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

title	published	routable	visible	lessons
Brainstorming for the "beyond" part	true	true	false	{slug brainstorming-3-parts} {order 3}

Lluvia de ideas para la parte "Más allá"

Un recordatorio de lo que se anuncia en las 3 partes del curso está en:
https://m3p2.com/fr/m3p2_pedagogy/m3p2-pedagogy

1 - Qué tenemos actualmente

Ya descrito en la lluvia de ideas de la parte "Principal".

2 - Lluvia de ideas para esta parte "más allá" del curso

Las diferentes propuestas están numeradas: comienzan con:
[BB-number] : Título
(BB para Brainstorming Beyond)

• Para reaccionar a una propuesta existente, agregue su comentario después de los demás comenzando con tres iniciales entre paréntesis que lo representan a usted: (CLM): comentario de tal.
  (mis iniciales): mi comentario

• Para agregar una propuesta, créala en o después (siguiendo la lógica de la propuesta) propuestas existentes. Para hacer esto, comience con un nuevo [BB-number] : Título.

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[BB-10] Elemento de estilo

(CLE) Los archivos del curso para cada una de las tres partes están diferenciados (libro de texto para la parte principal, hoja de referencia para la parte de resumen, anexo para la parte posterior). Por el momento no hay personalización, no se aplica una hoja de estilo específica a estos tipos de archivos. Podremos personalizar más el estilo de cada parte para diferenciarlas e identificarlas mejor. Pero ya, una idea para probar. En el primer prototipo, el fondo de esta parte posterior estaba ligeramente teñido de violeta claro. Para probar y elegir entre nosotros.

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[BB-20] Diferentes elementos

(CLE) Esta es la parte que permite:

• para decir todo lo que no tenemos tiempo para decir cursos clásicos, pero que soñaríamos con decir y que podría motivar más a los alumnos (más allá de la simple adquisición de conocimientos y habilidades).
• para reforzar las explicaciones sobre puntos particulares y puntuales del curso (pequeños vídeos ¿muy corto?)
• para proponer desafíos durante varios días, para que el alumno pueda pensar en un tema propuesto fuera del tiempo de aprendizaje frente a la pantalla (por ejemplo, pensar en ubicar una imagen reflejada, prestando atención a lo que ves reflejada en la ventana de un metro ...)
• proponer cuestionarios, o pequeños ejercicios de comprensión o que permitan adquirir reflejos (escritos en javascript por ejemplo)
• proponer problemas representativos de los exámenes.

Para organizar estos elementos, deben agruparse en un número reducido de categorías. fácilmente identificable.

Se ofrecen las siguientes categorías, pero para discusión, todas las ideas son bienvenidas:

!!!! PUNTO DIFICIL (contribuir o indicar un punto muy difícil de entender para los estudiantes)

!!! ¿DOMINAS? !!!

!!!

!!! Describe la prueba. !!!

!!! Prueba de texto (y sus imágenes, figuras)
!!! [la imagen de la prueba]
!!! Texto de la pregunta. !!! !!!

!!!

!!! Propuesta 1. !!!

!!!
!!! Texto de su respuesta.

!!!

!!! !!!

!!!

!!! Propuesta 2 !!!

!!!
!!! Texto de su respuesta.

!!!

!!! !!!

!! PARA IR MÁS ALLÁ

!! PUNTO CULTURAL

! TU DESAFÍO : Título del desafío ! ! [Una imagen que presenta el desafío] ! !

!

! Elegí este desafío. !

! El texto completo del desafío ! ! !

!

! Pregunta 1 !

! ! Respuesta 1 ! !

! !

!

! Pregunta 2 !

! ! Respuesta 2 ! !

! !

!

! Pregunta 3 !

! ! Respuesta 3 ! !

! !

!

! Pregunta 4 !

! ! Respuesta 4 ! !

! !

!

! Pregunta 5 !

! ! Respuesta 5 ! !

! !

!

! Pregunta 6 !

! ! Respuesta 6 ! !

! !

!

! Pregunta 7 !

! ! Respuesta 7 ! ! !

!

! Pregunta 8 !

! ! Respuesta 8 ! !

!

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Algunos ejemplos

! 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> ! ! <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. ! !

!

! Can this problem be treated by geometrical optics ? !

! !

!

! 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. !!