courses/01.curriculum/01.physics-chemistry-biology/03.niv3/02.geometrical-optics/05.paraxial-optics/02.paraxial-optics-simple-elements/03.lens/03.lens-beyond/annex.en.md

title	published	vissible	media_order
lensball : beyond	true	true	lentille-boule-orleans-1bis.jpg,lensball-brut-820-760.jpg,Einstein-ring-free.jpg

! YOUR CHALLENGE : Looking at a cathedral through a lensball. Can you predict your observation? ! ! ! ! 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. ! !

!

! Ready to answer M3P2 team questions ? !

! !

!

! 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) !

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!! BEYOND : The gravitationnal lensball (or Einstein's ring), due to a black hole or a galaxy. !! Similarities, and differences. !! !! !! !!

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!! To see !!

!! still to be done, in progress. !!

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!!! DO YOU MASTER ? !!!

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