---
title: 'virtual object : beyond'
published : true
visible : true
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! *YOUR  CHALLENGE*  :  Looking at a cathedral through a lensball. Can you predict your observation?
!
! ![](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 $DS1$ : $S_1C_1=+5\:cm$, $n_{ini}=1$ (air) and $n_{fin}=1.5$ (glass).<br>
!  Second spherical refrating surface $DS2$ : $S_1C_1=-5\:cm$, $n_{ini}=1.5$ (glass) and $n_{fin}=1$ (air)<br>
!  Distance between $DS1$ and $DS2$ vertices : $S_1S_2=+10\:cm$<br>
!  Object cathedral $AB$ :  $AB=90\;m$ and $S_1A=-400\;m$<br>
!  Let us write $A_1B_1$ 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 $A_1B_1$ 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>

!!!! *DIFFICULT POINT* (contribute, or indicate a difficult point of understanding)
!!!!

!! *CULTURAL POINT* (contributor)
!!

!!! *DO  YOU  MASTER ?*
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