Question 4
Durée : 8 mn
Note maximale : 8
Question
Posons \(\delta = \arrowvert r_{1} - r_{2}\arrowvert\) la différence de marche et \(\varphi = \frac{2 \pi \delta}{\lambda}\) la différence de phase correspondante. Etudier l'intensité \(I\) dans les cas suivants.
a. \(\delta = K \lambda\)
b. \(\delta = \left(K + \frac{1}{2}\right) \lambda\)
c. \(\delta = \left(K \pm \frac{1}{2} \right) \frac{\lambda}{2}\)
Solution
Posons : \(\varphi = \frac{2 \pi \delta}{\lambda} = \frac{2 \pi}{v T} \arrowvert r_{1} - r_{2}\arrowvert = \frac{\omega}{v}\arrowvert r_{1} - r_{2}\arrowvert\) et \(I = I_{1} + I_{2} + 2 \sqrt{I_{1}I_{2}} \cos \varphi\) ( 2 points )
si \(\delta = K \lambda\)
\(\varphi = 2 K \pi \qquad \cos \varphi = 1 \quad\textrm{ et }\quad I = I_{1} + I_{2} + 2 \sqrt{I_{1}I_{2}}\)
interférence constructive ( 2 points )
si \(\delta = \left(K + \frac{1}{2}\right) \lambda\)
\(\varphi = \left(2 K + 1 \right) \pi \quad \cos \varphi = -1 \quad \textrm{ et } \quad I = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}\)
interférence destructive ( 2 points )
si \(\delta = \left(K \pm \frac{1}{2}\right) \frac{\lambda}{2}\)
\(\varphi = \left(2 K \pm 1 \right) \frac{\pi}{2} \quad \cos \varphi = 0 \quad \textrm{ et } \quad I=I_{1} + I_{2}\)
addition des intensités ( 2 points )