|
|
|
@ -110,4 +110,390 @@ Se ofrecen las siguientes categorías, pero para discusión, todas las ideas son |
|
|
|
! respuesta |
|
|
|
! </details> |
|
|
|
|
|
|
|
------------------------------ |
|
|
|
|
|
|
|
##### 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*<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 |
|
|
|
! |
|
|
|
!  |
|
|
|
! |
|
|
|
! 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 $`\overline{S_2S_3}=-10\;cm`$ |
|
|
|
! |
|
|
|
! </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_ |
|
|
|
! |
|
|
|
!  |
|
|
|
! |
|
|
|
! *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> |
|
|
|
! $`\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}`$ |
|
|
|
! $`=\dfrac {arctan\left(\frac{-0.017}{-0.175}\right)} {arctan\left(\frac{90}{-400}\right)}`$ $`=\dfrac{0.097}{-0.221}=-0.44`$ |
|
|
|
! |
|
|
|
! <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. |
|
|
|
!! |
|
|
|
!!  |
|
|
|
!! |
|
|
|
!! <details markdown=1> |
|
|
|
!! <summary> |
|
|
|
!! To see |
|
|
|
!! </summary> |
|
|
|
!! still to be done, in progress. |
|
|
|
!! </details> |
|
|
|
|
|
|
|
|
