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Claude Meny 5 years ago
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      00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md

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00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md

@ -183,7 +183,8 @@ Sont proposées les catégories suivantes, mais à débattre, toutes les idées
! * The first spherical refracting surface $`DS1`$ encountered by the light has
! the follwing characteristics :<br>
! $`\overline{S_1C_1}=+|R|=+5\;cm`$ , $`n_{ini}=1`$ and $`n_{fin}=1.5`$
! * The second spherical refracting surface $DS2$ encountered by the light has the follwing characteristics :<br>
! * The second spherical refracting surface $DS2$ encountered by the light
! has the follwing characteristics :<br>
! $`\overline{S_2C_2}=-|R|=-5\;cm`$ , $`n_{ini}=1.5`$ and $`n_{fin}=1`$
!
! * Algebraic distance between $DS1$ and $DS2$ is : $`\overline{S_1S_2}=+10\;cm`$
@ -451,25 +452,25 @@ Sont proposées les catégories suivantes, mais à débattre, toutes les idées
!<summary>
! What is the apparent magnification of the cathedral ?
!</summary>
!
! <br>
! * "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>
! * 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>
! * 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>
! * You could obtained directly this algebraic value of $`M_A`$ by considering algebraic
! lengthes and angles values in the calculations :<br><br>
! $`\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}`$

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