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@ -110,4 +110,390 @@ Se ofrecen las siguientes categorías, pero para discusión, todas las ideas son |
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! respuesta |
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! respuesta |
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! </details> |
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! </details> |
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------------------------------ |
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##### Algunos ejemplos |
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! *YOUR CHALLENGE* : An object (a painting), a physical system (a lensball), how many scenarios and optical systems? |
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! |
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! _Skill tested : understanding of physical situations_ |
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! |
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!  |
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! |
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! *Discovery time : 30 minutes*<br> |
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! *Resolution time : 10 minutes* |
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! |
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! <details markdown=1> |
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! <summary> |
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! I choose it |
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! </summary> |
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! A lensball is a simple physical system: a sphere of glass of refractive index $`n=1.5`$ and of radius $`R=5\;cm`$. |
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! |
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! A ball lensball is placed in front of a painting. Depending on the position of the observer or the camera, |
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! the optical system (the sequence of simple optical elements crossed by light between the physical object |
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! and the observed image) that forms the image differs. |
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! |
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! Observe the 3 images of the painting given by the lensball : |
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! |
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! Image 1 |
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! |
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!  |
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! |
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! Images 2 (the smallest) and 3 |
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! |
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!  |
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! |
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! For each image of the painting, can you identify the optical system, then specify ` |
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! the characteristics of the various simple elements that constitute the system and their relative distances? |
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! |
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! * _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._ |
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! * _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._ |
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! |
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! <\details> |
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! <details markdown=1> |
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! <summary> |
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! Ready to answer M3P2 team questions for image 1? |
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! </summary> |
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! |
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! <details markdown=1> |
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! <summary> |
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! Where is the painting located? |
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! </summary> |
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! * The painting is located on the other side of the lens, in relation to you. |
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! </details> |
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! <details markdown=1> |
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! <summary> |
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! What is the optical system giving the image of the painting? |
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! </summary> |
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! <br> |
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! * The optical system is composed of two spherical refracting surfaces, centered on the same optical axis.<br> |
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! <br> |
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! </details> |
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! <details markdown=1> |
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! <summary> |
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! How do you characterize each of the single optical elements that make up this optical system, |
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! and their relative distances? |
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! </summary> |
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! <br> |
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! * The optical axis is oriented positively in the direction of light propagation |
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! (from the painting towards the lensball).<br> |
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! <br> |
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! * The first spherical refracting surface |
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! $`DS1`$ encountered by the light has |
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! the follwing characteristics :<br> |
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! $`\overline{S_1C_1}=+|R|=+5\;cm`$, |
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! $`n_{ini}=1`$ and $`n_{fin}=1.5`$. |
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! <br> |
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! * The second spherical refracting surface |
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! $DS2$ encountered by the light has the follwing characteristics :<br> |
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! $`\overline{S_2C_2}=-|R|=-5\;cm`$ , |
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! $`n_{ini}=1.5`$ and $`n_{fin}=1`$ |
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! |
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! * Algebraic distance between $DS1$ and $DS2$ is : $`\overline{S_1S_2}=+10\;cm`$ |
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! |
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! </details> |
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! <details markdown=1> |
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! <summary> |
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! If you had to determine the characteristics of the image (position, size), how |
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! would you handle the problem? |
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! </summary> |
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! |
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! * $`DS1`$ gives an image $`B_1`$ of an object $`B`$. This image $`B_1`$ for $`DS1`$ |
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! becomes the object for $`DS2`$. $`DS2`$ gives an image $`B'1`$ of the object $`B_1`$ |
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! |
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! </details> |
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! </details> |
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! <!--FOR IMAGES 2 & 3--> |
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! |
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! <details markdown=1> |
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! <summary> |
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! Ready to answer M3P2 team questions for images 2 and 3? |
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! </summary> |
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! |
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! <details markdown=1> |
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! <summary> |
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! Where is the painting located? |
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! </summary> |
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! |
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! * The painting is located on the same side of the lens as you, behind you. |
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! |
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! </details> |
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! <details markdown=1> |
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! <summary> |
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! |
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! What are the two optical systems at the origin of the two images of the painting? And |
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! can you characterize each of the single optical elements (+ their relative distances) |
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! that make up each of these optical systems ? |
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! </summary> |
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! |
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! * A first optical system $`OS1`$ is composed of a simple convexe spherical mirror |
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! (the object is reflected on the front face of the ball lensball). Keaping the optical |
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! axis positively oriented in the direction of the incident light propagation on the lensball, |
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! the algebraic value of the mirror radius is : $`\overline{SC}=+5\;c`$. |
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! |
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! * The second optical system $`OS2`$ is composed of three simple optical elements :<br><br> |
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! 1) The light crosses a spherical refracting surface $`DS1`$ with characteristics : |
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! $`\overline{S_1C_1}=+|R|=+5\;cm`$ , $`n_{ini}=1`$ and $`n_{fin}=1.5`$. |
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! |
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! 2) Then the light is reflected at the surface of the last lensball interface that |
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! acts like a spherical mirror of characteristics : $`\overline{S_2C_2}=-|R|=-5\;cm`$, |
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! $`n=1.5`$. |
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! |
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! 3) Finally the light crosses back the first interface of the lensball that acts |
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! like a spherical refracting surface those characteristics are : |
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! $`\overline{S_3C_3}=+|R|=+5\;cm`$ , $`n_{ini}=1.5$ and $n_{fin}=1`$. |
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! |
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! Relative algebraic distances between the different elements of $`OS2`$ are : |
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! |
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! $`\overline{S_1S_2}=+10\;cm`$ and $`\overline{S_2S_3}=-10\;cm`$ |
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! |
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! </details> |
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! <details markdown=1> |
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! <summary> |
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! Which image is associated with each of the optical systems? |
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! </summary> |
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! |
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! * It is difficult to be 100% sure before having made the calculations. |
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! |
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! </details> |
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! <details markdown=1> |
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! <summary> |
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! Why do we had to take the picture in the darkness, with only the painting |
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! illuminated behind the camera, to obtain images 2 and 3 ? |
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! </summary> |
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! |
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! * At a refracting interface, part of the light incident power is refracted, |
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! and part is reflected. For transparent material like glass and for visible light, |
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! the part of the reflected power is small. If the room had been homogeneously |
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! illuminated, the images 2 and 3 of the painting on the wall behind the camera would |
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! have been faintly visible compared to the image of the front wall through the lensball. |
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! </details> |
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! </details> |
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! </details> |
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! *YOUR CHALLENGE* : Looking at a cathedral through a lensball. Can you predict your observation? |
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! |
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! _Skill tested : to know how to carry out calculations_ |
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! |
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!  |
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! |
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! *Discovery time : 2 hours*<br> |
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! *Resolution time : 30 minutes* |
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! <details markdown=1> |
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! <summary> |
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! I choose it |
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! </summary> |
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! A lensball is a polished spherical ball of radius $`R=5 cm`$, made of glass of refractive |
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! index value $`n_{glass}=1.5`$. The cathedral is 90 meters high, and you stand with the lensball |
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! 400 meters from the cathedral. You look at the cathedral through the lens, your eye being at |
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! 20 cm from the center of the lens. What would you expect to see? |
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! |
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! * _The resolution time is the typical expected time to be allocated to this problem_ |
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! _if it was part of an examen for an optics certificate_. |
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! |
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! * _The discovery time is the expected time you require to prepare this challenge_ |
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! _if you don't have practice. However, this is just an indication, take as much time as_ |
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! _you need. The time to question yourself serenely about how to handle the problem,_ |
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! _about the method of resolution and its validity, about some possible approximations_ |
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! _if they can be justified, and the time you need to check the equations if you have_ |
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! _not previously memorised them and to perfom the calculation, are important._ |
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! |
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! <details markdown=1> |
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! <summary> |
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! Ready to answer questions ? |
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! </summary> |
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! <!--question 1--> |
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! <details markdown=1> |
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! <summary> |
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! What is the scientifical framework you choose to study this problem ? |
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! </summary> |
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! * All the characteristic sizes in this problem are much bigger than the wavelength of |
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! the visible radiation ($`\lambda\approx5\mu m`$), so I deal with this problem in the |
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! framework of geometrical optics, and in the paraxial approximation in order to |
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! characterize the image. |
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! |
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! * The cathedral sustains an angle of $`arctan\dfrac{90}{400}=13°`$ from the lensball. |
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! This value seems reasonable to justify at first order the use of the paraxial approximation |
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! (_we usually consider that angles of incidence would not exceed 10° on the various simple |
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! optical element encountered between the objet (here the cathedral) and the final image |
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! (retina of the eye or matrix sensor of a camera_). |
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! </details> |
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! <!--question 2--> |
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! <details markdown=1> |
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! <summary> |
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! Describe the optical system for this use of the lensball. |
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! </summary> |
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! |
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! * The lensball breaks down into two refracting spherical surfaces sharing the same |
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! centre of curvature C and of opposite radius (in algebraic values). |
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! |
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! </details> |
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! <!--question 3--> |
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! <details markdown=1> |
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! <summary> |
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! What is your method of resolution ? |
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! </summary> |
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! * You don't use general equations 3a and 3b for a thick lens, they are too complicated |
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! to remind, and you don't have in m3p2 to "use" but to "build a reasoning". And you don't |
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! know at this step how to handle with centered optical systems. |
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! |
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! * But this system is simple, so you will calculate the image of the cathedral by the |
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! first spherical refracting surface $`DS_1`$ encountered by the light from the cathedral $`DS_1`$. |
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! Then this image becomes the object for the second spherical refracting surface $`DS_2`$ |
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! and so I can determine position and size of the final image. |
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! |
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! * For a spherical refracting surface, general equations are :<br><br> |
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! $`\dfrac{n_{fin}}{\overline{SA_{ima}}}-\dfrac{n_{ini}}{\overline{SA_{obj}}}=\dfrac{n_{fin}-n_{ini}}{\overline{SC}}`$ |
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! for the position.<br> |
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! $`\overline{M_T}=\dfrac{n_{ini}\cdot\overline{SA_{ima}}}{n_{fin}\cdot\overline{SA_{obj}}}`$ |
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! for the transverse magnification. |
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! |
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! </details> |
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! <!--question 4--> |
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! <details markdown=1> |
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! <summary> |
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! How do you set down your calculations? |
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! </summary> |
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! |
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! * The optical axis is the straight line that joins the center C of the lens to my eye, |
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! positively oriented in the direction of light propagation light for that observation, |
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! so from the cathedral to my eye. |
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! * First spherical refrating surface $`DS1`$ : $`\overline{S_1C_1}=+5\:cm`$, $`n_{ini}=1`$ (air) |
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! and $`n_{fin}=1.5`$ (glass).<br> |
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! Second spherical refrating surface $`DS2`$ : $`\overline{S_1C_1}=-5\:cm`$, $`n_{ini}=1.5`$ (glass) |
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! and $`n_{fin}=1`$ (air)<br> |
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! Distance between $`DS1`$ and $`DS2`$ vertices : $`\overline{S_1S_2}=+10\:cm`$<br> |
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! Object cathedral $`AB`$ : $`\overline{AB}=90\;m`$ and $`\overline{S_1A}=-400\;m`$<br> |
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! Let us write $`\overline{A_1B_1}`$ the intermediate image (the image of the cathedral |
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! given by $`DS1`$. |
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! |
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! * Specific equations for $`DS1`$ are :<br><br> |
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! $`\dfrac{1.5}{\overline{S_1A_1}}-\dfrac{1}{\overline{S_1A}}=\dfrac{0.5}{\overline{S_1C_1}}`$ (équ. DS1a), |
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! and $`\overline{M_T}=\dfrac{\overline{S_1A_1}}{1.5\cdot\overline{S_1A}}`$ (équ. DS1b)<br><br> |
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! Specific equations for $`DS2`$ are :<br><br> |
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! $`\dfrac{1}{\overline{S_2A'}}-\dfrac{1.5}{\overline{S_2A_1}}=-\dfrac{0.5}{\overline{S_2C_2}}`$ (équ. DS2a), and |
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! $`\overline{M_T}=\dfrac{1.5\cdot\overline{S_2A'}}{\overline{S_2A_1}}`$ (équ. DS2b)<br><br> |
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! The missing link between these two sets of equations is :<br> |
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! $`\overline{S_2A_1}=\overline{S_2S_1}+\overline{S_1A_1}=\overline{S_1A_1}-\overline{S_1S_2}`$. |
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! |
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! </details> |
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! <!--question 5--> |
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! <details markdown=1> |
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! <summary> |
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! Do you see some approximation that can be done ? |
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! </summary> |
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! * In the visible range, refractive index values of transparent material are in the range [1 ; 2], |
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! then the focal lengthes of a spherical refractive surface (object as well as image) are |
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! expected to remain in the same order of magnitude than the radius of curvature, |
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! so a few centimeters in this case (we talk in absolute value here). |
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! |
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! * We can if we want just check this fact for $`DS1`$ ($`|S_1C_1|=5\;cm`$) using équation DS1 :<br> |
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! \- considering $`\overline{S_1A_1}\longrightarrow\infty`$ to obtain the object focal length |
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! $`\overline{S_1F_1}`$} we get :<br> |
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! $`-\dfrac{1}{\overline{S_1F_1}}=\dfrac{0.5}{\overline{S_1C_1}}`$ |
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! $`\Longrightarrow=\overline{S_1F_1}=-10\;cm`$<br><br> |
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! \- considering $`\overline{S_1A}\longrightarrow\infty`$ to obtain the image focal length |
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! $`\overline{S_1F'_1}`$ we get :<br> |
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! $`\dfrac{1.5}{\overline{S_1F'_1}}=\dfrac{0.5}{\overline{S_1C_1}}\Longrightarrow\overline{S_1F'_1}=+15\;cm`$ |
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! |
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! * The distance of the cathedral from the lensball $`|\overline{S_1A}|=90\;m`$ is huge |
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! compared to the object focal length $`|\overline{S_1F_1}|=10\;cm`$, we can consider |
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! that the cathedral is at infinity from the lensball and so the image $`\overline{A_1B_1}`$ |
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! of the cathedral stands quasi in the image focal plane of $`DS1`$ : |
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! $`\overline {S_1A_1}=\overline {S_1F'_1}=+15cm`$. So we can directly use equation DS2a with :<br> |
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! $`\overline{S_2A_1}=\overline{S_2F'_1}=\overline{S_2S_1}+\overline{S_1F'_1}`$ |
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! $`=\overline{S_1F'_1}-\overline{S_1S_2}=+15-10=+5\;cm`$.. |
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! |
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! </details> |
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! <details markdown=1> |
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! <summary> |
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! Where is the image and how tall it is ? |
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! </summary> |
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! |
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! * To perform calculation, you must choose a unic lenght unit in your calculation, |
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! here $`cm`$ or $`m`$. We choose $`m`$ below. |
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! * Equation DS1a gives :<br> |
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! $`\dfrac{1.5}{\overline{S_1A_1}}-\dfrac{1}{-400}=\dfrac{0.5}{0.05}`$ $`\Longrightarrow\overline{S_1A_1}=0.15\;m`$<br> |
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! With more than 2 significant figures, your calculator would tell you $`0.150037`$, |
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! which nearly exactly the value of $`\overline{S_1F'_1}=+0.15\;m`$, so the approximation |
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! $`\overline{S_1A_1}=\overline{S_1F'_1}`$ you could have done is fully justified. |
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! |
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! * Equation DS2a gives :<br> |
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! $`\dfrac{1}{\overline{S_2A'}}-\dfrac{1.5}{-0.1+0.15}=\dfrac{-0.5}{-0.05}`$ |
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! $`\Longrightarrow\overline{S_2A'}=0.025\;m`$ |
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! |
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! * The final image is real, and stands 2.5 cm in front of the lensball in the side |
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! of your eye. Do not bring your eye or camera too close of the lensball \! |
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! |
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! * The size of an image (transversally to the optical axis) is given by the transverse |
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! magnification $`M_T`$. By Definition $`M_T`$ is the ratio of the algebraic size of |
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! the final image $`\overline{A'B'}`$ to the algebraic size of the initial object $`\overline{AB}`$. |
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! With an intermediate image, it can be break down :<br><br> |
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! $`M_T=\dfrac{\overline{A'B'}}{\overline{AB}}`$ |
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! $`=\dfrac{\overline{A'B'}}{\overline{A_1B_1}}\times\dfrac{A_1B_1}{\overline{AB}}`$<br><br> |
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! It is the product of the transverse magnifications of the cathedral introduced |
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! by the two spherical refracting surfaces of the lensball. <br><br> |
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! $`\overline{M_T}`$ introduced by $`DS1`$ is |
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! $`\overline{M_T}=\dfrac{\overline{S_1A_1}}{1.5\cdot\overline{S_1A}}`$ |
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! $`=\dfrac{+0.15}{1.5\times(-400)}=-0.00025`$<br><br> |
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! $`\overline{M_T}`$ introduced by $`DS2`$ is |
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! $`\overline{M_T}=\dfrac{1.5\cdot\overline{S_2A'}}{\overline{S_2A_1}}`$ |
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! $`=\dfrac{1.5\cdot\overline{S_2A'}}{\overline{S_1A_1}-\overline{S_1S_2}}`$ |
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! $`=\dfrac{1.5\cdot0.025}{+0.15-0.10} =0.75`$<br><br> |
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! So $`\overline{M_T}`$ introduced by the lensball is :<br><br> |
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! $`\overline{M_T}=-0.00025\times0.75`$ $`=-0.00019\approx-1.9\cdot10^{-4}`$<br><br> |
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! The image is $`\dfrac{1}{-1.9\cdot10^{-4}}\approx5300`$ smaller than the cathedral.<br><br> |
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! $`M_T=\dfrac{\overline{A'B'}}{\overline{AB}}\approx8\cdot10^{-4}`$ |
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! $`\Longrightarrow\overline{A'B'}=\overline{AB} \times M_T`$ |
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! $`=1.9\cdot10^{-4} \times 90\;m=-0.017\;m`$<br><br> |
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! The image is 1.7 cm height and it is reversed. |
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!</details> |
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! |
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!<!--question 7--> |
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!<details markdown=1> |
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!<summary> |
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! What is the apparent magnification of the cathedral ? |
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!</summary> |
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! * Apparent magnification = angular magnification = magnifying power. |
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! |
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! * As calculated previously, standing 400 metres from the cathedral, the 90 m heigh |
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! cathedral sustends the apparent angles of $`\alpha=arctan\left(\dfrac{90}{400}\right)=0.221\;rad=12.7°`$ |
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! at your eye.<br> |
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! <br> |
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! * The image of the cathedral is 1.7 cm heigth and is located between the lens |
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! (from its vertex $`S2`$) and your eyes and at 2.5cm from the lens. If your eye is |
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! 20cm away from the lens, so the distance eye-image is 17.5 cm (we use no algebraic values). |
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! Thus the image of the catedral subtends the apparent angle |
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! $`\alpha'=arctan\left(\dfrac{1.7}{17.5}\right)=0.097\;rad=5.6°`$ at your eye.<br> |
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! <br> |
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! * The apparent magnification $`M_A`$ of the cathedral throught the lensball for my |
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! eye in that position is<br> |
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! $`M_A=\dfrac{\alpha'}{\alpha}=\dfrac{0.097}{0.221}=0.44`$.<br><br> |
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! Taking into account that the image is reversed, the algebraic value of the apparent |
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! magnification is $`\overline{M_A}=-0.44`$.<br> |
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! <br> |
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! * You could obtained directly this algebraic value of $`M_A`$ by considering algebraic |
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! lengthes and angles values in the calculations :<br><br> |
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! $`\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}`$ |
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! $`=\dfrac {arctan\left(\frac{-0.017}{-0.175}\right)} {arctan\left(\frac{90}{-400}\right)}`$ $`=\dfrac{0.097}{-0.221}=-0.44`$ |
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! |
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! <br> |
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! _Cathedral of Orleans (France)_ |
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! </details> |
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! </details> |
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! </details> |
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! </details> |
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!! *BEYOND* : The gravitationnal lensball (or Einstein's ring), due to a black hole or a galaxy. |
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!! Similarities, and differences. |
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!! |
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!!  |
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!! |
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!! <details markdown=1> |
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!! <summary> |
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!! To see |
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!! </summary> |
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!! still to be done, in progress. |
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!! </details> |
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