# Finding the phase difference of two interfering waves

A system illustrated in the given figure consists of two coherent point sources $$1$$ and $$2$$ located in a certain plane so that their dipole moments are oriented at right angles to that plane. The sources are separated by a distance d, the radiation wavelength is equal to $$\lambda$$. Taking into account that the oscillations of source $$2$$ lag in phase behind the oscillations of source $$1$$ by $$\phi$$ ($$\phi<\pi$$), find the angles $$\theta$$ at which the radiation intensity is maximum: This is how I solved: From the figure given above the two waves interfering at P have covered different distances $$S_1P=x$$ and $$S_2P=x+\Delta x$$. The electric fields at P due to the two waves may be written as $$E_1=E_0\sin(kx-\omega t)$$ $$and\enspace \enspace ^*E_2=E_0\sin(k(x+\Delta x)-\omega t)$$ $$^*$$But we know that $$E_2$$ is initially behind $$E_1$$ by a phase $$\phi$$.So the actual equation of electric field at point P becomes: $$E_2=E_0\sin(k(x+\Delta x)-\omega t-\phi)$$ $$\implies E_2=E_0\sin(kx-\omega t+k\Delta x-\phi)$$ Hence the net phase difference at P will be $$\phi_{net}=(kx-\omega t+k\Delta x-\phi)-(kx-\omega t)$$ $$\phi_{net}=k\Delta x -\phi$$.But we the that $$\Delta x=S_2A=d\sin(\frac{\pi}{2}-\theta)=d\cos(\theta)$$ $$\phi_{net}=\left(\frac{2\pi }{\lambda}\right)d\cos(\theta)-\phi$$ For the intensity maxima $$\phi_{net}=2n\pi$$ $$\implies cos(\theta)=\frac{n+\frac{\phi}{2\pi}}{\frac{d}{\lambda}}$$ But my answer is slightly different from the answer provided in the text. I cannot figure out any flaw in my method. Following in the author's solution:

The phase difference produced due to the extra path is $$\phi_1=\frac{2\pi}{\lambda}d\cos(\theta)$$.
But the interfering waves has an additional phase difference given as $$\phi_2=\phi$$. Hence $$\phi_{net}=\phi_1+\phi_2$$ $$for\enspace maxima\enspace 2n\pi=\phi_1+\phi_2$$ $$\implies 2n\pi=\left(\frac{2\pi }{\lambda}\right)d\cos(\theta)+\phi$$ $$\implies cos(\theta)=\frac{n-\frac{\phi}{2\pi}}{\frac{d}{\lambda}}$$

Here the author's explanation does't seem logical to me as he has not considered the fact ray $$S_2P$$ is lagging behind the ray$$S_1P$$ and rather just added the two phase angles.
I would be glad to know if I am correct or wrong, with an explanation for it.

• "the ray 𝑆2𝑃 will be ahead of 𝑆1𝑃" No; it will be behind, as it has to travel a longer path. Jun 14, 2021 at 13:32

So the phase differences add. You have made a sign error when introducing the phase difference $$\phi$$ because $$t$$ is already negative. So you want to have $$-\omega(t - T)$$ in the second ray so that it is delayed by a time $$T$$. Therefore $$\phi = (-\omega)(-T)$$ is positive.