Question 3
Durée : 5 mn
Note maximale : 7
Question
Calculer la dérivée partielle \(\frac{\delta A}{\delta i}\)
\(\sin i = n\sin r~~~~~\color{blue}(1)\\\color{black}\sin i' = n\sin r' ~~\color{blue}(2)\\\color{black}r+r'=A~~~~~~~~~~\color{blue}(3)\\\color{black}D=i+i'-A~~~~\color{blue}(4)\)
\(n,\) indice du prisme
Solution
Pour \(A\) et \(n\) constants, nous obtenons de même :
\(\color{blue}\begin{array}{r c l c}\cos i~di&=&n\cos r~dr&~~\color{red}\textrm{(1 point)}\\\cos i'di'&=&n\cos r'dr'&~~\color{red}\textrm{(1 point)}\\0&=&dr+dr'&~~\color{red}\textrm{(1 point)}\\dD&=&di+di'&~~\color{red}\textrm{(1 point)}\end{array}\)
d'où
\(\color{blue}\frac{\delta D}{\delta i} = 1-\frac{\cos r'\cos i}{\cos r\cos i'}~~\color{red}\textrm{(3 points)}\)
\(\sin i = n\sin r~~~~~\color{blue}(1)\\\color{black}\sin i' = n\sin r' ~~\color{blue}(2)\\\color{black}r+r'=A~~~~~~~~~~\color{blue}(3)\\\color{black}D=i+i'-A~~~~\color{blue}(4)\)
\(n,\) indice du prisme