--- 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 :
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* ------------- **[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. ----------------- **[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

!!!
!!!
!!! !!! Choix de réponse 2 !!! !!! Texte si la réponse 2 est choisie

!!!
!!!
!! *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 !
------------------------------ ##### 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*
! *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 ! ! ![](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> !
! ! Ready to answer M3P2 team questions for image 1? ! ! !
! ! Where is the painting located? ! ! * The painting is located on the other side of the lens, in relation to you. !
!
! ! What is the optical system giving the image of the painting? ! !
! * The optical system is composed of two spherical refracting surfaces, centered on the same optical axis.
!
!
!
! ! How do you characterize each of the single optical elements that make up this optical system, ! and their relative distances? ! !
! * The optical axis is oriented positively in the direction of light propagation ! (from the painting towards the lensball).
!
! * The first spherical refracting surface ! $`DS1`$ encountered by the light has ! the follwing characteristics :
! $`\overline{S_1C_1}=+|R|=+5\;cm`$, ! $`n_{ini}=1`$ and $`n_{fin}=1.5`$. !
! * The second spherical refracting surface ! $DS2$ encountered by the light has the follwing characteristics :
! $`\overline{S_2C_2}=-|R|=-5\;cm`$ , ! $`n_{ini}=1.5`$ and $`n_{fin}=1`$ ! ! * Algebraic distance between $DS1$ and $DS2$ is : $`\overline{S_1S_2}=+10\;cm`$ ! !
!
! ! If you had to determine the characteristics of the image (position, size), how ! would you handle the problem? ! ! ! * $`DS1`$ 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 $`B_1`$ ! !
!
! ! !
! ! Ready to answer M3P2 team questions for images 2 and 3? ! ! !
! ! Where is the painting located? ! ! ! * The painting is located on the same side of the lens as you, behind you. ! !
!
! ! ! 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 ? ! ! ! * 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 :

! 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_ ! ! ![](lensball-brut-820-760.jpg) ! ! *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 :

! $`\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}`$ ! $`=\dfrac {arctan\left(\frac{-0.017}{-0.175}\right)} {arctan\left(\frac{90}{-400}\right)}`$ $`=\dfrac{0.097}{-0.221}=-0.44`$ ! ! ![](lentille-boule-orleans-1bis.jpg)
! _Cathedral of Orleans (France)_ !
!
!
! !! *BEYOND* : The gravitationnal lensball (or Einstein's ring), due to a black hole or a galaxy. !! Similarities, and differences. !! !! ![](Einstein-ring-free.jpg) !! !!
!! !! To see !! !! still to be done, in progress. !!