ can be $\mathrm{J}, \mathrm{J}-1, \ldots,-\mathrm{J}+1,-\mathrm{J}, \mathrm{J}$ is the total angular quantum number, $\mathrm{g}_{\mathrm{k}}$ is the Landé factor, $\mu_{B}=\frac{h e}{4 \pi m_{e}}$ is Bohr magneton, $\mathrm{h}$ is the Plank constant, e is the electron charge, $m_{e}$ is the electron mass, B is the magnetic field. As shown in Fig. 3, the D1 spectral line is emitted when Sodium atoms jump from the energy level ${ }^{2} \mathrm{P}_{1 / 2}$ down to ${ }^{2} \mathrm{~S}_{1 / 2}$. We have $J=\frac{1}{2}$ for both ${ }^{2} \mathrm{P}_{1 / 2}$ and ${ }^{2} \mathrm{~S}_{1 / 2}$. Therefore, in the magnetic field, each energy level will be split into two levels. We define the energy gap of two splitting levels as $\Delta \mathrm{E}_{1}$ for ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ for ${ }^{2} \mathrm{~S}_{1 / 2}$ respectively $\left(\Delta \mathrm{E}_{1}\right.$ $<\Delta \mathrm{E}_{2}$ ). As a result, the D1 line is split into 4 spectral lines (a, b, c, and d), as showed in Fig. 3. Please write down the expression of the frequency $(v)$ of four lines $a, b, c$, and $d$.

![](https://cdn.mathpix.com/cropped/2023_12_21_aeb204c8d13bb4047ac3g-1.jpg?height=525&width=640&top_left_y=1085&top_left_x=731)

Figure 3

Solution:
The frequency of D1 line $\left({ }^{2} \mathrm{P}_{1 / 2}\right.$ to $\left.{ }^{2} \mathrm{~S}_{1 / 2}\right)$ is given by: $v_{0}=c / \lambda(\lambda=589.6 \mathrm{~nm})$

When magnetic field $\mathrm{B}$ is applied, the frequency of the line a,b,c,d are expressed as:

1) ${ }^{2} \mathrm{P}_{1 / 2}(\mathrm{mj}=-1 / 2) \rightarrow{ }^{2} \mathrm{~S}_{1 / 2}(\mathrm{mj}=1 / 2)$ : frequency of (a)): $v_{a}=v_{0}-\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$; 
2) ${ }^{2} \mathrm{P}_{1 / 2}(\mathrm{mj}=1 / 2) \rightarrow{ }^{2} \mathrm{~S}_{1 / 2}(\mathrm{mj}=1 / 2)$ : frequency of $(\mathrm{b}): v_{b}=v_{0}-\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$; 
3) ${ }^{2} \mathrm{P}_{1 / 2}(\mathrm{mj}=-1 / 2) \rightarrow{ }^{2} \mathrm{~S}_{1 / 2}(\mathrm{mj}=-1 / 2)$ : frequency of (c): $v_{c}=v_{0}+\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$; 
4) ${ }^{2} \mathrm{P}_{1 / 2}(\mathrm{mj}=1 / 2) \rightarrow{ }^{2} \mathrm{~S}_{1 / 2}(\mathrm{mj}=-1 / 2)$ : frequency of $(\mathrm{d}): v_{d}=v_{0}+\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$;

Final answer: $v_{a}=v_{0}-\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$,$v_{b}=v_{0}-\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{c}=v_{0}+\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{d}=v_{0}+\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$
Subfield: Optics


Question: (e)  As shown in Fig. 4, when the magnetic field is turned on, each fringe of the D1 line will split into four sub-fringes $(1,2,3$, and 4$)$. The diameter of the four sub-fringes near the center is measured as $D_{1}, D_{2}, D_{3}$, and $D_{4}$. Please give the expression of the splitting energy gap $\Delta \mathrm{E}_{1}$ of ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ of ${ }^{2} \mathrm{~S}_{1 / 2}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_aeb204c8d13bb4047ac3g-1.jpg?height=539&width=808&top_left_y=2055&top_left_x=630)

Figure 4

Solution:
$\quad \theta_{m}<<1, \cos \theta_{m}=1-\frac{\theta_{m}^{2}}{2}$, (e1)



$2 n t \cos \theta_{m}=m \lambda, \quad 1-\frac{\theta_{m}^{2}}{2}=\frac{m \lambda}{2 n t}$, (e2)



$\lambda \rightarrow \lambda+\Delta \lambda, \theta_{m} \rightarrow \theta_{m}^{\prime}$

$1-\frac{\theta_{m}^{\prime 2}}{2}=\frac{m(\lambda+\Delta \lambda)}{2 n t}$,

$\frac{\theta_{m}^{2}-\theta_{m}^{\prime 2}}{2}=\frac{m \Delta \lambda}{2 n t}$ (e3)



$2 f \theta_{m}=D_{m}, \quad \frac{D_{m}^{2}-D_{m}^{\prime 2}}{8 f^{2}}=\frac{m \Delta \lambda}{2 n t}=\frac{\Delta \lambda}{\lambda}$

$\Delta \lambda=\lambda \frac{D_{m}^{2}-D_{m}^{\prime 2}}{8 f^{2}}$ (e4) 



The lines a, b, c, and d correspond to sub-fringe 1, 2, 3, and 4. From Question (d), we have.

The wavelength difference of the spectral line $a$ and $b$ is given by:

$\Delta \lambda_{1}=\lambda \frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$



$\Delta E_{1}=h\left(v_{b}-v_{a}\right), \Delta E_{2}=h\left(v_{d}-v_{b}\right)$

or $\Delta E_{1}=h\left(v_{d}-v_{c}\right), \Delta E_{2}=h\left(v_{c}-v_{a}\right)$ (e5)

The wavelength difference of the spectral line $a$ and $b$ is given by:

$\Delta \lambda_{1}=\lambda \frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$

Then we obtain

$\Delta E_{1}=h \Delta v_{1}=\left|-\frac{h c}{\lambda} \bullet \frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}\right|=\frac{h c}{\lambda} \bullet \frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$

( or $\Delta E_{1}=h \Delta v_{1}=\left|-\frac{h c}{\lambda} \bullet \frac{D_{4}^{2}-D_{3}^{2}}{8 f^{2}}\right|=\frac{h c}{\lambda} \bullet \frac{D_{4}^{2}-D_{3}^{2}}{8 f^{2}}$ ) (e6)

Similarly, for $\Delta \mathrm{E}_{2}$, we get

$\Delta \lambda_{2}=\lambda \frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$

$\Delta E_{2}=h \Delta v_{2}=\left|-\frac{h c}{\lambda} \bullet \frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}\right|=\frac{h c}{\lambda} \bullet \frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$

( or $\Delta E_{1}=h \Delta v_{1}=\left|-\frac{h c}{\lambda} \bullet \frac{D_{3}^{2}-D_{1}^{2}}{8 f^{2}}\right|=\frac{h c}{\lambda} \bullet \frac{D_{3}^{2}-D_{1}^{2}}{8 f^{2}}$ ) (e7)

Final answer: $\Delta E_{1}=\frac{h c}{\lambda} \cdot \frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$,$\Delta E_{2}=\frac{h c}{\lambda} \cdot \frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$