Съдържание

Условие Physical situation Geometry Optical path length Experimental data Отговори Решение

IPhO 2006, Теория — Задача 1. Gravity in a Neutron Interferometer

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Всички материали от IPhO 2006

Theory Question 1: Gravity in a Neutron Interferometer · 8 юли 2006 г. · 10 т.

Внимание: Бележка към темата

The original PDF and Word editions contain the same single problem, seven connected requests, two interferometer diagrams and three supplied Answer Script pages. The printed answer form divides the 1.0 point for 1.4 into 0.8 for the expression and 0.2 for the numerical volume. Its blank response spaces are retained as unfilled form entries; they are not additional tasks. Native spelling “Collela” and “which tell us that” are retained.

Условие

Enter all your answers into the Answer Script.

Symbolic diagram of the interferometer: the incoming beam IN hits the left beam splitter BS, splits into two paths at angle 2θ that reflect off mirrors M at the top and bottom and recombine at the right beam splitter BS, from which beams go to OUT2 (upper) and OUT1 (lower). Dotted vertical lines mark horizontal distances a, a above and below. Legend: BS - Beam Splitters, M - Mirror. — Figure 1a

Perspective sketch of the interferometer tilted around the axis of the incoming beam IN by angle φ (marked with dotted lines and a right-angle mark); the diamond is lifted out of the horizontal plane and the outgoing beams OUT1 and OUT2 are shown. — Figure 1b

Physical situation

We consider the situation of the famous neutron-interferometer experiment by Collela, Overhauser and Werner, but idealize the set-up inasmuch as we shall assume perfect beam splitters and mirrors within the interferometer. The experiment studies the effect of the gravitational pull on the de Broglie waves of neutrons.

The symbolic representation of this interferometer in analogy to an optical interferometer is shown in Figure 1a. The neutrons enter the interferometer through the IN port and follow the two paths shown. The neutrons are detected at either one of the two output ports, OUT1 or OUT2. The two paths enclose a diamond-shaped area, which is typically a few $\mathrm{cm}^2$ in size.

The neutron de Broglie waves (of typical wavelength of $10^{-10}\ \mathrm{m}$ ) interfere such that all neutrons emerge from the output port OUT1 if the interferometer plane is horizontal. But when the interferometer is tilted around the axis of the incoming neutron beam by angle $\phi$ (Figure 1b), one observes a $\phi$ dependent redistribution of the neutrons between the two output ports OUT1 and OUT2.

Geometry

For $\phi = 0^{\circ}$ the interferometer plane is horizontal; for $\phi = 90^{\circ}$ the plane is vertical with the output ports above the tilt axis.

1.1 (1.0) How large is the diamond-shaped area $A$ enclosed by the two paths of the interferometer? [1 т.]

1.2 (1.0) What is the height $H$ of output port OUT1 above the horizontal plane of the tilt axis? [1 т.]

Express $A$ and $H$ in terms of $a$ , $\theta$ , and $\phi$ .

Optical path length

The optical path length $N_{\mathrm{opt}}$ (a number) is the ratio of the geometrical path length (a distance) and the wavelength $\lambda$ . If $\lambda$ changes along the path, $N_{\mathrm{opt}}$ is obtained by integrating $\lambda^{-1}$ along the path.

1.3 (3.0) What is the difference $\Delta N_{\mathrm{opt}}$ in the optical path lengths of the two paths when the interferometer has been tilted by angle $\phi$ ? Express your answer in terms of $a$ , $\theta$ , and $\phi$ as well as the neutron mass $M$ , the de Broglie wavelength $\lambda_0$ of the incoming neutrons, the gravitational acceleration $g$ , and Planck’s constant $h$ . [3 т.]

1.4 (1.0) Introduce the volume parameter

V = \frac{h^2}{gM^2}

and express $\Delta N_{\mathrm{opt}}$ solely in terms of $A$ , $V$ , $\lambda_0$ , and $\phi$ . State the value of $V$ for $M = 1.675\times10^{-27}\,\mathrm{kg}$ , $g = 9.800\ \mathrm{m\,s^{-2}}$ , and $h = 6.626\times10^{-34}\ \mathrm{J\,s}$ . [1 т.]

1.5 (2.0) How many cycles — from high intensity to low intensity and back to high intensity — are completed by output port OUT1 when $\phi$ is increased from $\phi = -90^{\circ}$ to $\phi = 90^{\circ}$ ? [2 т.]

Experimental data

The interferometer of an actual experiment was characterized by $a = 3.600\ \mathrm{cm}$ and $\theta = 22.10^{\circ}$ , and 19.00 full cycles were observed.

1.6 (1.0) How large was $\lambda_0$ in this experiment? [1 т.]

1.7 (1.0) If one observed 30.00 full cycles in another experiment of the same kind that uses neutrons with $\lambda_0 = 0.2000\ \mathrm{nm}$ , how large would be the area $A$ ? [1 т.]

Hint: If $|\alpha x| \ll 1$ , it is permissible to replace $(1+x)^{\alpha}$ by $1+\alpha x$ .

Отговори

Решение

Geometry

Each side of the diamond has length $L = \dfrac{a}{\cos\theta}$ and the distance between parallel sides is $D = \dfrac{a}{\cos\theta}\sin(2\theta) = 2a\sin\theta$ . The area is the product thereof, $A = LD$ , giving

1.1

A = 2a^2\tan\theta\,.

The height $H$ by which a tilt of $\phi$ lifts OUT1 above IN is $H = D\sin\phi$ or

1.2

H = 2a\sin\theta\,\sin\phi\,.

Optical path length

Only the two parallel lines for IN and OUT1 matter, each having length $L$ . With the de Broglie wavelength $\lambda_0$ on the IN side and $\lambda_1$ on the OUT1 side, we have

\Delta N_{\mathrm{opt}} = \frac{L}{\lambda_0} - \frac{L}{\lambda_1} = \frac{a}{\lambda_0\cos\theta}\left(1 - \frac{\lambda_0}{\lambda_1}\right).

The momentum is $h/\lambda_0$ or $h/\lambda_1$ , respectively, and the statement of energy conservation reads

\frac{1}{2M}\left(\frac{h}{\lambda_0}\right)^2 = \frac{1}{2M}\left(\frac{h}{\lambda_1}\right)^2 + MgH\,,

which implies

\frac{\lambda_0}{\lambda_1} = \sqrt{1 - 2\frac{gM^2}{h^2}\lambda_0^2 H}\,.

Upon recognizing that $(gM^2/h^2)\lambda_0^2 H$ is of the order of $10^{-7}$ , this simplifies to

\frac{\lambda_0}{\lambda_1} = 1 - \frac{gM^2}{h^2}\lambda_0^2 H\,,

and we get

\Delta N_{\mathrm{opt}} = \frac{a}{\lambda_0\cos\theta}\frac{gM^2}{h^2}\lambda_0^2 H

1.3

\Delta N_{\mathrm{opt}} = 2\frac{gM^2}{h^2}a^2\lambda_0\tan\theta\,\sin\phi\,.

A more compact way of writing this is

1.4

\Delta N_{\mathrm{opt}} = \frac{\lambda_0 A}{V}\sin\phi\,,

where

1.4

V = 0.1597\times10^{-13}\ \mathrm{m^3} = 0.1597\ \mathrm{nm\,cm^2}

is the numerical value for the volume parameter $V$ .

There is constructive interference (high intensity in OUT1) when the optical path lengths of the two paths differ by an integer, $\Delta N_{\mathrm{opt}} = 0, \pm1, \pm2, \ldots$ , and we have destructive interference (low intensity in OUT1) when they differ by an integer plus half, $\Delta N_{\mathrm{opt}} = \pm\frac{1}{2}, \pm\frac{3}{2}, \pm\frac{5}{2}, \ldots$ . Changing $\phi$ from $\phi = -90^{\circ}$ to $\phi = 90^{\circ}$ gives

\Delta N_{\mathrm{opt}}\Big|_{\phi=-90^{\circ}}^{\phi=90^{\circ}} = \frac{2\lambda_0 A}{V}\,,

which tell us that

1.5

\sharp\ \text{of cycles} = \frac{2\lambda_0 A}{V}\,.

Experimental data

For $a = 3.6\ \mathrm{cm}$ and $\theta = 22.1^{\circ}$ we have $A = 10.53\ \mathrm{cm}^2$ , so that

1.6

\lambda_0 = \frac{19\times0.1597}{2\times10.53}\ \mathrm{nm} = 0.1441\ \mathrm{nm}\,.

And 30 full cycles for $\lambda_0 = 0.2\ \mathrm{nm}$ correspond to an area

1.7

A = \frac{30\times0.1597}{2\times0.2}\ \mathrm{cm}^2 = 11.98\ \mathrm{cm}^2\,.

Answer Script — Geometry

Country Code · Student Code · Question Number: 1

For Examiners Use Only

1.1	The area is $A =$	1.0
1.2	The height is $H =$	1.0

Answer Script — Optical path length

Country Code · Student Code · Question Number: 1

For Examiners Use Only

1.3	In terms of $a$ , $\theta$ , $\phi$ , $M$ , $\lambda_0$ , $g$ , and $h$ : $\Delta N_{\mathrm{opt}} =$	3.0
1.4	In terms of $A$ , $V$ , $\lambda_0$ , and $\phi$ : $\Delta N_{\mathrm{opt}} =$	0.8
	The numerical value of $V$ is $V =$	0.2
1.5	The number of cycles is $\#\text{ of cycles} =$	2.0

Answer Script — Experimental data

Country Code · Student Code · Question Number: 1

For Examiners Use Only

1.6	The de Broglie wavelength was $\lambda_0 =$	1.0
1.7	The area is $A =$	1.0

Оригинал в Архива: IPhO_2006_Theo_question 1.pdf · официални решения: 1_sol.pdf · допълнителни данни: 1.doc