Results after Problem 8

The list of the contest leaders after the eighth problem:

 Points Name Country School Physics teacher 20.055 Szabó Attila Hungary Leőwey Klára High School, Pécs Simon Péter, Dr Kotek László 18.0512 Nikita Sopenko Russia Lyceum No.14, Tambov Valeriy Vladimirovich Biryukov 14.8787 Jakub Šafin Slovak Pavol Horov Secondary, Michalovce Jozef Smrek 12.8105 Ilie Popanu Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 12.5541 Lars Dehlwes Germany Ohm-Gymnasium Erlangen Martin Perleth 10.4591 Alexandra Vasileva Russia Lyceum "Second School", Moscow A.R. Zilberman, G.F. Lvovskaya, G.Z. Arabuly 9.7732 Ivan Tadeu Ferreira Antunes Filho Brazil Colégio Objetivo, Lins, São Paulo 8.5043 Ion Toloaca Moldova liceul "Mircea Eliade" Igor Iurevici Nemtov; Andrei Simboteanu 8.0645 Cristian Zanoci Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 7.7917 Nadezhda Vartanian Russia Smolensk Pedagogical Lyceum Mishchenko Andrei Anatolievich 7.4407 Dinis Cheian Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 6.8529 Jakub Supeł Poland 14th School of Stanisław Staszic, Warsaw Włodzimierz Zielicz 6.5644 Kohei Kawabata Japan Nada High School 5.8911 Papimeri Dumitru Moldova Lyceuum "Orizont", Chisinau Igor Evtodiev 5.6695 Brahim Saadi Algeria Preparatory School for Science & Technology of Annaba Derradji Nasreddine 5.014 Petar Tadic Montenegro Gimnazija ,,Stojan Cerovic" Niksic Ana Vujacic 4.2254 Ng Fei Chong Malaysia SMJK Chung Ling, Penang 3.7517 Luís Gustavo Lapinha Dalla Stella Brazil Colégio Integrado Objetivo, Barueri, Brazil Ronaldo Fogo 3.4205 Bharadwaj Rallabandi India Narayana Jr. College, Basheer Bagh, India Vyom Sekhar Singh 3.1554 Selver Pepić Bosnia-Herzegovina Fourth Gymnasium Ilidža, Sarajevo Rajfa Musemić 2.7716 Adrian Nugraha Utama Indonesia SMA Sutomo 1 Medan Manaek Nababan 2.5937 Mikhail Shirkin Russia Gymnasium of  Ramenskoye Petrova Elena Georgyevna﻿ 2.374 Lorenzo Comoglio Italy Liceo Scientifico del Cossatese e Valle Strona Chiara Bandini 2.2747 Kai-Chi Huang Taiwan Taipei Municipal Chien-Kuo High School Shun-Ju Liu 2.1 Krzysztof Markiewicz Poland XIV Highschool in Warsaw Robert Stasiak 2 Teoh Yee Seng Malaysia SMJK HENG EE,Penang; SMJK CHUNG LING,Penang 2 Jaan Toots Estonia Tallinn Secondary Science School Toomas Reimann 1.7444 Andrew Zhao United States Webster Thomas High School Dykstra, William 1.4641 Hideki Yukawa Japan Nada high school 1.4641 Bruno Bento Barros de Araújo Brazil Ari de Sá Cavalcante Edney Melo 1.21 Midhul Varma India Vidyadham Junior, Hyderabad Manikanta Kumar 1 Task Ohmori Japan Nada High School T.Hamaguchi 1 Sharad Mirani India Prakash Higher Secondary School Ruchi Sadana, Sunil Sharma 1 Lev Ginzburg Russia Advanced Educational Scientific Center, MSU, Moscow I.V. Lukjanov, S.N. Oks 1 José Luciano de Morais Neto Brazil Colégio Ari de Sá Cavalcante Leonardo Bruno 0.9801 Mekan Toyjanow Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshkun 0.9 Jôhanes Sebástian Paiva Melo Brazil Colégio Ari de Sá Cavalcante Eduardo Kilder; Italo Reann 0.81 Meylis Malikov Turkmenistan Turgut Ozal Turkmen Turkish High School Halit Coshkun 0.81 Liara Guinsberg Brazil Colégio Integrado Objetivo, São Paulo, Brazil Ronaldo Fogo 0.792 Ulysse Lojkine France Lycée Henri IV, Paris M. Lacas 0.72 Rajat Sharma India Pragati Vidya Peeth,Gwalior Mr. Rakesh Ranjan

Points for problem 8:

 3.3027 Jakub Šafin 2.5022 Alexandra Vasileva 1.8799 Ivan Tadeu Ferreira Antunes Filho 1.7363 Lars Dehlwes 1.6226 Ilie Popanu 1.4495 Nikita Sopenko 1.1859 Szabó Attila 1.1392 Ng Fei Chong 0.9801 Cristian Zanoci 0.7175 Petar Tadic

Correct solutions (ordered according to the arrival time):

1.-2. Ilie Popanu (Moldova)

1.-2. Jakub Šafin (Slovak)

3.-5. Alexandra Vasileva (Russia)

3.-5. Lars Dehlwes (Germany)

3.-5. Ng Fei Chong (Malaysia)

6. Nikita Sopenko (Russia)

7. Ivan Tadeu Ferreira Antunes Filho (Brazil)

8. Szabó, Attila (Hungary)

9. Cristian Zanoci (Moldova)

Additionally, Petar Tadic (Montenegro) has solved only Part A (his was 5th correct solution for Part A ).

Regarding these lists, few comments are needed. The order numbers of the arrival list are found by averaging the order numbers for the Part A, and for the Part B. For instance, Ilie Popanu was the fastest with part A, and second fastest with part B. Meanwhile, Jakub was the fastest with part B and second fastest with Part A. So, overall they share the first and second place. The solution of Cristian Zanoci received additional penalty of 0.81, due to very late arrival.

The best solution bonus goes to Jakub Šafin (50%), Aleksandra Vasileva (25%), and Ivan Tadeu Ferreira Antunes Filho (25%).

The number of registered participants: 263 from 45 countries.

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The hints of 22nd April were in the form of additional images, which were made using a green laser, with two different beam diameters, two different distances between the camera and the bright spot, and two different surfaces onto which the laser beam was directed (to create the bright spot). These images are only for helping you in testing your hypothesis; these are not to be used for taking measurements (use only the red laser images provided in the problem text!), because (a) the camera was different (the pixel size on this page is different from the pixel size on the page of Problem 8), (b) the distances were not the same, (c) the beam diameters were different from what has been used for taking the original images. NB! the exposure time for the images below was not kept constant, so the average image brightness is not an indicative measure.

First image: distance between the lens and the bright spot is La(value not disclosed); surface onto which the laser beam was directed: flat matte metal surface (of a caliper), surface height fluctuations of the order of a couple of micrometers (seems perfectly smooth for a naked eye), laser beam diameter La (value not disclosed).

Second image: distance between the lens and the bright spot is La; surface onto which the laser beam was directed: white laser printer paper (A4), surface height fluctuations ca 10 times larger than for the matte metal surface, laser beam diameter La.

Third image: distance between the lens and the bright spot is La; surface onto which the laser beam was directed: white laser printer paper (A4), laser beam diameter Lb (value not disclosed).

Fourth image: distance between the lens and the bright spot is Lb (value not disclosed); surface onto which the laser beam was directed: matte metal surface, laser beam diameter La.

Fifth image: distance between the lens and the bright spot is Lb ; surface onto which the laser beam was directed: white laser printer paper (A4), laser beam diameter La.

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The hints of 29th April. In order to solve the problem, you need to understand, why the circles on pictures are filled with randomly distributed dark and bright spots.

To begin with, let us notice that for all the considered surfaces, the surface height fluctuations exceed the light wavelength. Why so? Well, because all these surfaces are matte, ie. scatter light in all directions. Such a scattering means that under a microscope, there are surface fragments of different orientations, and the size of these fragments needs to be not less that the wavelength (otherwise the light would "feel" the average slope of the fragment). These two requirements (different orientations + minimal size) imply automatically that the surface height variations are not less than the wavelength. And, of course, these height fluctuations are random (by no means regular like a diffraction grating).

Second, let us recall the Huygens' principle: a wavefront can be substituted with an array of sources. The surface scatters the light: small pieces (with a  diameter of the order of wavelength) work as small mirrors, and there are mirrors of all the possible orientations. So, all these surface points which are lit by laser work as sources. A sensor point receives light from all these sources, with essentially random phases. In terms of vector diagram, vectors of random directions are added up. At some sensor points, many  Huygens sources happen to contribute in a similar phase and a bright dot is recorded. For another point, the contributions of Huygens sources almost cancel out and a dark dot is recorded.

Finally, regarding the case of violet laser on a sheet of laser printer paper: apparently, something described above fails here: the dark spots completely disappear! It may be helpful to have a look on the image where a violet laser beam hits the edge of the sheet: both the white wall and the sheet are illuminated by a half of the beam. NB! Here, the beam diameter is different from what has been used for other photos.

And here is another photo of the same thing:

And finally, beam hitting only the wall (exposure time and f-stop are the same as above).

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The hints of 6th May.
For part (A). Study how the first minimum of the single slit diffraction is obtained in MIT lecture notes by S. Liao et al, module 14, page 14.
For part (B). Study the difference between the appearance of the sites where the violet laser beam hits the wall, and the paper (the last three images).

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The hints of 9th May.
For part (A). When you study the diffraction pattern on the sensor created by the light scattered from the bright spot on the sheet of paper, bear in mind pt. VII-14 of the Formula sheet
For part (B). Pay attention to the fact that as long as a coherent light is scattered from a bright spot on a sheet of paper, there is a diffraction pattern on the sensor. Indeed, according to the first five images on this page, the effect of the surface roughness is minimal (larger roughness makes only the dark stripes slightly less dark). The violet and green wavelengths are not that different (404 nm vs 532 nm), and definitely cannot explain the complete disappearance of the diffraction pattern. Another thing you need to ask yourself is, why is the spot of the violet laser beam on a paper so much brighter than on a white wall (see the last three photos). Bear in mind that both are white, ie. in both cases, almost all the incident light is scattered back.

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The hints of 14th May.

For par (A). If you consider the surface of the paper sheet as a set of Huygens sources of light, the light from all these sources arrives to a point on the sensor at an essential random phase, due to the height fluctuations of the paper. This can be depicted on a vector diagram: each Huygens source contributes a small vector, and all these vectors are added. At some sensor points, these vectors happen to be more aligned and so the sum will be larger – this is a centre of a bright speckle on the image. Now, divide the bright spot on the paper sheet into two halves, and correspondingly, the set of Huygens sources into two groups. Suppose that at a sensor point N, the vector sums for the both group are parallel (ie. the corresponding waves are in the same phase). The net sum over all the Huygens sources is the sum of these two vector sums, and will be larger when compared with the case when these two vectors were antiparallel; let this happen at another (near-by) sensor point M. Note that if we move from the point N to a neighbouring point on the sensor, all these small vectors will rotate slightly, because the optical path length will change according to the new position on the sensor. The rotation of those vectors which correspond to neighbouring Huygens sources will be almost equal, because the change of the optical path lengths will be almost identical. The largest relative rotation is observed for the most separated pair of Huygens sources. Now, let us return to the vectors representing the two halves of the bright spot: if the displacement on the sensor is small, the relative rotation of the vectors within each group is small, so the change of the length of the vector representing one single half of the bright spot is small. Meanwhile, the relative rotation of the vectors representing the two halves is larger than the average relative rotation within a group (corresponding to the distance of the respective Huygens sources). When we move from N to M, this rotation angle will be of the order of $\pi$.

For part (B). If you have a money checker or a UV ("black light") tube, compare the appearance of several white objects under the UV light: white paper, a white cloth, a white plastic card, and observe the difference.

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The hints of 28th May.

For part (A). Let us bring the earlier hints together and expand slightly. The paper surface is a partially reflecting rough (not perfectly flat) surface. Once illuminated by a laser beam, according to the Huygens principle, each surface point can be considered as a source of electromagnetic waves. This is almost exactly the same situation as what is observed in the case of a diffraction grating or a single-slit diffraction. Let us consider a single slit diffraction. According to the Huygens principle, the wavefront will evolve after the grating in the same way what would happen if there were small wave sources (much smaller than the wave length) placed along the open slit, see figure.

The incoming wave has a flat wavefront, which corresponds to a parallel beam: the wave phase is equal for all the points on a plane, perpendicular to the beam. Because of that, all the Huygens sources at the slit oscillate (ie. generate waves) at the same phase. Now, let us study the interference of the waves generated by these small sources at a point $D$ on a screen; it is assumed that the screen is in a focal plane of a lens. To this end, we divide the slit into two halves, and find condition under which all the light cancels out (ie. we have a completely dark spot on the screen).  To begin with, let us require that at $D$, the light from the leftmost source of left group (point $A$ in fig.), and the light from the leftmost source of the right group (point $B$ in fig.) cancel out. The light from the Huygens sources travels according to the rules of geometrical optics and the respective ray trajectories are drawn as blue and red lines. First we draw a perpendicular $BC$ to the rays and notice that the optical paths from $C$ to $D$ and from $B$ to $D$ are equal. Indeed, blue and red lines can be considered to be the rays coming from an infinitely remote point $P$, and according to the Fermat' principle, the valid ray paths from $P$ to $D$ obey minimal optical path length; since both blue and red lines are valid ray paths, both need to be minimal and hence, equal. So, the optical path difference from the Huygens sources at $A$ and $B$ to the point $D$ is given by the segment length $AC$; for a minimum, $\frac d2\sin\alpha=(n+\frac 12)\lambda$, where $n$ is an integer and $d$ – the slit width. If these two rays cancel out, all the similar ray pairs (such as the one marked with dashed lines) cancel out, too, and we can conclude that for $\sin\alpha=(2n+1)\lambda$, there is a minimum. Who said we can divide the slit only int two halves? We can divide it also into four quarters, into eight equal pieces etc; if we do so and repeat the previous analysis, we can conclude that the minima are also observed for $\sin\alpha=2^m(2n+1)\lambda$, where $m$ is another positive integer. The expression $2^m(2n+1)$ can take all the integer values except for 0, so the condition of the diffraction minimum can be equivalently written as $\sin\alpha=n\lambda,\;n\ne 0$.

Let us return to the case of the bright laser spot on a rough surface of a paper sheet. In order to study the condition for the diffraction minima, we need to put the Huygens sources on the paper surface at the region, illuminated by the laser. Unlike in the case of the single slit diffraction, these sources are not on the original wave front (which is flat – unlike the paper surface at which the sources are situated) and hence, each source needs to be characterized by its phase – which is determined by the height of the given paper surface point. This situation is depicted in the figure below,

So, the the phase difference with which the rays scattered at the paper surface points $B$ and $L$ would meet at an ininitely remote screen is given by the length difference between the segments $ABCD$ and $KLM$. If we scan different angles for the outgoing rays (and keep the range of angles small), the lengths of the segments $ABC$ and $KLM$ remain nearly constant (assuming that $AK\gg AB, KL$); what will vary mostly is the length $CD$. Since the paper surface has an irregular shape, the rays will add up with random phases; in terms of a vector diagram (where the vector lengths represent the amplitudes of the waves, and the directions are given by the phases), this means adding randomly oriented vectors: both groups of vectors (the ones representing the Huygens sources marked with blue, and the ones corresponding to red) will add up in a vector of a random length and orientation. The length is not completely random, however – typically it is not longer than a certain limit value. Now, suppose that for a certain point $P$ at the screen, the light intensity is nearly zero. This means that the two vectors representing the net amplitudes of the two groups of Huygens sources need to cancel out, ie. to be antiparallel. If we move on the screen away from that point of minimal intensity, the two vectors start rotating with respect to each other, due to the phase difference accumulated on the segment $CD$ (note that this length is equal for all the pairs of "blue" and "red" Huygens sources). At a certain point $Q$, they are parallel – this is the center of a bright speckle on the screen – and later they will be again antiparallel (this will happen at the next minimum).

Note that what has been described above is somewhat simplifying; most importantly, the vector representing the sum of waves created by the set of "red" (or "blue") Huygens sources evolves similarly to the vector representing the sum of waves created by all the Huygens sources (we can divide the red group into two subgroups and apply the above  discussion recursively). However, since the subgroups of Huygens sources have twice smaller physical length, the rotation of the respective wave amplitude vectors is twice slower, and can be neglected for the analysis of the transition from the minimal intensity point $P$ to the nearest maximal intensity point $Q$.

For part (B). In the case of a violet laser, the light originating from the bright spot on the paper sheet cannot be a scattered laser light, because the laser light would be still coherent, and some speckle pattern would be unavoidably seen. Your task is to figure out, would it be possible to have a non-coherent light being emitted (not scattered) by the paper sheet itself.