Subfield: Mechanics

A The Extremum Principle in Mechanics

Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts).

<img_4336>

Figure 1
Context question:
A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$.
Context answer:
\boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$}


Context question:
A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$.
Context answer:
\boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$}


Extra Supplementary Reading Materials:

We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA).
Context question:
A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant <br> potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and <br> $\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have <br> coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the <br> expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates.
Context answer:
\boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$}


Extra Supplementary Reading Materials:

B The Extremum Principle in Optics

A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time.

<img_4508>

Figure 2
Context question:
B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle.
Context answer:
\boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$}


Extra Supplementary Reading Materials:

Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height.

<img_4478>

Figure 3: Tank of Sugar Solution
Context question:
B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$.
Context answer:
\boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$}


Context question:
B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0]

You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or <br> $\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant
Context answer:
\boxed{$x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$}


Context question:
B.4 Obtain the value of $x_{0}$, the point where the beam meets the bottom of the tank. Take $y_{0}=10.0 \mathrm{~cm}$, <br> $n_{0}=1.50, k=0.050 \mathrm{~cm}^{-1}\left(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\right)$.
Context answer:
\boxed{24.0}


Extra Supplementary Reading Materials:

C The Extremum Principle and the Wave Nature of Matter 

We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from $\mathrm{O}$ to $\mathrm{P}$ can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves.
Context question:
C.1 As the particle moves along its trajectory by an infinitesimal distance $\Delta s$, relate the change $\Delta \varphi$ in the phase <br> of its de Broglie wave to the change $\Delta A$ in the action and the Planck constant.
Context answer:
\boxed{$\Delta \phi=\frac{2 \pi \Delta A}{h}$}


Context question:
C.2 Recall the problem from part A where the particle traverses from $\mathrm{O}$ to $\mathrm{P}$ (see Fig. 4). Let an opaque partition be placed at the boundary $\mathrm{AB}$ between the two regions. There is a small opening $\mathrm{CD}$ of width $d$ in $\mathrm{AB}$ such that $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$.

Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta \phi_{C D}$ between the two paths to first order.

<img_4466>

Figure 4
Context answer:
\boxed{$\Delta \phi_{C D}=0$}


Extra Supplementary Reading Materials:

D Matter Wave Interference

Consider an electron gun at $\mathrm{O}$ which directs a collimated beam of electrons to a narrow slit at $\mathrm{F}$ in the opaque partition $\mathrm{A}_{1} \mathrm{~B}_{1}$ at $x=x_{1}$ such that OFP is a straight line. $\mathrm{P}$ is a point on the screen at $x=x_{0}$ (see Fig. 5). The speed in I is $v_{1}=2.0000 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$ and $\theta=10.0000^{\circ}$. The potential in II is such that speed $v_{2}=$ $1.9900 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$. The distance $x_{0}-x_{1}$ is $250.00 \mathrm{~mm}$ $\left(1 \mathrm{~mm}=10^{-3} \mathrm{~m}\right)$. Ignore electron-electron interaction.

<img_4338>

Figure 5
Context question:
D.1If the electrons at $\mathrm{O}$ have been accelerated from rest, calculate the accelerating potential $U_{1}$.
Context answer:
\boxed{$1.139 \times 10^{3}$}


Context question:
D.2Another identical slit $\mathrm{G}$ is made in the partition $\mathrm{A}_{1} \mathrm{~B}_{1}$ at a distance of $215.00 \mathrm{~nm}\left(1 \mathrm{~nm}=10^{-9} \mathrm{~m}\right)$ below slit <br> $\mathrm{F}$ (Fig. 5). If the phase difference between de Broglie waves arriving at $\mathrm{P}$ through the slits $\mathrm{F}$ and $\mathrm{G}$ is $2 \pi \beta$, calculate $\beta$.
Context answer:
\boxed{5.13}


Question: D.3 What is the smallest distance $\Delta y$ from $\mathrm{P}$ at which null (zero) electron detection maybe expected on the <br> screen? [Note: you may find the approximation $\sin (\theta+\Delta \theta) \approx \sin \theta+\Delta \theta \cos \theta$ useful]

Solution:
<img_4379>

From previous part for null (zero) electron detection $\Delta \phi=5.5 \times 2 \pi$

$$
\begin{aligned}
& \therefore m v_{1} \frac{d \sin \theta}{h}-\frac{m v_{2} d \sin (\theta+\Delta \theta)}{h}=5.5 \\
& \sin (\theta+\Delta \theta)=\frac{\frac{m v_{1} d \sin \theta}{h}-5.5}{\frac{m v_{2} d}{h}} \\
& =\frac{v_{1}}{v_{2}} \sin \theta-\frac{h}{m} \frac{5.5}{v_{2} d} \\
& =\frac{2}{1.99} \sin 10^{\circ}-\frac{5.5}{1374.78 \times 1.99 \times 10^{7} \times \times 2.15 \times 10^{-7}} \\
& =0.174521-0.000935
\end{aligned}
$$

This yields $\Delta \theta=-0.0036^{\circ}$

The closest distance to $\mathrm{P}$ is

$$
\begin{aligned}
\Delta y & =\left(x_{0}-x_{1}\right)(\tan (\theta+\Delta \theta)-\tan \theta) \\
& =250(\tan 9.9964-\tan 10) \\
& =-0.0162 m m \\
& =-16.2 \mu \mathrm{m}
\end{aligned}
$$

The negative sign means that the closest minimum is below $\mathrm{P}$.

Approximate Calculation for $\theta$ and $\Delta y$

Using the approximation $\sin (\theta+\Delta \theta) \approx \sin \theta+\Delta \theta \cos \theta$

The phase difference of $5.5 \times 2 \pi$ gives

$$
m v_{1} \frac{d \sin 10^{\circ}}{h}-m v_{2} \frac{d\left(\sin 10^{\circ}+\Delta \theta \cos 10^{\circ}\right)}{h}=5.5
$$

From solution of the previous part

$$
m v_{1} \frac{d \sin 10^{\circ}}{h}-m v_{2} \frac{d \sin 10^{\circ}}{h}=5.13
$$



Therefore

$$
m v_{2} \frac{d \Delta \theta \cos 10^{\circ}}{h}=0.3700
$$

This yields $\Delta \theta \approx 0.0036^{\circ}$

$\Delta y=-0.0162 \mathrm{~mm}=-16.2 \mu \mathrm{m}$ as before

Final answer: $-16.2 $ $\mu \mathrm{m}$