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 )

  1. 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 )

  2. 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 )

  3. 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 )