Update annex.es.md

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.es.md

@@ -80,14 +80,14 @@ media_order: 'lentille-boule-orleans-1bis.jpg,lensball-brut-820-760.jpg,Einstein
 !</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>
+! * 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>
-! $\overline{S_2A_1}=\overline{S_2F'_1}=\overline{S_2S_1}+\overline{S_1F'_1}$ $=\overline{S_1F'_1}-\overline{S_1S_2}=+15-10=+5\;cm$..
+! * 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-->
@@ -95,27 +95,27 @@ media_order: 'lentille-boule-orleans-1bis.jpg,lensball-brut-820-760.jpg,Einstein
 !<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.
+! * 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.
+! $`\dfrac{1.5}{\overline{S_1A_1}}-\dfrac{1}{-400}=\dfrac{0.5}{0.05}`$ $`\Longrightarrow\overline{S_1A_1}=0.15\;m`$<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>
-! $\dfrac{1}{\overline{S_2A'}}-\dfrac{1.5}{-0.1+0.15}=\dfrac{-0.5}{-0.05}$ $\Longrightarrow\overline{S_2A'}=0.025\;m$
+! $`\dfrac{1}{\overline{S_2A'}}-\dfrac{1.5}{-0.1+0.15}=\dfrac{-0.5}{-0.05}`$ $`\Longrightarrow\overline{S_2A'}=0.025\;m`$
 !
 ! * 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>
+! * 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>
+! $`\overline{M_T}`$ 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>
 !
@@ -127,15 +127,15 @@ media_order: 'lentille-boule-orleans-1bis.jpg,lensball-brut-820-760.jpg,Einstein
 ! 
 ! * "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.
+! * 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 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 $\overline{M_A}=-0.44$.
+! * 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> $\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}$ $=\dfrac  {arctan\left(\frac{-0.017}{-0.175}\right)} {arctan\left(\frac{90}{-400}\right)}$ $=\dfrac{0.097}{-0.221}=-0.44$
+! * 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>