Update textbook.fr.md

12.temporary_ins/02.arithmetic-number-theory/10.n1/10.integers-and-their-representation/10.main/textbook.fr.md

@@ -69,6 +69,40 @@ Il sait que si il ne comprends pas un mots, il y a des explications de vocabulai
 !!!!! ![](graduated-axis_L800.jpg)
 !!!!! </details>
 
+!<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> 
+! $`\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}`$
+! $`=\dfrac  {arctan\left(\frac{-0.017}{-0.175}\right)} {arctan\left(\frac{90}{-400}\right)}`$ $`=\dfrac{0.097}{-0.221}=-0.44`$
+!
+! ![](graduated-axis_L800.jpg)<br>
+! _Cathedral of Orleans (France)_
+! </details>
+! </details>
+! </details>
+! </details>
+