All Problems

Interference of Light

Problem 5.66

A certain oscillation results from the addition of coherent oscillations of the same direction ξk=acos⁡[ωt+(k−1)φ]\xi_{k}=a \cos [\omega t+(k-1) \varphi], where kk is the number of the osciliation (k=1,2,…,N),φ(k=1,2, \ldots, N), \varphi is the phase difference between the kth and (k−1)(k-1) th oscillations. Find the amplitude of the resultant oscillation.

Reveal Answer
5.66.5.66 . Let us represent the kk th oscillation in the complex form ξk=aθi[ωt+(k−1)φ]=ak∗eiωt \xi_{k}=a \theta^{i[\omega t+(k-1) \varphi]}=a_{k}^{*} \mathrm{e}^{i \omega t} where ak∗=aei(k−1)φa_{k}^{*}=a e^{i(k-1) \varphi} is the complex amplitude. Then the complex amplitude of the resulting oscillation is A∗=∑k=1Naθi(k−1)=a[1+eiΦ+ei2φ+…+ei(N−1)φ]==a(etφN−1)/(eiφ−1) \begin{aligned} A^{*}=\sum_{k=1}^{N} a \theta^{i(k-1)} &=a\left[1+\mathrm{e}^{i \Phi}+\mathrm{e}^{i 2 \varphi}+\ldots+\mathrm{e}^{i(N-1) \varphi}\right]=\\ &=a\left(\mathrm{e}^{t \varphi N}-1\right) /\left(\mathrm{e}^{\mathrm{i} \varphi}-1\right) \end{aligned} Multiplying A∗A^{*} by the complex conjugate value and extracting the square root, we obtain the real amplitude A=a1−cos⁡Nφ1−cos⁡φ=asin⁡(Nφ/2)sin⁡(φ/2) A=a \sqrt{\frac{1-\cos N \varphi}{1-\cos \varphi}}=a \frac{\sin (N \varphi / 2)}{\sin (\varphi / 2)}
Prev
Next