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: enter image description here

This is how I solved:

enter image description here

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.

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

1 Answer 1


The author is correct. The second ray has two phase differences:

  1. Lag from additional distance traveled
  2. Lag from initial phase difference

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.

This seems counterintuitive, but is actually true, because going forward in space in the direction of propagation is equivalent to finding the waves that left earlier (i.e. going backward in time). So an increase in distance has the same effect as adding a time delay. Therefore #1 can also be seen as a time delay, thus you have two time delays which add up.

  • $\begingroup$ @Vincet Thacker. Thank you very much for providing me with a deep insight on how different rays lag or lead with respect to other waves. $\endgroup$ Jun 17, 2021 at 5:33

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