Update annex.fr.md

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

@@ -249,6 +249,228 @@ Sont proposées les catégories suivantes, mais à débattre :
 ! </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 M3P2 team 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>
+!
+! <!--question 6-->
+! <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.
+!
+! * 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<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 $`\overline{M_A}=-0.44`$.
+!
+! * You could obtained directly this algebraic value of $`M_A`$ by considering algebraic
+! lengthes and angles values in the calculations :<br><br> 
+! $`\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}`$
+! $`=\dfrac  {arctan\left(\frac{-0.017}{-0.175}\right)} {arctan\left(\frac{90}{-400}\right)}`$ $`=\dfrac{0.097}{-0.221}=-0.44`$
+!  </details>
+!
+! ![](lentille-boule-orleans-1bis.jpg)<br>
+! _Cathedral of Orleans (France)_
+!
+! </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>
+