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IPhO 2015, Теория — Задача 2. The Extremum Principle

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46th International Physics Olympiad 2015 Mumbai - India — Q T-2 — The Extremum Principle · 10 т.

Условие

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 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 O with speed $v_1$ along a line making an angle $\theta_1$ with the x-axis. It reaches point 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).

¹ Manoj Harbola (IIT-Kanpur) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.

A1 Obtain an expression for $v_2$ in terms of $m$ , $v_1$ and $V_0$ . [0,2 т.]

A2 Express $v_2$ in terms of $v_1, \theta_1$ and $\theta_2$ . [0,3 т.]

The x-y plane with origin O. Region I (shaded, V = 0) lies between the y-axis and the vertical dotted line AB at x = x₁; region II (V = V₀) lies to the right. An arrow from O makes angle θ₁ with the x-axis; in region II an arrow towards P makes angle θ₂ with the horizontal. A point marked × on AB is at height α above B, shown by a double arrow. — Figure 1

We define a quantity called action $A = m\int v(s)ds$ , where $ds$ 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 $2\pi mRv$ . 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).

A3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and $(x_0, y_0)$ respectively and the boundary point where the particle transits from region I to region II have coordinates $(x_1, \alpha)$ . Note that $x_1$ is fixed and the action depends on the coordinate $\alpha$ only. State the expression for the action $A(\alpha)$ . Use PLA to obtain the relationship between $v_1/v_2$ and these coordinates. [1 т.]

B The Extremum Principle in Optics

Refraction of a light ray from O: medium I (shaded, n₁) below a horizontal dotted boundary, medium II (n₂) above. The ray crosses the boundary at x = α, making angle i₁ with the vertical dashed normal in medium I and i₂ in medium II, and reaches P at (x₀, y₀). Axes x and y. — Figure 2

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.

B1 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. [0,5 т.]

A tank seen from the side: a horizontal laser beam enters at the origin (0,0) at the top-left, the y-axis points up and a dashed x-axis runs horizontally; the beam curves downward and meets the bottom of the tank at (x₀, −y₀). — Figure 3: Tank of Sugar Solution

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.

B2 Assume that the refractive index $n(y)$ depends only on $y$ . Use the equation obtained in B1 to obtain the expression for the slope $dy/dx$ of the beam’s path in terms of refractive index $n_0$ at $y = 0$ and $n(y)$ . [1,5 т.]

B3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_0$ from the bottom of the tank as shown in figure 3. Take $n(y) = n_0 - ky$ where $n_0$ and $k$ are positive constants. Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.

You may use: $\int \sec\theta d\theta = \ln(\sec\theta + \tan\theta) + \mathrm{constant}$ , where $\sec\theta = 1/\cos\theta$ or

\int \frac{dx}{\sqrt{x^2-1}} = \ln\left(x + \sqrt{x^2-1}\right) + \mathrm{constant}

[1,2 т.]

B4 Obtain the value of $x_0$ , the point where the beam meets the bottom of the tank. Take $y_0 = 10.0$ cm, $n_0 = 1.50$ , $k = 0.050\ \mathrm{cm}^{-1}$ (1 cm = $10^{-2}$ m). [0,8 т.]

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.

C1 As the particle moves along its trajectory by an infinitesimal distance $\Delta s$ , relate the change $\Delta\varphi$ in the phase of its de Broglie wave to the change $\Delta A$ in the action and the Planck constant. [0,6 т.]

C2 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 AB between the two regions. There is a small opening CD of width $d$ in AB such that $d \ll (x_0 - x_1)$ and $d \ll x_1$ .

Region I (V = 0) and region II (V = V₀) separated by the partition AB at x = x₁ with an opening CD of width d (CD=d). Two paths from O: OCP through C and ODP through D, both reaching P at (x₀,y₀). Axes x and y. — Figure 4

Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta\varphi_{\mathrm{CD}}$ between the two paths to first order. [1,2 т.]

D Matter Wave Interference

Consider an electron gun at $\mathrm{O}$ which directs a collimated beam of electrons to a narrow slit at F in the opaque partition $\mathrm{A_1B_1}$ at $x = x_1$ such that OFP is a straight line. P is a point on the screen at $x = x_0$ (see Fig. 5). The speed in I is $v_1 = 2.0000 \times 10^{7}\ \mathrm{m\,s^{-1}}$ and $\theta = 10.0000^{\circ}$ . The potential in II is such that speed $v_2 = 1.9900 \times 10^{7}\ \mathrm{m\,s^{-1}}$ . The distance $x_0 - x_1$ is $250.00\ \mathrm{mm}$ $(1\mathrm{mm} = 10^{-3}\mathrm{m})$ . Ignore electron-electron interaction.

Electron beam arrows from O at 10° to the x-axis towards the opaque partition A₁B₁ at x = x₁, with slit F and a second slit G below it (separation 215.00 nm marked by a double arrow). A dashed straight line OFP continues to point P on the dotted screen at x = x₀; regions I and II on either side of the partition; 250 mm between x₁ and x₀. — Figure 5

D1 If the electrons at $\mathrm{O}$ have been accelerated from rest, calculate the accelerating potential $U_1$ . [0,3 т.]

D2 Another identical slit G is made in the partition $\mathrm{A_1B_1}$ at a distance of $215.00\ \mathrm{nm}$ $(1\mathrm{nm} = 10^{-9}\mathrm{m})$ below slit F (Fig. 5). If the phase difference between de Broglie waves arriving at P through the slits F and G is $2\pi\beta$ , calculate $\beta$ . [0,8 т.]

D3 What is the smallest distance $\Delta y$ from P at which null (zero) electron detection maybe expected on the screen? [Note: you may find the approximation $\sin(\theta + \Delta\theta) \approx \sin\theta + \Delta\theta\cos\theta$ useful] [1,2 т.]

D4 The beam has a square cross section of $500\mathrm{nm} \times 500\mathrm{nm}$ and the setup is 2 m long. What should be the minimum flux density $I_{min}$ (number of electrons per unit normal area per unit time) if, on an average, there is at least one electron in the setup at a given time? [0,4 т.]

Отговори

Решение

(A1) Solution:

From the principle of Conservation of Mechanical Energy

\frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + V_0

v_2 = (v_1^2 - \frac{2V_0}{m})^{1/2}

(A2) Solution:

At the boundary there is an impulsive force ( $\propto dV/dx$ ) in the $-x$ direction. Hence only the velocity component in the $x-$ direction $v_{1x}$ suffers change . The component in the $y-$ direction remains unchanged. Therefore

$v_{1y} = v_{2y}$

$v_1\sin\theta_1 = v_2\sin\theta_2$

(A3) Solution:

By definition $A(\alpha)$ from $O$ to $P$ is

A(\alpha) = mv_1\sqrt{x_1^2 + \alpha^2} + mv_2\sqrt{(x_0 - x_1)^2 + (y_0 - \alpha)^2}

Differentiating w.r.t. $\alpha$ and setting the derivative of $A(\alpha)$ to zero

\frac{v_1\alpha}{(x_1^2 + \alpha^2)^{1/2}} - \frac{v_2(y_0 - \alpha)}{[(x_0 - x_1)^2 + (y_0 - \alpha)^2]^{1/2}} = 0

\therefore \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}}

Note this is the same as A2, namely $v_1\sin\theta_1 = v_2\sin\theta_2$ .

(B1) Solution:

The speed of light in medium I is $c/n_1$ and in medium II is $c/n_2$ , where $c$ is the speed of light in vacuum. Let the two media be separated by the fixed line $y = y_1$ . Then time $T(\alpha)$ for light to travel from origin (0,0) and ( $x_0$ , $y_0$ ) is

T(\alpha) = n_1(\sqrt{y_1^2 + \alpha^2})/c + n_2(\sqrt{(x_0 - \alpha)^2 + (y_0 - y_1)^2})/c

Differentiating w.r.t. $\alpha$ and setting the derivative of $T(\alpha)$ to zero

\frac{n_1\alpha}{(y_1^2 + \alpha^2)^{1/2}} - \frac{n_2(y_0 - \alpha)}{[(x_0 - \alpha)^2 + (y_0 - y_1)^2]^{1/2}} = 0

\therefore \quad n_1\sin i_1 = n_2\sin i_2

[Note: Derivation is similar to A3. This is Snell’s law.]

(B2) Solution:

From Snell’s law $n_0\sin i_0 = n(y)\sin i$

Then, $\dfrac{dy}{dx} = -\cot i$

n_0\sin i_0 = \frac{n(y)}{\sqrt{1 + (\frac{dy}{dx})^2}}

\frac{dy}{dx} = -\sqrt{\left(\frac{n(y)}{n_0\sin i_0}\right)^2 - 1}

(B3) Solution:

\int \frac{dy}{\sqrt{(\frac{n_0 - ky}{n_0\sin\ i_0})^2 - 1}} = -\int dx

Note $i_0 = 90^{o}$ so $\sin i_0 = 1$ .

Method I We employ the substitution

\xi = \frac{n_0 - ky}{n_0}

\int \frac{d\xi(-\frac{n_0}{k})}{\sqrt{\xi^2 - 1}} = -\int dx

Let $\xi = \sec\theta$ . Then

\frac{n_0}{k}\ln(\sec\theta + \tan\theta) = x + c

Or METHOD II

We employ the substitution

\xi = \frac{n_0 - ky}{n_0}

\int \frac{d\xi(-\frac{n_0}{k})}{\sqrt{\xi^2 - 1}} = -\int dx

-\frac{n_0}{k}\ln\left(\frac{n_0 - ky}{n_0} + \sqrt{(\frac{n_0 - ky}{n_0})^2 - 1}\right) = -x + c

Now continuing

Considering the substitutions and boundary condition, $x = 0$ for $y = 0$ we obtain that the constant $c = 0$ .

Hence we obtain the following trajectory:

x = \frac{n_0}{k}\ln\left(\frac{n_0 - ky}{n_0} + \sqrt{(\frac{n_0 - ky}{n_0})^2 - 1}\right)

(B4) Solution:

Given $y_0 = 10.0$ cm. $\quad n_0 = 1.50 \quad k = 0.050\ \mathrm{cm}^{-1}$

From (B3)

x_0 = \frac{n_0}{k}\ln\left[\left(\frac{n_0 - ky}{n_0}\right) + \left(\left(\frac{n_0 - ky}{n_0}\right)^2 - 1\right)^{1/2}\right]

Here $y = -y_0$

x_0 = \frac{n_0}{k}\ln\left[\frac{(n_0 + ky_0)}{n_0} + \left(\frac{(n_0 + ky_0)^2}{n_0^2} - 1\right)^{1/2}\right]

= 30\ln\left[\frac{2}{1.5} + \left(\left(\frac{2}{1.5}\right)^2 - 1\right)^{1/2}\right]

= 30\ln\left[\frac{4}{3} + \left(\frac{7}{9}\right)^{1/2}\right]

= 30\ln\left[\frac{4}{3} + 0.88\right]

= 24.0\ \mathrm{cm}

(C1) Solution:

From the de Broglie hypothesis

\lambda \to \lambda_{dB} = h/mv

where $\lambda$ is the de Broglie wavelength and the other symbols have their usual meaning

\Delta\phi = \frac{2\pi}{\lambda}\Delta s

= \frac{2\pi}{h}mv\Delta s

= \frac{2\pi\Delta A}{h}

Diagram for C2: y-axis and x-axis from O, partition A–B at x₁ separating regions I and II. Paths O→C→P and O→D→P, with C above B at the partition and D above C. Perpendicular foot E on OD from C and F on CP from D, marked with right angles; θ₁ at O and θ₂ at C measured from a dashed horizontal.

(C2) Solution:

Consider the extreme trajectories $OCP$ and $ODP$ of (C1)

The geometrical path difference is $ED$ in region I and $CF$ in region II.

This implies (note: $d \ll (x_0 - x_1)$ and $d \ll x_1$ )

\Delta\phi_{CD} = \frac{2\pi d\sin\theta_1}{\lambda_1} - \frac{2\pi d\sin\theta_2}{\lambda_2}

\Delta\phi_{CD} = \frac{2\pi mv_1 d\sin\theta_1}{h} - \frac{2\pi mv_2 d\sin\theta_2}{h}

= 2\pi\frac{md}{h}(v_1\sin\theta_1 - v_2\sin\theta_2)

= 0 \qquad \text{(from A2 or B1)}

Thus near the classical path there is invariably constructive interference.

(D1) Solution:

qU_1 = \frac{1}{2}\,mv^2

= \frac{9.11 \times 10^{-31} \times 4 \times 10^{14}}{2}J

= 2 \times 9.11 \times 10^{-17}J

= \frac{2 \times 9.11 \times 10^{-17}}{1.6 \times 10^{-19}}\ eV

= 1.139 \times 10^{3}\ eV \quad (\simeq 1100\ eV)

U_1 = 1.139 \times 10^{3}\ V

(D2) Solution: Phase difference at $P$ is

\Delta\phi_P = \frac{2\pi d\sin\theta}{\lambda_1} - \frac{2\pi d\sin\theta}{\lambda_2}

= 2\pi(v_1 - v_2)\frac{md}{h}\sin 10^{\circ} = 2\pi\beta

\beta = 5.13

Diagram for D3: y-axis and x-axis from O, partition A₁–B₁ at x₁ separating regions I and II, slits F and G on it separated by 215 nm (double arrow). Lines O→F→P and O→G→P with perpendiculars drawn to show the path differences.

(D3) Solution:

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

\therefore mv_1\frac{d\sin\theta}{h} - \frac{mv_2 d\sin(\theta + \Delta\theta)}{h} = 5.5

sin(\theta + \Delta\theta) = \frac{\frac{mv_1 d\sin\theta}{h} - 5.5}{\frac{mv_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

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

The closest distance to P is

\Delta y = (x_0 - x_1)(\tan(\theta + \Delta\theta) - \tan\theta)

= 250(\tan 9.9964 - \tan 10)

= -0.0162mm

= -16.2\mu\mathrm{m}

The negative sign means that the closest minimum is below 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

mv_1\frac{d\sin 10^{\circ}}{h} - mv_2\frac{d(\sin 10^{\circ} + \Delta\theta\cos 10^{\circ})}{h} = 5.5

From solution of the previous part

mv_1\frac{d\sin 10^{\circ}}{h} - mv_2\frac{dsin10^{\circ}}{h} = 5.13

Therefore

mv_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

(D4) Solution: The product of the speed of the electrons and number of electron per unit volume on an average yields the intensity.

Thus $N = 1 =$ Intensity $\times$ Area $\times$ Length/ Electron Speed

$= I_{min} \times 0.25 \times 10^{-12} \times 2/2 \times 10^{7}$

This gives $I_{min} = 4\times 10^{19}\ \mathrm{m^{-2}\ s^{-1}}$

R. Bach, D. Pope, Sy-H Liou and H. Batelaan, New J. of Physics Vol. 15, 033018 (2013).

Оригинал в Архива: 2.pdf · официални решения: 2_sol.pdf