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@ -169,10 +169,10 @@ You know $`\overline{SA_{obj}}`$, $`n_{inc}`$ and $`n_{eme}`$, you have previous |
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Positions of object focal point F and image focal point F’ are easily obtained from the conjunction equation (equ. 1). |
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* Image focal length $`\overline{OF'}`$ : $`\left(|\overline{OA_{obj}}|\rightarrow\infty\Rightarrow A_{ima}=F'\right)`$<br> |
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(equ.1)$`\Longrightarrow\dfrac{n_{eme}}{\overline{SF'}}=\dfrac{n_{eme}-n_{inc}}{\overline{SC}}\Longrightarrow\overline{SF'}=\dfrac{n_{eme}\cdot\overline{SC}}{n_{eme}-n_{inc}}`$ |
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(equ.1)$`\Longrightarrow\dfrac{n_{eme}}{\overline{SF'}}=\dfrac{n_{eme}-n_{inc}}{\overline{SC}}`$$`\Longrightarrow\overline{SF'}=\dfrac{n_{eme}\cdot\overline{SC}}{n_{eme}-n_{inc}}`$ |
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* Object focal length $`\overline{OF}`$ : $`\left(|\overline{OA_{ima}}|\rightarrow\infty\Rightarrow A_{obj}=F\right)`$<br> |
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(equ.1) $`\Longrightarrow-\dfrac{n_{inc}}{\overline{SF}}=\dfrac{n_{eme}-n_{inc}}{\overline{SC}}\Longrightarrow\overline{SF}=-\dfrac{n_{inc}\cdot\overline{SC}}{n_{eme}-n_{inc}} |
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(equ.1) $`\Longrightarrow-\dfrac{n_{inc}}{\overline{SF}}=\dfrac{n_{eme}-n_{inc}}{\overline{SC}}`$$`\Longrightarrow\overline{SF}=-\dfrac{n_{inc}\cdot\overline{SC}}{n_{eme}-n_{inc}} |
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`$ |
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##### 2 - Thin spherical refracting surface representation |
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