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Réponse à la question 2.8 #5

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b7d24f2
réponse optique géométrique 2.8
jeher23 Mar 10, 2020
b9579b9
réponse Q2.8
jeher23 Mar 10, 2020
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16 changes: 16 additions & 0 deletions Optique/002-Optique Geometrique/Q002-008.md
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Expand Up @@ -8,3 +8,19 @@ Pour une lentille biconvexe diélectrique épaisse d’indice n et de rayons de

### Réponse

La matrice générale d'une lentille biconvexe est de la forme :
$$\begin{pmatrix}
1+\frac{d(1-n)}{nR_1}&\frac{d}{n}\\
(n-1)\left(\frac{1}{R_2}-\frac{1}{R_1}-\frac{d(n-1)}{nR_1R_2} \right)&\frac{d(n-1)}{nR_2} + 1
\end{pmatrix}$$

Dans notre cas, $R_1 = -R_2 = R$.
La focale dans la matrice ABCD précédente est $f = \frac{-1}{C}$. Si nous voulons $|{f}| = \infty$ nous avons alors:
$$\begin{align*}
C &= 0\\
0 &= (n-1)\left(\frac{1}{R_2}-\frac{1}{R_1}-\frac{d(n-1)}{nR_1R_2} \right)\\
0 &=\frac{-2}{R}-\frac{d(n-1)}{-nR^2}\\
\frac{2}{R} &= \frac{d(n-1)}{nR^2}\\
d &= \frac{2nR}{n-1}
\end{align*}$$