---
title: Brainstorming partie "au-delà"
published: true
routable: true
visible: false
lessons:
    - 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 :<br>
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 :<br>
*[BB-numéro] : Titre* <br>
_(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 :<br>
(CLM) : commentaire d'untel<br>
*(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*

-------------

**[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.<br>
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.

-----------------

**[BB-20] Différents éléments**.

(CLE) C'est la partie qui permet :<br>
  * 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 ?*
!!! <details markdown=1>
!!! <summary>
!!! Describe the test.
!!! </summary>
!!! Texte du test (et ses images, figures, audio, video, etc ...)<br>
!!! [a figure for the test]<br>
!!! Texte de la question
!!! <details markdown=2>
!!! <summary>
!!! Choix de réponse 1
!!! </summary>
!!! Texte si la réponse 1 est choisie<br><br>
!!! </details markdown=2>
!!! <details markdown=3>
!!! <summary>
!!! Choix de réponse 2
!!! </summary>
!!! Texte si la réponse 2 est choisie<br><br>
!!! </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._
!
! <details markdown=1>
! <summary>
! INDICE "key word"
! </summary>
! diffusion
! </details>
! <details markdown=1>
! <summary>
! ANSWER
! </summary>
! diffusion
! </details>

------------------------------

##### 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_
!
! ![](physical-system-versus-optical-system_L650.gif)
!
! *Discovery time : 30 minutes*<br>
! *Resolution time : 10 minutes*
!
! <details markdown=1>
! <summary>
! I choose it
! </summary>
! 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
!
! ![](lentille-boule-1-transparence_L650.jpg)
!
! Images 2 (the smallest) and 3
!
! ![](lentille-boule-2-reflexions_L650.jpg)
!
! 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 $`\overline{S_2S_3}=-10\;cm`$
!
! </details>
! <details markdown=1>
! <summary>
! Which image is associated with each of the optical systems?
! </summary>
! 
! * It is difficult to be 100% sure before having made the calculations. 
!
! </details>
! <details markdown=1>
! <summary>
! 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 ?
! </summary>
!
! * 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.
! </details>
! </details>
! </details>

! *YOUR  CHALLENGE*  :  Looking at a cathedral through a lensball. Can you predict your observation?
!
! _Skill tested : to know how to carry out calculations_
!
! ![](lensball-brut-820-760.jpg)
!
! *Discovery time : 2 hours*<br>
! *Resolution time : 30 minutes*
! <details markdown=1>
! <summary>
! I choose it
! </summary>
! 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._
!
! <details markdown=1>
! <summary>
! Ready to answer questions ?
! </summary>
! <!--question 1-->
! <details markdown=1>
! <summary>
! What is the scientifical framework you choose to study this problem ?
! </summary>
! * 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_).
! </details>
! <!--question 2-->
! <details markdown=1>
! <summary>
! Describe the optical system for this use of the lensball.
! </summary>
!
! * The lensball breaks down into two refracting spherical surfaces sharing the same 
! centre of curvature C and of opposite radius (in algebraic values).
!
! </details>
! <!--question 3-->
! <details markdown=1>
! <summary>
! What is your method of resolution ?
! </summary>
! * 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 :<br><br>
! $`\dfrac{n_{fin}}{\overline{SA_{ima}}}-\dfrac{n_{ini}}{\overline{SA_{obj}}}=\dfrac{n_{fin}-n_{ini}}{\overline{SC}}`$ 
! for the position.<br>
! $`\overline{M_T}=\dfrac{n_{ini}\cdot\overline{SA_{ima}}}{n_{fin}\cdot\overline{SA_{obj}}}`$ 
! for the transverse magnification.
!
! </details>
! <!--question 4-->
! <details markdown=1>
! <summary>
! How do you set down your calculations?
! </summary>
!
! * 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).<br>
!  Second spherical refrating surface $`DS2`$ : $`\overline{S_1C_1}=-5\:cm`$, $`n_{ini}=1.5`$ (glass) 
! and $`n_{fin}=1`$ (air)<br>
!  Distance between $`DS1`$ and $`DS2`$ vertices : $`\overline{S_1S_2}=+10\:cm`$<br>
!  Object cathedral $`AB`$ :  $`\overline{AB}=90\;m`$ and $`\overline{S_1A}=-400\;m`$<br>
!  Let us write $`\overline{A_1B_1}`$ the intermediate image (the image of the cathedral
! given by $`DS1`$.
!
!  * Specific equations for $`DS1`$ are :<br><br> 
! $`\dfrac{1.5}{\overline{S_1A_1}}-\dfrac{1}{\overline{S_1A}}=\dfrac{0.5}{\overline{S_1C_1}}`$ (équ. DS1a), 
! and $`\overline{M_T}=\dfrac{\overline{S_1A_1}}{1.5\cdot\overline{S_1A}}`$ (équ. DS1b)<br><br>
! Specific equations for $`DS2`$ are :<br><br> 
! $`\dfrac{1}{\overline{S_2A'}}-\dfrac{1.5}{\overline{S_2A_1}}=-\dfrac{0.5}{\overline{S_2C_2}}`$  (équ. DS2a),  and 
! $`\overline{M_T}=\dfrac{1.5\cdot\overline{S_2A'}}{\overline{S_2A_1}}`$ (équ. DS2b)<br><br>
! The missing link between these two sets of equations is :<br>
! $`\overline{S_2A_1}=\overline{S_2S_1}+\overline{S_1A_1}=\overline{S_1A_1}-\overline{S_1S_2}`$.
!
! </details>
! <!--question 5-->
! <details markdown=1>
! <summary>
! Do you see some approximation that can be done ? 
! </summary>
! * 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 :<br>
! \- considering $`\overline{S_1A_1}\longrightarrow\infty`$ to obtain the object focal length
! $`\overline{S_1F_1}`$} we get :<br>
! $`-\dfrac{1}{\overline{S_1F_1}}=\dfrac{0.5}{\overline{S_1C_1}}`$ 
! $`\Longrightarrow=\overline{S_1F_1}=-10\;cm`$<br><br>
! \- considering $`\overline{S_1A}\longrightarrow\infty`$ to obtain the image focal length 
! $`\overline{S_1F'_1}`$ we get :<br>
! $`\dfrac{1.5}{\overline{S_1F'_1}}=\dfrac{0.5}{\overline{S_1C_1}}\Longrightarrow\overline{S_1F'_1}=+15\;cm`$
! 
! * 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 :<br>
! $`\overline{S_2A_1}=\overline{S_2F'_1}=\overline{S_2S_1}+\overline{S_1F'_1}`$ 
! $`=\overline{S_1F'_1}-\overline{S_1S_2}=+15-10=+5\;cm`$..
!
! </details>
! <details markdown=1>
! <summary>
! Where is the image and how tall it is ?
! </summary>
!
! * To perform calculation, you must choose a unic lenght unit in your calculation, 
! here $`cm`$ or $`m`$. We choose $`m`$ below.
! * Equation DS1a gives :<br>
! $`\dfrac{1.5}{\overline{S_1A_1}}-\dfrac{1}{-400}=\dfrac{0.5}{0.05}`$ $`\Longrightarrow\overline{S_1A_1}=0.15\;m`$<br>
! 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 :<br>
! $`\dfrac{1}{\overline{S_2A'}}-\dfrac{1.5}{-0.1+0.15}=\dfrac{-0.5}{-0.05}`$ 
! $`\Longrightarrow\overline{S_2A'}=0.025\;m`$
!
! * 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 :<br><br>
!  $`M_T=\dfrac{\overline{A'B'}}{\overline{AB}}`$ 
! $`=\dfrac{\overline{A'B'}}{\overline{A_1B_1}}\times\dfrac{A_1B_1}{\overline{AB}}`$<br><br>
! It is the product of the transverse magnifications of the cathedral introduced 
! by the two spherical refracting surfaces of the lensball. <br><br>
! $`\overline{M_T}`$ introduced by $`DS1`$ is 
! $`\overline{M_T}=\dfrac{\overline{S_1A_1}}{1.5\cdot\overline{S_1A}}`$
! $`=\dfrac{+0.15}{1.5\times(-400)}=-0.00025`$<br><br>
! $`\overline{M_T}`$ introduced by $`DS2`$ is 
! $`\overline{M_T}=\dfrac{1.5\cdot\overline{S_2A'}}{\overline{S_2A_1}}`$ 
! $`=\dfrac{1.5\cdot\overline{S_2A'}}{\overline{S_1A_1}-\overline{S_1S_2}}`$
! $`=\dfrac{1.5\cdot0.025}{+0.15-0.10} =0.75`$<br><br>
! So $`\overline{M_T}`$ introduced by the lensball is :<br><br>
! $`\overline{M_T}=-0.00025\times0.75`$ $`=-0.00019\approx-1.9\cdot10^{-4}`$<br><br>
! The image is  $`\dfrac{1}{-1.9\cdot10^{-4}}\approx5300`$ smaller than the cathedral.<br><br> 
! $`M_T=\dfrac{\overline{A'B'}}{\overline{AB}}\approx8\cdot10^{-4}`$ 
! $`\Longrightarrow\overline{A'B'}=\overline{AB} \times M_T`$
! $`=1.9\cdot10^{-4} \times 90\;m=-0.017\;m`$<br><br>
! The image is 1.7 cm height and it is reversed.
!</details>
!
!<!--question 7-->
!<details markdown=1>
!<summary>
! What is the apparent magnification of the cathedral ?
!</summary>
! * 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.<br>
! <br>
! * 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.<br>
! <br>
! * The apparent magnification $`M_A`$ of the cathedral throught the lensball for my 
! eye in that position is<br>
!  $`M_A=\dfrac{\alpha'}{\alpha}=\dfrac{0.097}{0.221}=0.44`$.<br><br>
!  Taking into account that the image is reversed, the algebraic value of the apparent
! magnification is $`\overline{M_A}=-0.44`$.<br>
! <br>
! * You could obtained directly this algebraic value of $`M_A`$ by considering algebraic
! lengthes and angles values in the calculations :<br><br> 
! $`\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}`$
! $`=\dfrac  {arctan\left(\frac{-0.017}{-0.175}\right)} {arctan\left(\frac{90}{-400}\right)}`$ $`=\dfrac{0.097}{-0.221}=-0.44`$
!
! ![](lentille-boule-orleans-1bis.jpg)<br>
! _Cathedral of Orleans (France)_
! </details>
! </details>
! </details>
! </details>

!! *BEYOND* : The gravitationnal lensball (or Einstein's ring), due to a black hole or a galaxy.
!! Similarities, and differences.
!!
!! ![](Einstein-ring-free.jpg)
!!
!! <details  markdown=1>
!! <summary>
!! To see
!! </summary>
!! still to be done, in progress. 
!! </details>





  



