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

3. The circular restricted three-body problem

In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.

Two-body problem

Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
Context question:
(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.
Context answer:
$$
\begin{aligned}
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
\end{aligned}
$$


Context question:
(b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.
Context answer:
\boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$}


Extra Supplementary Reading Materials:

Circular restricted three-body problem

Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$.

The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$.
Context question:
(a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.
Context answer:
\boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$}


Question: (b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given.

Solution:
$$
\begin{gathered}
\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}} \\
\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}
\end{gathered}
$$

Final answer: $\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}}$ , $\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}$
Subfield: Mechanics


Question: (a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.)
![](https://cdn.mathpix.com/cropped/2023_12_21_2ce10c2f629ebbdfd870g-1.jpg?height=166&width=876&top_left_y=1708&top_left_x=622)

Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u.

Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ?

Solution:
We can always look the infinitesimal process in the comoving frame such that the rocket is at rest in it. Then from conservation of momentum:

$$
\mathbf{u} \Delta m+m \Delta \mathbf{v}=0
\tag{11}
$$

Divided by the infinitesimal time $\Delta t$ we get

$$
\frac{\Delta \mathbf{v}}{\Delta t}=-\frac{\mathbf{u}}{m} \frac{\Delta m}{\Delta t}
\tag{12}
$$

As a result

$$
\mathbf{a}=\frac{d \mathbf{v}}{d t}=-\frac{\mu}{m} \mathbf{u}
\tag{13}
$$

Final answer: $\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$
Subfield: Mechanics

4. The fundamental rocket equation

In this problem, we will investigate the acceleration of rockets.

Rocket repulsion

In empty space, an accelerating rocket must "throw" something backward to gain speed from repulsion. Assume there is zero gravity.

The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out.
<img_4415>

Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$.

(Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.)

Question: (a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.)
<img_4337>

Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u.

Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ?

Solution:
We can always look the infinitesimal process in the comoving frame such that the rocket is at rest in it. Then from conservation of momentum:

$$
\mathbf{u} \Delta m+m \Delta \mathbf{v}=0
\tag{11}
$$

Divided by the infinitesimal time $\Delta t$ we get

$$
\frac{\Delta \mathbf{v}}{\Delta t}=-\frac{\mathbf{u}}{m} \frac{\Delta m}{\Delta t}
\tag{12}
$$

As a result

$$
\mathbf{a}=\frac{d \mathbf{v}}{d t}=-\frac{\mu}{m} \mathbf{u}
\tag{13}
$$

Final answer: $\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$
Subfield: Mechanics

4. The fundamental rocket equation

In this problem, we will investigate the acceleration of rockets.

Rocket repulsion

In empty space, an accelerating rocket must "throw" something backward to gain speed from repulsion. Assume there is zero gravity.

The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out.
<img_4415>

Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$.

(Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.)
Context question:
(a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.)
<img_4337>

Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u.

Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ?
Context answer:
\boxed{$\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$}

Question: (b) Suppose that the rocket is stationary at time $t=0$. The empty rocket has mass $m_{0}$ and the rocket full of fuel has mass $9 m_{0}$. The engine is turned on at time $t=0$, and fuel is ejected at the rate $\mu$ and relative velocity $\mathbf{u}$ described previously. What will be the speed of the rocket when it runs out of fuel, in terms of $u=|\mathbf{u}|$ ?



(Hint: useful integral formula: $\int_{a}^{b} \frac{1}{x} d x=\ln \left(\frac{b}{a}\right)$, and $\ln (10) \approx 2.302, \ln (9) \approx 2.197$.)

Solution:
From part a) we can get a differential equation:

$$
d v=-u \frac{d m}{m}
\tag{14}
$$



Notice that we already get rid of all the vectors and express everything in scalar. It might be tricky to get the sign correct. By integrating it, we find

$$
\int_{0}^{v_{f}} d v=-u \int_{10 m_{0}}^{m_{0}} \frac{d m}{m}
\tag{15}
$$

As a result, the final speed of the rocket is

$$
v_{f}=u \ln \frac{10 m_{0}}{m_{0}}=u \ln 10 \approx 2.302 u
\tag{16}
$$



Final answer: $u \ln 10$
Subfield: Mechanics

4. The fundamental rocket equation

In this problem, we will investigate the acceleration of rockets.

Rocket repulsion

In empty space, an accelerating rocket must "throw" something backward to gain speed from repulsion. Assume there is zero gravity.

The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out.
<img_4415>

Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$.

(Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.)
Context question:
(a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.)
<img_4337>

Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u.

Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ?
Context answer:
\boxed{$\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$}


Context question:
(b) Suppose that the rocket is stationary at time $t=0$. The empty rocket has mass $m_{0}$ and the rocket full of fuel has mass $9 m_{0}$. The engine is turned on at time $t=0$, and fuel is ejected at the rate $\mu$ and relative velocity $\mathbf{u}$ described previously. What will be the speed of the rocket when it runs out of fuel, in terms of $u=|\mathbf{u}|$ ?



(Hint: useful integral formula: $\int_{a}^{b} \frac{1}{x} d x=\ln \left(\frac{b}{a}\right)$, and $\ln (10) \approx 2.302, \ln (9) \approx 2.197$.)
Context answer:
$u \ln 10$


Question: (c) Discuss the factors that limit the final speed of the rocket.

Solution:
The ratio of the mass of the rocket it self (the spaceship and the fuel container) to the mass of the fuel it can carry.

The fuel eject speed $u$.
Subfield: Mechanics

4. The fundamental rocket equation

In this problem, we will investigate the acceleration of rockets.

Rocket repulsion

In empty space, an accelerating rocket must "throw" something backward to gain speed from repulsion. Assume there is zero gravity.

The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out.
<img_4415>

Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$.

(Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.)
Context question:
(a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.)
<img_4337>

Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u.

Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ?
Context answer:
\boxed{$\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$}


Context question:
(b) Suppose that the rocket is stationary at time $t=0$. The empty rocket has mass $m_{0}$ and the rocket full of fuel has mass $9 m_{0}$. The engine is turned on at time $t=0$, and fuel is ejected at the rate $\mu$ and relative velocity $\mathbf{u}$ described previously. What will be the speed of the rocket when it runs out of fuel, in terms of $u=|\mathbf{u}|$ ?



(Hint: useful integral formula: $\int_{a}^{b} \frac{1}{x} d x=\ln \left(\frac{b}{a}\right)$, and $\ln (10) \approx 2.302, \ln (9) \approx 2.197$.)
Context answer:
$u \ln 10$


Context question:
(c) Discuss the factors that limit the final speed of the rocket.
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Chemical rockets 

Nearly all rockets get energy from the chemical reaction of the fuel (burning with the oxidant) they carry. This is called a combustion reaction. For example, the combustion reaction for hydrogen is

$$
2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}
\tag{8}
$$

Question: (a) Suppose that the product of combustion is just one species of molecule $\left(\mathrm{H}_{2} \mathrm{O}\right.$, for example) with mass $m_{p}$ and average kinetic energy $E$. What is the upper limit of exhaust speed $u$ when these molecules are ejected from the rocket?

Solution:
$$
E=\frac{1}{2} m_{p} u^{2} \Longrightarrow u=\sqrt{\frac{2 E}{m_{p}}}
\tag{17}
$$

Final answer: $\sqrt{\frac{2 E}{m_{p}}}$
Subfield: Mechanics


Question: (b) To manufacture fast rockets, would you recommend using hydrogen as the fuel? Assume that we only use oxygen as oxidant. If so, please give your reasoning. If not, what fuel would you suggest and why? (There are no numerical values here, so please give the best reasoning based on your own knowledge. Any reasonable answer is acceptable.)

Solution:
Yes, I would recommend using (liquid) hydrogen. The reason is that the energy $E$ produced by any chemical reaction is about the same magnitude. Therefore, to get larger ejection speed of the fuel, we hope the product of the reaction to be as light a possible. Since we fix the oxidant to be oxygen, the lightest product we can get is water $\mathrm{H}_{2} \mathrm{O}$ from the burning of hydrogen.

Final answer: 开放性回答
Subfield: Modern Physics


Question: a - Determine the positions and widths of the principal maximum on the screen. The width is defined as the distance between the minima on either side of the maxima.

Solution:
a - Consider first the $\mathrm{x}$-direction. If waves coming from neighbouring slits (with separation $d_{1}$ ) traverse paths of lengths that differ by:

$$
\Delta_{1}=n_{1} \cdot \lambda
$$

where $\mathrm{n}_{1}$ is an integer, then a principal maximum occurs. The position on the screen (in the $\mathrm{x}$-direction) is:

$$
x_{n_{1}}=\frac{n_{1} \cdot \lambda \cdot L}{d_{1}}
$$

since $\mathrm{d}_{1}<<\mathrm{d}_{2}$.

The path difference between the middle slit and one of the slits at the edge is then:

$$
\Delta_{\left(\frac{N_{1}}{2}\right)}=\frac{N_{1}}{2} \cdot n_{1} \cdot \lambda
$$

If on the other hand this path difference is:

$$
\Delta_{\left(\frac{N_{1}}{2}\right)}=\frac{N_{1}}{2} \cdot n_{1} \cdot \lambda+\frac{\lambda}{2}
$$

then the first minimum, next to the principal maximum, occurs. The position of this minimum on the screen is given by:

$$
\begin{gathered}
x_{n_{1}}+\Delta x=\frac{\left(\frac{N_{1}}{2} \cdot n_{1} \cdot \lambda+\frac{\lambda}{2}\right) \cdot L}{\frac{N_{1}}{2} \cdot d_{1}}=\frac{n_{1} \cdot \lambda \cdot L}{d_{1}}+\frac{\lambda \cdot L}{N_{1} \cdot d_{1}} \\
\rightarrow \quad \Delta x=\frac{\lambda \cdot L}{N_{1} \cdot d_{1}}
\end{gathered}
$$

The width of the principal maximum is accordingly:

$$
2 . \Delta x=2 \cdot \frac{\lambda . L}{N_{1} \cdot d_{1}}
$$

A similar treatment can be made for the $y$-direction, in which there are $\mathrm{N}_{2}$ slits with separation $\mathrm{d}_{2}$. The positions and widths of the principal maximal are:

$$
\begin{gathered}
\left(x_{n_{1},} y_{n_{2}}\right)=\left(\frac{n_{1} \cdot \lambda \cdot L}{d_{1}}, \frac{n_{2} \cdot \lambda \cdot L}{d_{2}}\right) \\
2 \cdot \Delta x=2 \cdot \frac{\lambda \cdot L}{N_{1} \cdot d_{1}} ; \quad 2 \cdot \Delta y=2 \cdot \frac{\lambda \cdot L}{N_{2} \cdot d_{2}}
\end{gathered}
$$

Final answer: $(x_{n_{1},} y_{n_{2}})=(\frac{n_{1} \cdot \lambda \cdot L}{d_{1}}, \frac{n_{2} \cdot \lambda \cdot L}{d_{2}})$;
$2 \cdot \Delta x=2 \cdot \frac{\lambda \cdot L}{N_{1} \cdot d_{1}} ; \quad 2 \cdot \Delta y=2 \cdot \frac{\lambda \cdot L}{N_{2} \cdot d_{2}}$
Subfield: Modern Physics


Question: b - Calculate the position and width of the maxima as a function of the angle $\Theta$ (for small $\Theta)$.

Solution:
b - In the x-direction the beam 'sees' a grid with spacing a, so that in this direction we have:

$$
x_{n_{1}}=\frac{n_{1} \cdot \lambda \cdot L}{a} \quad \Delta x=2 \cdot \frac{\lambda \cdot L}{N_{0} \cdot a}
$$



In the y-direction, the beam 'sees' a grid with effective spacing a. $\cos (\Theta)$.

Analogously, we obtain:

$$
y_{n_{2}}=\frac{n_{2} \cdot \lambda \cdot L}{a \cdot \cos (\theta)} \quad \Delta y=2 \cdot \frac{\lambda \cdot L}{N_{0} \cdot a \cdot \cos (\theta)}
$$

In the $z$-direction, the beam 'sees' a grid with effective spacing a.sin $(\Theta)$. This gives rise to principal maxima with position and width:

$$
y_{n_{3}}^{\prime}=\frac{n_{3} \cdot \lambda \cdot L}{a \cdot \sin (\theta)} \quad \Delta y^{\prime}=2 \cdot \frac{\lambda \cdot L}{N_{1} \cdot a \cdot \sin (\theta)}
$$

This pattern is superimposed on the previous one. Since $\sin (\Theta)$ is very small, only the zeroth-order pattern will be seen, and it is very broad, since $\mathrm{N}_{1} \cdot \sin (\Theta)<<\mathrm{N}_{0}$. The diffraction pattern from a plane wave falling on a thin plate of a cubic crystal, at a small angle of incidence to the normal, will be almost identical to that from a two-dimensional grid.

Final answer: position is $(x_{n_{1}},y_{n_{2}},y_{n_{3}}^{\prime})=(\frac{n_{1} \lambda L}{a},\frac{n_{2} \lambda L}{a \cos (\theta)},\frac{n_{3} \lambda L}{a \sin (\theta)})$, and the width $\Delta x=2 \frac{\lambda L}{N_{0} a}$, $\Delta y=2 \frac{\lambda L}{N_{0} a \cos (\theta)}$, $\Delta y^{\prime}=2 \frac{\lambda L}{N_{1} a \sin (\theta)}$
Subfield: Modern Physics

X-ray Diffraction from a crystal.

We wish to study X-ray diffraction by a cubic crystal lattice. To do this we start with the diffraction of a plane, monochromatic wave that falls perpendicularly on a 2-dimensional grid that consists of $\mathrm{N}_{1} \times \mathrm{N}_{2}$ slits with separations $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$. The diffraction pattern is observed on a screen at a distance $L$ from the grid. The screen is parallel to the grid and $\mathrm{L}$ is much larger than $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$.
Context question:
a - Determine the positions and widths of the principal maximum on the screen. The width is defined as the distance between the minima on either side of the maxima.
Context answer:
$(x_{n_{1},} y_{n_{2}})=(\frac{n_{1} \cdot \lambda \cdot L}{d_{1}}, \frac{n_{2} \cdot \lambda \cdot L}{d_{2}})$;
$2 \cdot \Delta x=2 \cdot \frac{\lambda \cdot L}{N_{1} \cdot d_{1}} ; \quad 2 \cdot \Delta y=2 \cdot \frac{\lambda \cdot L}{N_{2} \cdot d_{2}}$


Extra Supplementary Reading Materials:

We consider now a cubic crystal, with lattice spacing a and size $\mathrm{N}_{0} \cdot a \times \mathrm{N}_{0} \cdot a \times \mathrm{N}_{1} \cdot a \cdot \mathrm{N}_{1}$ is much smaller than $\mathrm{N}_{0}$. The crystal is placed in a parallel X-ray beam along the z-axis at an angle $\Theta$ (see Fig. 1). The diffraction pattern is again observed on a screen at a great distance from the crystal.

<img_4346>

Figure 1 Diffraction of a parallel X-ray beam along the z-axis. The angle between the crystal and the $y$-axis is $\Theta$.
Context question:
b - Calculate the position and width of the maxima as a function of the angle $\Theta$ (for small $\Theta)$.
Context answer:
position is $(x_{n_{1}},y_{n_{2}},y_{n_{3}}^{\prime})=(\frac{n_{1} \lambda L}{a},\frac{n_{2} \lambda L}{a \cos (\theta)},\frac{n_{3} \lambda L}{a \sin (\theta)})$, and the width $\Delta x=2 \frac{\lambda L}{N_{0} a}$, $\Delta y=2 \frac{\lambda L}{N_{0} a \cos (\theta)}$, $\Delta y^{\prime}=2 \frac{\lambda L}{N_{1} a \sin (\theta)}$


Extra Supplementary Reading Materials:

The diffraction pattern can also be derived by means of Bragg's theory, in which it is assumed that the X-rays are reflected from atomic planes in the lattice. The diffraction pattern then arises from interference of these reflected rays with each other.

Question: c - Show that this so-called Bragg reflection yields the same conditions for the maxima as those that you found in $b$.

Solution:
c - In Bragg reflection, the path difference for constructive interference between neighbouring planes:

$$
\Delta=2 \cdot a \cdot \sin (\phi) \approx 2 \cdot a \cdot \phi=n \cdot \lambda \quad \rightarrow \quad \frac{x}{L} \approx 2 \cdot \phi \approx \frac{n \cdot \lambda}{a} \quad \rightarrow \quad x \approx \frac{n \cdot \lambda \cdot L}{a}
$$

Here $\phi$ is the angle of diffraction.

This is the same condition for a maximum as in section $b$.