From 5bd923353b248aab2183faa63286457e962ccc0b Mon Sep 17 00:00:00 2001 From: Claude Meny Date: Wed, 3 Feb 2021 15:19:53 +0100 Subject: [PATCH] Update annex.fr.md --- .../instructions-for-levels/30.beyond/annex.fr.md | 11 ++++++----- 1 file changed, 6 insertions(+), 5 deletions(-) diff --git a/00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md b/00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md index b53d46f8c..ee80a3652 100644 --- a/00.brainstorming-pedagogical-teams/45.synthesis-structuring/instructions-for-levels/30.beyond/annex.fr.md +++ b/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 :
! $`\overline{S_1C_1}=+|R|=+5\;cm`$ , $`n_{ini}=1`$ and $`n_{fin}=1.5`$ -! * The second spherical refracting surface $DS2$ encountered by the light has the follwing characteristics :
+! * The second spherical refracting surface $DS2$ encountered by the light +! has the follwing characteristics :
! $`\overline{S_2C_2}=-|R|=-5\;cm`$ , $`n_{ini}=1.5`$ 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 ! ! 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 :

! $`\overline{M_A}=\dfrac{\overline{\alpha'}}{\overline{\alpha}}`$