Съдържание

IPhO 1987, Теория и експеримент — Задача 4. Refractive indices

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Всички материали от IPhO 1987

Problems of the 18th International Physics Olympiad (Jena, 1987) · 20 т.

Внимание: Бележка към темата

Several hand-made figures are missing from the source file (blank space is left in their place): the diagram of the focussing points referred to in problem 2, the LC-grid circuit of problem 3, and several solution figures (toroid coordinates, LC-grid unit, the prism set-ups of problem 4).

Увод

Gunnar Friege & Gunter Lind $^1$

Leibniz-Institute for Science Education (IPN) at the University of Kiel, Germany

Abstract

The 18th International Physics Olympiad took place in 1987 in the German Democratic Republic (GDR). This article contains the competition problems, their solutions and also a (rough) grading scheme.

Introduction

The 18th international Physics olympics in 1987 was the second International Physics Olympiad hosted by the German Democratic Republic (GDR) . The organisation was lead by the ministry for education and the problems were formulated by a group of professors of different universities. However, the main part of the work was done by the physics department of the university of Jena. The company Carl-Zeiss and a special scientific school in Jena were involved also.

In the competition three theoretical and one experimental problem had to be solved. The theoretical part was quite difficult. Only the first of the three problems (“ascending moist air”) had a medium level of difficulty. The points given in the markings were equal distributed. Therefore, there were lots of good but also lots of unsatisfying solutions. The other two theoretical problems were rather difficult. About half of the pupils even did not find an adequate start in solving these problems. The third problem (“infinite LC-grid") revealed quite a few complete solutions. The high level of difficulty can probably be explained with the fact that many pupils nearly had no experience with the subject. Concerning the second problem (“electrons in a magnetic field”) only a few pupils worked on the last part 3 (see below).

The experimental problem (“refracting indices”) was much more easier than the theoretical problems. There were lots of different possibilities of solution and most of the pupils managed to come up with partial or complete solutions. Over the half of all teams got more points in the experimental part than in the theoretical part of the competition.

The problems and their solutions are based on the original German and English versions of the competition problems. Only minor changes have been made. Despite the fact that nowadays almost all printed figures are generated with the aid of special computer programmes, the original hand-made figures are published here.

$^1$ Contact: Leibniz-Institute for Science Education (IPN) at the University of Kiel
Olshausenstrasse 62, 24098 Kiel, Germany ipho@ipn.uni-kiel.de

Условие

Find the refractive indices of a prism, $n_p$ , and a liquid, $n_l$ . Ignore dispersion.

a) Determine the refractive index $n_p$ of a single prism by two different experimental methods.

Illustrate your solution with accurate diagrams and deduce the relations necessary to calculate the refractive index. (One prism only should be used). [10 т.]

b) Use two identical prisms to determine the refractive index $n_L$ of a liquid with $n_L < n_p$ . Illustrate your solution with accurate diagrams and deduce the relations necessary to calculate the refractive index. [10 т.]

Apparatus:

Two identical prisms with angles of 30°, 60° and 90°; a set square, a glass dish, a round table, a liquid, sheets of graph paper, other sheets of paper and a pencil.

Formulae: $\sin(\alpha \pm \beta) = \sin\alpha \cdot \cos\beta \pm \cos\alpha \cdot \sin\beta$

Additional remarks: You may mark the opaque sides of the prisms with a pencil. The use of the lamp is optional.

Отговори

Решение

a) Calculation of the refractive index of the prism

First method:

Draw a straight line A – B on a sheet of paper and let this be your line of sight. Place the prism with its rectangular edge facing you onto the line (at point P on the line). Now turn the prism in the direction of the arrow until the dark edge of total reflection which can be seen in the short face of the prism coincides with the 90° edge of the prism. Mark a point M and measure the length $c_1$ . Measure also the length of the short face of the prism.

The following equations apply:

\sin\alpha_T = \frac{1}{n_p} \qquad (1)

\frac{\sin\alpha}{\sin\beta} = n_p \qquad (2)

\beta = 60^{\circ} - \alpha_T \qquad (3)

\gamma = 30^{\circ} + \alpha \qquad (4)

\frac{\sin\gamma}{\sin(90^{\circ} - \alpha)} = \frac{a}{c_1} \qquad (5)

From eq. (5) follows with eq. (4) and the given formulae:

\frac{a}{c_1} \cdot \cos\alpha = \sin(30^{\circ} + \alpha) = \frac{1}{2} \cdot \cos\alpha + \frac{1}{2} \cdot \sqrt{3} \cdot \sin\alpha

\sin\alpha = \frac{2a - c_1}{2 \cdot \sqrt{a^2 - a \cdot c_1 + c_1^{\,2}}} \qquad (6)

From eqs. (2), (3) and (1) follows:

\sin\alpha = n_p \cdot \sin(60^{\circ} - \alpha_T) = \frac{n_p}{2} \cdot \left(\sqrt{3} \cdot \cos\alpha_T - \sin\alpha_T\right)

n_p = +\left\{\frac{1}{3} \cdot (2 \cdot \sin\alpha + 1)^2 + 1\right\}^{1/2} \qquad (7)

When measuring $c_1$ and a one notices that within the error limits of ± 1 mm a equals $c_1$ .

Hence: $\sin\alpha = \frac{1}{2}$ and $n_p = 1.53$ . $\qquad (8)$

Second method:

Place edge C of the prism on edge A of a sheet of paper and look along the prism hypotenuse at edge A so that your direction of sight B-A and the table surface form an angle of 60°. Then shift the prism over the edge of the paper into the position shown, such that prism edge C can be seen inside the prism collinear with edge A of the paper outside the prism. The direction of sight must not be changed while the prism is being displaced.

The following equations apply:

\left.\begin{aligned} \tan\beta &= \frac{h}{c} \\ \tan 60^{\circ} &= \sqrt{3} = \frac{h}{b} \end{aligned}\right\} \Rightarrow h = b \cdot \sqrt{3} = \frac{c \cdot \sin\beta}{\sqrt{1 - \sin^2\beta}} \qquad (9)

\sin\beta = \sin 60^{\circ} \cdot \frac{1}{n_p} = \frac{\sqrt{3}}{2 \cdot n_p} \qquad (10)

n_p = \frac{1}{2} \cdot \sqrt{\left(\frac{c}{b}\right)^2 + 3} \qquad (11)

With the measured values c = 29 mm and b = 11.5 mm, it follows

n_p = 1.53. \qquad (12)

b) Determination of the refractive index of the liquid by means of two prisms

Place the two prisms into a glass dish filled with water as shown in the figure above. Some water will rise between the hypotenuse surfaces. By pressing and moving the prisms slightly against each other the water can be made to cover the whole surface. Look over the 60° edges of the prisms along a line of sight L (e.g. in the direction of a fixed point on an illuminated wall). Turn the glass dish together with the two prisms in such a way that the dark shadow of total reflection which can be seen in the short face of prism 1 coincides with the 60° edge of that prism (position shown in the figure below).

While turning the arrangement take care to keep the 60° edge (point K) on the line of sight L. In that position measure the length $b_1$ with a ruler (marking, reading). The figure below illustrates the position described.

If the refractive index of the prism is known (see part a) the refractive index of the liquid may be calculated as follows:

\sin\alpha = \frac{a}{\sqrt{a^2 + b_1^{\,2}}} \qquad (13)

\beta = \alpha - 30^{\circ}; \quad \gamma_1 = 30^{\circ} - \beta = 60^{\circ} - \alpha \qquad (14, 15)

$\frac{\sin\gamma_1}{\sin\gamma_2} = n_p$ refraction at the short face of prism 1. $\qquad (16)$

The angle of total reflection $\delta_t$ at the hypotenuse surface of prism 1 in the position described is:

\frac{\pi}{2} - \delta_t = 30^{\circ} - \gamma_2 \qquad (17)

\delta_t = 60^{\circ} + \arcsin\left(\frac{\sin\gamma_1}{n_p}\right) \qquad (18)

From this we can easily obtain $n_1$ :

n_1 = n_p \cdot \sin\delta_t = n_p \cdot \sin\left\{60^{\circ} + \arcsin\frac{\sin\gamma_1}{n_p}\right\} \qquad (19)

Numerical example for water as liquid:

$b_1 = 1.9\ \mathrm{cm}$ ; $\alpha = 55.84^{\circ}$ ; $\gamma_1 = 4.16^{\circ}$ ; $\delta_t = 62.77^{\circ}$ ; $a = 2.8\ \mathrm{cm}$ ; with $n_p = 1.5$ follows

n_1 = 1.33. \qquad (20)

Grading Scheme

Problem 4: Refractive indices
part a, first method	5
part a, second method	5
part b	10
	20

Оригинал в Архива: ipho87.pdf