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@ -172,7 +172,7 @@ Sont proposées les catégories suivantes, mais à débattre, toutes les idées |
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! What is the optical system giving the image of the painting? |
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! </summary> |
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! <br> |
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! * The optical system is composed of two spherical refracting surfaces, centered on the same optical axis. |
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! * The optical system is composed of two spherical refracting surfaces, centered on the same optical axis.<br> |
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! <br> |
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! </details> |
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! <details markdown=1> |
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@ -187,11 +187,13 @@ Sont proposées les catégories suivantes, mais à débattre, toutes les idées |
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!* The first spherical refracting surface |
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! $`DS1`$ encountered by the light has |
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! the follwing characteristics :<br> |
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! $`\overline{S_1C_1}=+|R|=+5\;cm`$ , $`n_{ini}=1`$ and $`n_{fin}=1.5`$ |
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! $`\overline{S_1C_1}=+|R|=+5\;cm`$ , |
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! $`n_{ini}=1`$ and $`n_{fin}=1.5`$<br> |
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! <br> |
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! * The second spherical refracting surface $DS2$ |
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! encountered by the light has the follwing characteristics :<br> |
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! $`\overline{S_2C_2}=-|R|=-5\;cm`$ , $`n_{ini}=1.5`$ and $`n_{fin}=1`$ |
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! * The second spherical refracting surface |
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! $DS2$ encountered by the light has the follwing characteristics :<br> |
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! $`\overline{S_2C_2}=-|R|=-5\;cm`$ , |
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! $`n_{ini}=1.5`$ and $`n_{fin}=1`$ |
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! |
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! * Algebraic distance between $DS1$ and $DS2$ is : $`\overline{S_1S_2}=+10\;cm`$ |
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! |
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@ -458,24 +460,23 @@ Sont proposées les catégories suivantes, mais à débattre, toutes les idées |
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!<summary> |
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! What is the apparent magnification of the cathedral ? |
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!</summary> |
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! <br> |
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! * "apparent magnification" = "angular magnification" = "magnifying power". |
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! * apparent magnification = angular magnification = magnifying power. |
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! |
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! * As calculated previously, standing 400 metres from the cathedral, the 90 m heigh |
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! cathedral sustends the apparent angles of $`\alpha=arctan\left(\dfrac{90}{400}\right)=0.221\;rad=12.7°`$ |
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! at your eye. |
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! at your eye.<br> |
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! <br> |
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! * The image of the cathedral is 1.7 cm heigth and is located between the lens |
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! (from its vertex $`S2`$) and your eyes and at 2.5cm from the lens. If your eye is |
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! 20cm away from the lens, so the distance eye-image is 17.5 cm (we use no algebraic values). |
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! Thus the image of the catedral subtends the apparent angle |
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! $`\alpha'=arctan\left(\dfrac{1.7}{17.5}\right)=0.097\;rad=5.6°`$ at your eye. |
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! $`\alpha'=arctan\left(\dfrac{1.7}{17.5}\right)=0.097\;rad=5.6°`$ at your eye.<br> |
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! <br> |
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! * The apparent magnification $`M_A`$ of the cathedral throught the lensball for my |
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! eye in that position is<br> |
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! $`M_A=\dfrac{\alpha'}{\alpha}=\dfrac{0.097}{0.221}=0.44`$.<br><br> |
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! Taking into account that the image is reversed, the algebraic value of the apparent |
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! magnification is $`\overline{M_A}=-0.44`$. |
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! magnification is $`\overline{M_A}=-0.44`$.<br> |
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! <br> |
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! * You could obtained directly this algebraic value of $`M_A`$ by considering algebraic |
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! lengthes and angles values in the calculations :<br><br> |
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