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IPhO 1975, Теория и експеримент — Задача 2. Thick lens

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Problems of the 8th International Physics Olympiad (Güstrow, 1975) · 7 юли 1975 г. · 10 т.

Увод

Gunnar Friege $^1$ & Gunter Lind

Introduction

The 8th International Physics Olympiad took place from the 7.7. to the 12.7. 1975 in Güstrow, in the German Democratic Republic (GDR). Altogether, 9 countries with 45 pupils participated. The teams came from Bulgaria, the German Democratic Republic, the Federal Republic of Germany (FRG), France, Poland, Rumania, Tchechoslowakia, Hungary and the USSR. The entire event took place in the pedagogic academy of Güstrow. Pupils and leaders were accommodated inside the university academy complex. On the schedule there was the competition and receptions as well as excursions to Schwerin, Rostock, and Berlin were offered. The delegation of the FRG reported of a very good organisation of the olympiad.

The problems and solutions of the 8th International Physics Olympiad were created by a commission of university physics professors and lecturers. The same commission set marking schemes and conducted the correction of the tests. The correction was carried out very quickly and was considered as righteous and, in cases of doubt, as very generous.

The main competition consisted of a 5 hour test in theory and a 4.5 hour experimental test. The time for the theoretical part was rather short and for the experimental part rather long. The problems originated from central areas of classical physics. The theoretical problems were relatively difficult, although solvable with good physics knowledge taught at school. The level of difficulty of the experimental problem was adequate. There were no additional devices necessary for the solution of the problems. Only basic formula knowledge was requested, and could be demanded from all pupils. Critics were only uttered concerning the second theoretical problem (thick lens). This problem requested relatively little physical understanding, but tested the mathematical skills and the routine in approaching problems (e.g. correct distinction of cases). However, it is also difficult to find substantial physics problems in the area of geometrical optics.

Altogether 50 points were the maximum to achieve; 30 in the theoretical test and 20 in the experimental test. The best contestant came from the USSR and had 43 points. The first prize (gold medal) was awarded with 39 points, the second prize (silver medal) with 34 points, the third prize (bronze medal) with 28 points and the fourth prize (honourable mention) with 22 points. Among the 45 contestants, 7 I. prizes, 9 II. prizes, 12 III. prizes and 8 IV. prizes were awarded, meaning that 80 % of all contestants were awarded.

The following problem descriptions and solution are based mainly on a translation of the original German version from 1975. Because the original drafts are not well preserved, some new sketches were drawn. We also gave the problems headlines and the solutions are in more detail.

$^1$ Remark: This article was written due to the special request to us by Dr. W. Gorzkowski, in order to close one of the last few gaps in the IPhO-report collection.
Contact: Dr. Gunnar Friege, Leibniz-Institute for Science Education (IPN) at the University of Kiel, Olshausenstr. 62, 24098 Kiel, Germany, friege@ipn.uni-kiel.de

Условие

Theoretical problem 2: “Thick lens”

The focal length f of a thick glass lens in air with refractive index $n$ , radius curvatures $r_1$ , $r_2$ and vertex distance $d$ (see figure) is given by:

f = \frac{n r_1 r_2}{(n-1)\left[n(r_2 - r_1) + d(n-1)\right]}

Thick biconvex lens (hatched, labelled n) on the optical axis with vertices S1 and S2, thickness d between them; the centre of curvature M2 lies far to the left and M1 far to the right; dimension arrows show r2 < 0 from S2 to M2 and r1 > 0 from S1 to M1.

Remark: $r_i > 0$ means that the central curvature point $\mathrm{M_i}$ is on the right side of the aerial vertex $\mathrm{S_i}$ , $r_i < 0$ means that the central curvature point $\mathrm{M_i}$ is on the left side of the aerial vertex $\mathrm{S_i}$ (i = 1,2).

For some special applications it is required, that the focal length is independent from the wavelength.

a) For how many different wavelengths can the same focal length be achieved? [1 т.]

b) Describe a relation between $r_i$ (i = 1,2), $d$ and the refractive index $n$ for which the required wavelength independence can be fulfilled and discuss this relation.

Sketch possible shapes of lenses and mark the central curvature points $\mathrm{M_1}$ and $\mathrm{M_2}$ . [6 т.]

c) Prove that for a given planconvex lens a specific focal length can be achieved by only one wavelength. [1 т.]

d) State possible parameters of the thick lens for two further cases in which a certain focal length can be realized for one wavelength only. Take into account the physical and the geometrical circumstances. [2 т.]

Отговори

Решение

Solution of problem 2:

a) The refractive index $n$ is a function of the wavelength $\lambda$ , i.e. $n = n(\lambda)$ . According to the given formula for the focal length $f$ (see above) which for a given f yields to an equation quadratic in $n$ there are at most two different wavelengths (indices of refraction) for the same focal length.

b) If the focal length is the same for two different wavelengths, then the equation

f(\lambda_1) = f(\lambda_2) \quad \textit{or} \quad f(n_1) = f(n_2) \qquad (1)

holds. Using the given equation for the focal length it follows from equation (1):

\frac{n_1 r_1 r_2}{(n_1 - 1)\left[n_1(r_2 - r_1) + d(n_1 - 1)\right]} = \frac{n_2 r_1 r_2}{(n_2 - 1)\left[n_2(r_2 - r_1) + d(n_2 - 1)\right]}

Algebraic calculations lead to:

\mathrm{r_1} - \mathrm{r_2} = \mathrm{d}\cdot\left(1 - \frac{1}{\mathrm{n_1 n_2}}\right) \qquad (2).

If the values of the radii $r_1$ , $r_2$ and the thickness satisfy this condition the focal length will be the same for two wavelengths (indices of refraction). The parameters in this equation are subject to some physical restrictions: The indices of refraction are greater than 1 and the thickness of the lens is greater than 0 m. Therefore, from equation (2) the relation

d > r_1 - r_2 > 0 \qquad (3)

is obtained.

The following table shows a discussion of different cases:

$r_1$	$r_2$	condition	centre of curvature
$r_1 > 0$	$r_2 > 0$	$0 < r_1 - r_2 < d$ or $r_2 < r_1 < d + r_2$	$M_2$ is always right of $M_1$ . $\overline{M_1 M_2} < \overline{S_1 S_2}$
$r_1 > 0$	$r_2 < 0$	$r_1 + \lvert r_2 \rvert < \mathrm{d}$	Order of points: $S_1 M_1 M_2 S_2$
$r_1 < 0$	$r_2 > 0$	never fulfilled
$r_1 < 0$	$r_2 < 0$	$0 < \lvert r_2 \rvert - \lvert r_1 \rvert < d$ or $\lvert r_1 \rvert < \lvert r_2 \rvert < d + \lvert r_1 \rvert$	$M_2$ is always right of $M_1$ . $\overline{M_1 M_2} < \overline{S_1 S_2}$

c) The radius $r_1$ or the radius $r_2$ is infinite in the case of the planconvex lens. In the following it is assumed that $r_1$ is infinite and $r_2$ is finite.

\lim_{r_1 \to \infty} f = \lim_{r_1 \to \infty} \frac{n r_2}{(n-1)\left[n\left(\frac{r_2}{r_1} - 1\right) + (n-1)\frac{d}{r_1}\right]} = \frac{r_2}{1-n} \qquad (4)

Equation (4) means, that for each wavelength (refractive index) there exists a different value of the focal length.

d) From the given formula for the focal length (see problem formulation) one obtains the following quadratic equation in $n$ :

A\cdot n^2 + B\cdot n + C = 0 \qquad (5)

with $A = (r_2 - r_1 + d)\cdot f$ , $B = -\left[f\cdot(r_2 - r_1) + 2\cdot f\cdot d + r_1\cdot r_2\right]$ and $C = f\cdot d$ .

Solutions of equation (5) are:

n_{1,2} = -\frac{B}{2\cdot A} \pm \sqrt{\frac{B^2}{4\cdot A^2} - \frac{C}{A}} \qquad (6).

Equation (5) has only one physical correct solution, if...

I) $A = 0$ (i.e., the coefficient of $n^2$ in equation (5) vanishes)

In this case the following relationships exists:

\mathrm{r_1} - \mathrm{r_2} = \mathrm{d} \qquad (7),

n = \frac{f\cdot d}{f\cdot d + r_1\cdot r_2} > 1 \qquad (8).

II) $B = 0$ (i.e. the coefficient of $n$ in equation (5) vanishes)

In this case the equation has a positive and a negative solution. Only the positive solution makes sense from the physical point of view. It is:

f\cdot(r_2 - r_1) + 2\cdot f\cdot d + r_1\cdot r_2 = 0 \qquad (9),

n^2 = -\frac{C}{A} = -\frac{d}{(r_2 - r_1 + d)} > 1 \qquad (10),

III) $B^2 = 4\,AC$

In this case two identical real solutions exist. It is:

\left[f\cdot(r_2 - r_1) + 2\cdot f\cdot d + r_1\cdot r_2\right]^2 = 4\cdot(r_2 - r_1 + d)\cdot f^2\cdot d \qquad (11),

n = -\frac{B}{2\cdot A} = \frac{f\cdot(r_2 - r_1) + 2\cdot f\cdot d + r_1\cdot r_2}{2\cdot f(r_2 - r_1 + d)} > 1 \qquad (12).

Marking scheme

Problem 2: “Thick lens” (10 points)

Part a	1 point
Part b – equation (1), equation (2)	2 points
– physical restrictions, equation (3)	1 point
– discussion of different cases	2 points
– shapes of lenses	1 point
Part c – discussion and equation (4)	1 point
Part d	2 point

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