courses/00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md

title	published	routable	visible	lessons
Brainstorming partie "au-delà"	true	true	false	{slug brainstorming-3-parts} {order 3}

Brainstorming partie "Au-delà"

Un rappel de ce qui est annoncé sur les 3 parties de cours est à :
https://m3p2.com/fr/m3p2_pedagogy/m3p2-pedagogy

1 - Ce dont nous disposons actuellement

Déjà décrit dans le brainstorming de la partie "Principal".

2 - Brainstorming pour cette partie "au-delà" de cours

Les différents propositions sont numérotées : elles commencent par :
[BB-numéro] : Titre
(BB pour Brainstorming Beyond)

• Pour réagir à une proposition existante, rajouter votre commentaire à la suite des autres en commençant par trois initiales entre parenthèse vous représentants :
  (CLM) : commentaire d'untel
  (mes initiales) : mon commentaire

• Pour ajouter une proposition, créez là dans ou à la suite (suivant la logique de la proposition) des propositions existantes. Pour cela, commencer par un nouveau [BB-numéro] : Titre

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[BB-10] Élément de style.

(CLE) Les fichiers de cours de chacune des trois parties sont différenciés (textbook pour la partie princiale, cheatsheet pour la partie synthèse, annex pour la partie au-delà). Pour l'instant aucune personnalisation, aucune feuille de style spécifique n'est appliquée à ces types de fichier.
Nous pourrons personnaliser plus le style de chaque partie pour mieux les différencier et identifier. Mais déjà, une idée à tester. Sur le premier prototype, le fond de cette partie au-delà était très légèrement colorée d'un violet clair. A tester et choisir entre nous.

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[BB-20] Différents éléments.

(CLE) C'est la partie qui permet :

• de dire tout ce qu'on a pas le temps de dire dans des cours classique, mais qu'on rêverait de dire et qui pourrait motiver plus les étudiants (au-delà de la simple acquisition de connaissances et compétences).
• de renforcer les explications sur des points particuliers et aïgus de cours (petites vidéos très courtes?)
• de proposer des défis sur plusieurs jours, pour que l'apprenant puisse réfléchir à un sujet proposé en dehors du temps d'apprentissage devant l'écran (par exemple réfléchir à la localisation d'une image en réflexion, en faisant attention à ce qu'on voit en réflexion dans la vitre d'un métro...)
• proposer des questionnaires, ou petits exercices de compréhension ou permettant d'acquérir des réflexes (écrit en javascript par exemple)
• proposer des problèmes représentatifs des examens.

Pour organiser ces éléments, il faut les regrouper dans un petit nombre de catégories facilement identifiables.

Sont proposées les catégories suivantes, mais à débattre, toutes les idées sont bienvenues :

!!!! DIFFICULT POINT (contribute, or indicate a difficult point of understanding) !!!!

!!! DO YOU MASTER ? !!!

!!!

!!! Describe the test. !!!

!!! Texte du test (et ses images, figures, audio, video, etc ...)
!!! [a figure for the test]
!!! Texte de la question !!!

!!!

!!! Choix de réponse 1 !!!

!!! Texte si la réponse 1 est choisie

!!! </details markdown=2> !!!

!!!

!!! Choix de réponse 2 !!!

!!! Texte si la réponse 2 est choisie

!!! </details markdown=3> !!! </details markdown=1>

!! BEYOND (to contribute) !!

!! CULTURAL POINT (contributor) !!

! YOUR CHALLENGE : look at the picture, and think of the right answers to the following questions ! ! Do not look at the answer, take time to think, a few days if necessary. The time to build your mental representation of the phenomenon, to formulate it in words is important, a thousand times more important than the ephemeral instant where you read the fews words of the solution. ! !

!

! INDICE "key word" !

! diffusion !

!

!

! ANSWER !

! diffusion !

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Quelques exemples

! YOUR CHALLENGE : An object (a painting), a physical system (a lensball), how many scenarios and optical systems? ! ! Skill tested : understanding of physical situations ! ! ! ! Discovery time : 30 minutes
! Resolution time : 10 minutes ! !

!

! I choose it !

! A lensball is a simple physical system: a sphere of glass of refractive index $n=1.5$ and of radius $R=5\;cm$. ! ! A ball lensball is placed in front of a painting. Depending on the position of the observer or the camera, ! the optical system (the sequence of simple optical elements crossed by light between the physical object ! and the observed image) that forms the image differs. ! ! Observe the 3 images of the painting given by the lensball : ! ! Image 1 ! ! ! ! Images 2 (the smallest) and 3 ! ! ! ! For each image of the painting, can you identify the optical system, then specify ! the characteristics of the various simple elements that constitute the system and their relative distances? ! ! * _The resolution time is the typical expected time to be allocated to this problem if it was part of an examen for an optics certificate._ ! * _The discovery time is the expected time required to prepare this challenge if you don't have practice. But take as much time as you need._ ! ! <\details> ! <details markdown=1> ! <summary> ! Ready to answer M3P2 team questions for image 1? ! </summary> ! ! <details markdown=1> ! <summary> ! Where is the painting located? ! </summary> ! * The painting is located on the other side of the lens, in relation to you. ! </details> ! <details markdown=1> ! <summary> ! What is the optical system giving the image of the painting? ! </summary> ! <br> ! * The optical system is composed of two spherical refracting surfaces, centered on the same optical axis.<br> ! <br> ! </details> ! <details markdown=1> ! <summary> ! How do you characterize each of the single optical elements that make up this optical system, ! and their relative distances? ! </summary> ! <br> ! * The optical axis is oriented positively in the direction of light propagation ! (from the painting towards the lensball).<br> ! <br> ! * The first spherical refracting surface ! $DS1$ encountered by the light has ! the follwing characteristics :<br> ! $\overline{S_1C_1}=+|R|=+5;cm$, ! $n_{ini}=1$ and $n_{fin}=1.5$. ! <br> ! * The second spherical refracting surface ! $DS2$ encountered by the light has the follwing characteristics :<br> ! $\overline{S_2C_2}=-|R|=-5;cm$ , ! $n_{ini}=1.5$ and $n_{fin}=1$ ! ! * Algebraic distance between $DS1$ and $DS2$ is : $\overline{S_1S_2}=+10;cm$ ! ! </details> ! <details markdown=1> ! <summary> ! If you had to determine the characteristics of the image (position, size), how ! would you handle the problem? ! </summary> ! ! * $DS1$ gives an image $B_1$ of an object $B$. This image $B_1$ for $DS1$ ! becomes the object for $DS2$. $DS2$ gives an image $B'1$ of the object $B_1$ ! ! </details> ! </details> ! <!--FOR IMAGES 2 & 3--> ! ! <details markdown=1> ! <summary> ! Ready to answer M3P2 team questions for images 2 and 3? ! </summary> ! ! <details markdown=1> ! <summary> ! Where is the painting located? ! </summary> ! ! * The painting is located on the same side of the lens as you, behind you. ! ! </details> ! <details markdown=1> ! <summary> ! ! What are the two optical systems at the origin of the two images of the painting? And ! can you characterize each of the single optical elements (+ their relative distances) ! that make up each of these optical systems ? ! </summary> ! ! * A first optical system $OS1$ is composed of a simple convexe spherical mirror ! (the object is reflected on the front face of the ball lensball). Keaping the optical ! axis positively oriented in the direction of the incident light propagation on the lensball, ! the algebraic value of the mirror radius is : $\overline{SC}=+5;c$. ! ! * The second optical system $OS2$ is composed of three simple optical elements :<br><br> ! 1) The light crosses a spherical refracting surface $DS1$ with characteristics : ! $\overline{S_1C_1}=+|R|=+5;cm$ , $n_{ini}=1$ and $n_{fin}=1.5$. ! ! 2) Then the light is reflected at the surface of the last lensball interface that ! acts like a spherical mirror of characteristics : $\overline{S_2C_2}=-|R|=-5;cm$, ! $n=1.5$. ! ! 3) Finally the light crosses back the first interface of the lensball that acts ! like a spherical refracting surface those characteristics are : ! $\overline{S_3C_3}=+|R|=+5;cm$ , $n_{ini}=1.5$ and $n_{fin}=1$. ! ! Relative algebraic distances between the different elements of $OS2$ are : ! ! $\overline{S_1S_2}=+10;cm$ and $\overline{S_2S_3}=-10;cm`$ ! !

!

!

! Which image is associated with each of the optical systems? !

! ! * It is difficult to be 100% sure before having made the calculations. ! !

!

!

! Why do we had to take the picture in the darkness, with only the painting ! illuminated behind the camera, to obtain images 2 and 3 ? !

! ! * At a refracting interface, part of the light incident power is refracted, ! and part is reflected. For transparent material like glass and for visible light, ! the part of the reflected power is small. If the room had been homogeneously ! illuminated, the images 2 and 3 of the painting on the wall behind the camera would ! have been faintly visible compared to the image of the front wall through the lensball. !

!

!

! YOUR CHALLENGE : Looking at a cathedral through a lensball. Can you predict your observation? ! ! Skill tested : to know how to carry out calculations ! ! ! ! Discovery time : 2 hours
! Resolution time : 30 minutes !

!

! I choose it !

! A lensball is a polished spherical ball of radius $R=5 cm$, made of glass of refractive ! index value $n_{glass}=1.5$. The cathedral is 90 meters high, and you stand with the lensball ! 400 meters from the cathedral. You look at the cathedral through the lens, your eye being at ! 20 cm from the center of the lens. What would you expect to see? ! ! * The resolution time is the typical expected time to be allocated to this problem ! if it was part of an examen for an optics certificate. ! ! * The discovery time is the expected time you require to prepare this challenge ! if you don't have practice. However, this is just an indication, take as much time as ! you need. The time to question yourself serenely about how to handle the problem, ! about the method of resolution and its validity, about some possible approximations ! if they can be justified, and the time you need to check the equations if you have ! not previously memorised them and to perfom the calculation, are important. ! !

!

! Ready to answer questions ? !

! !

!

! What is the scientifical framework you choose to study this problem ? !

! * All the characteristic sizes in this problem are much bigger than the wavelength of ! the visible radiation ($\lambda\approx5\mu m$), so I deal with this problem in the ! framework of geometrical optics, and in the paraxial approximation in order to ! characterize the image. ! ! * The cathedral sustains an angle of $arctan\dfrac{90}{400}=13°$ from the lensball. ! This value seems reasonable to justify at first order the use of the paraxial approximation ! (we usually consider that angles of incidence would not exceed 10° on the various simple ! optical element encountered between the objet (here the cathedral) and the final image ! (retina of the eye or matrix sensor of a camera). !

! !

!

! Describe the optical system for this use of the lensball. !

! ! * The lensball breaks down into two refracting spherical surfaces sharing the same ! centre of curvature C and of opposite radius (in algebraic values). ! !

! !

!

! What is your method of resolution ? !

! * You don't use general equations 3a and 3b for a thick lens, they are too complicated ! to remind, and you don't have in m3p2 to "use" but to "build a reasoning". And you don't ! know at this step how to handle with centered optical systems. ! ! * But this system is simple, so you will calculate the image of the cathedral by the ! first spherical refracting surface $DS_1$ encountered by the light from the cathedral $DS_1$. ! Then this image becomes the object for the second spherical refracting surface $DS_2$ ! and so I can determine position and size of the final image. ! ! * For a spherical refracting surface, general equations are :

! $\dfrac{n_{fin}}{\overline{SA_{ima}}}-\dfrac{n_{ini}}{\overline{SA_{obj}}}=\dfrac{n_{fin}-n_{ini}}{\overline{SC}}$ ! for the position.
! $\overline{M_T}=\dfrac{n_{ini}\cdot\overline{SA_{ima}}}{n_{fin}\cdot\overline{SA_{obj}}}$ ! for the transverse magnification. ! !

! !

!

! How do you set down your calculations? !

! ! * The optical axis is the straight line that joins the center C of the lens to my eye, ! positively oriented in the direction of light propagation light for that observation, ! so from the cathedral to my eye. ! * First spherical refrating surface $DS1$ : $\overline{S_1C_1}=+5\:cm$, $n_{ini}=1$ (air) ! and $n_{fin}=1.5$ (glass).
! Second spherical refrating surface $DS2$ : $\overline{S_1C_1}=-5\:cm$, $n_{ini}=1.5$ (glass) ! and $n_{fin}=1$ (air)
! Distance between $DS1$ and $DS2$ vertices : $\overline{S_1S_2}=+10\:cm$
! Object cathedral $AB$ : $\overline{AB}=90\;m$ and $\overline{S_1A}=-400\;m$
! Let us write $\overline{A_1B_1}$ the intermediate image (the image of the cathedral ! given by $DS1$. ! ! * Specific equations for $DS1$ are :

! $\dfrac{1.5}{\overline{S_1A_1}}-\dfrac{1}{\overline{S_1A}}=\dfrac{0.5}{\overline{S_1C_1}}$ (équ. DS1a), ! and $\overline{M_T}=\dfrac{\overline{S_1A_1}}{1.5\cdot\overline{S_1A}}$ (équ. DS1b)

! Specific equations for $DS2$ are :

! $\dfrac{1}{\overline{S_2A'}}-\dfrac{1.5}{\overline{S_2A_1}}=-\dfrac{0.5}{\overline{S_2C_2}}$ (équ. DS2a), and ! $\overline{M_T}=\dfrac{1.5\cdot\overline{S_2A'}}{\overline{S_2A_1}}$ (équ. DS2b)

! The missing link between these two sets of equations is :
! $\overline{S_2A_1}=\overline{S_2S_1}+\overline{S_1A_1}=\overline{S_1A_1}-\overline{S_1S_2}$. ! !

! !

!

! Do you see some approximation that can be done ? !

! * In the visible range, refractive index values of transparent material are in the range [1 ; 2], ! then the focal lengthes of a spherical refractive surface (object as well as image) are ! expected to remain in the same order of magnitude than the radius of curvature, ! so a few centimeters in this case (we talk in absolute value here). ! ! * We can if we want just check this fact for $DS1$ ($|S_1C_1|=5\;cm$) using équation DS1 :
! - considering $\overline{S_1A_1}\longrightarrow\infty$ to obtain the object focal length ! $\overline{S_1F_1}$} we get :
! $-\dfrac{1}{\overline{S_1F_1}}=\dfrac{0.5}{\overline{S_1C_1}}$ ! $\Longrightarrow=\overline{S_1F_1}=-10\;cm$

! - considering $\overline{S_1A}\longrightarrow\infty$ to obtain the image focal length ! $\overline{S_1F'_1}$ we get :
! $\dfrac{1.5}{\overline{S_1F'_1}}=\dfrac{0.5}{\overline{S_1C_1}}\Longrightarrow\overline{S_1F'_1}=+15\;cm$ ! ! * The distance of the cathedral from the lensball $|\overline{S_1A}|=90\;m$ is huge ! compared to the object focal length $|\overline{S_1F_1}|=10\;cm$, we can consider ! that the cathedral is at infinity from the lensball and so the image $\overline{A_1B_1}$ ! of the cathedral stands quasi in the image focal plane of $DS1$ : ! $\overline {S_1A_1}=\overline {S_1F'_1}=+15cm$. So we can directly use equation DS2a with :
! $\overline{S_2A_1}=\overline{S_2F'_1}=\overline{S_2S_1}+\overline{S_1F'_1}$ ! $=\overline{S_1F'_1}-\overline{S_1S_2}=+15-10=+5\;cm$.. ! !

!

!

! Where is the image and how tall it is ? !

! ! * To perform calculation, you must choose a unic lenght unit in your calculation, ! here $cm$ or $m$. We choose $m$ below. ! * Equation DS1a gives :
! $\dfrac{1.5}{\overline{S_1A_1}}-\dfrac{1}{-400}=\dfrac{0.5}{0.05}$ $\Longrightarrow\overline{S_1A_1}=0.15\;m$
! With more than 2 significant figures, your calculator would tell you $0.150037$, ! which nearly exactly the value of $\overline{S_1F'_1}=+0.15\;m$, so the approximation ! $\overline{S_1A_1}=\overline{S_1F'_1}$ you could have done is fully justified. ! ! * Equation DS2a gives :
! $\dfrac{1}{\overline{S_2A'}}-\dfrac{1.5}{-0.1+0.15}=\dfrac{-0.5}{-0.05}$ ! $\Longrightarrow\overline{S_2A'}=0.025\;m$ ! ! * The final image is real, and stands 2.5 cm in front of the lensball in the side ! of your eye. Do not bring your eye or camera too close of the lensball ! ! ! * The size of an image (transversally to the optical axis) is given by the transverse ! magnification $M_T$. By Definition $M_T$ is the ratio of the algebraic size of ! the final image $\overline{A'B'}$ to the algebraic size of the initial object $\overline{AB}$. ! With an intermediate image, it can be break down :

! $M_T=\dfrac{\overline{A'B'}}{\overline{AB}}$ ! $=\dfrac{\overline{A'B'}}{\overline{A_1B_1}}\times\dfrac{A_1B_1}{\overline{AB}}$

! It is the product of the transverse magnifications of the cathedral introduced ! by the two spherical refracting surfaces of the lensball.

! $\overline{M_T}$ introduced by $DS1$ is ! $\overline{M_T}=\dfrac{\overline{S_1A_1}}{1.5\cdot\overline{S_1A}}$ ! $=\dfrac{+0.15}{1.5\times(-400)}=-0.00025$

! $\overline{M_T}$ introduced by $DS2$ is ! $\overline{M_T}=\dfrac{1.5\cdot\overline{S_2A'}}{\overline{S_2A_1}}$ ! $=\dfrac{1.5\cdot\overline{S_2A'}}{\overline{S_1A_1}-\overline{S_1S_2}}$ ! $=\dfrac{1.5\cdot0.025}{+0.15-0.10} =0.75$

! So $\overline{M_T}$ introduced by the lensball is :

! $\overline{M_T}=-0.00025\times0.75$ $=-0.00019\approx-1.9\cdot10^{-4}$

! The image is $\dfrac{1}{-1.9\cdot10^{-4}}\approx5300$ smaller than the cathedral.

! $M_T=\dfrac{\overline{A'B'}}{\overline{AB}}\approx8\cdot10^{-4}$ ! $\Longrightarrow\overline{A'B'}=\overline{AB} \times M_T$ ! $=1.9\cdot10^{-4} \times 90\;m=-0.017\;m$

! The image is 1.7 cm height and it is reversed. !

! ! !

!

! What is the apparent magnification of the cathedral ? !

! * Apparent magnification = angular magnification = magnifying power. ! ! * As calculated previously, standing 400 metres from the cathedral, the 90 m heigh ! cathedral sustends the apparent angles of $\alpha=arctan\left(\dfrac{90}{400}\right)=0.221\;rad=12.7°$ ! at your eye.
!
! * The image of the cathedral is 1.7 cm heigth and is located between the lens ! (from its vertex $S2$) and your eyes and at 2.5cm from the lens. If your eye is ! 20cm away from the lens, so the distance eye-image is 17.5 cm (we use no algebraic values). ! Thus the image of the catedral subtends the apparent angle ! $\alpha'=arctan\left(\dfrac{1.7}{17.5}\right)=0.097\;rad=5.6°$ at your eye.
!
! * The apparent magnification $M_A$ of the cathedral throught the lensball for my ! eye in that position is
! $M_A=\dfrac{\alpha'}{\alpha}=\dfrac{0.097}{0.221}=0.44$.

! Taking into account that the image is reversed, the algebraic value of the apparent ! magnification is $\overline{M_A}=-0.44$.
!
! * You could obtained directly this algebraic value of $M_A$ by considering algebraic ! lengthes and angles values in the calculations :

! $\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}$ ! $=\dfrac {arctan\left(\frac{-0.017}{-0.175}\right)} {arctan\left(\frac{90}{-400}\right)}$ $=\dfrac{0.097}{-0.221}=-0.44$ ! !
! Cathedral of Orleans (France) !

!

!

!

!! BEYOND : The gravitationnal lensball (or Einstein's ring), due to a black hole or a galaxy. !! Similarities, and differences. !! !! !! !!

!!

!! To see !!

!! still to be done, in progress. !!