Coherent and Incoherent Addition of Waves

What this subtopic covers

• NCERT 10.4. The superposition principle, coherence, the derivation of $I=4I_0\cos^2(\phi/2)$, and what happens instead when the sources are incoherent. Short in the textbook, but it supplies the algebra used in every YDSE intensity question.

Superposition principle

• At a particular point in a medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves separately.
• Everything in interference follows from this one statement.

Coherent sources — the definition to write

• Coherent sources are two sources that emit continuous light waves of the same frequency (and therefore the same wavelength), and are either exactly in phase or maintain a constant phase difference between them at all times.
• Two conditions, if the question asks for them as a list: (a) same frequency/wavelength; (b) constant (or zero) phase difference maintained at all times.
• The physical demonstration NCERT uses: two needles $S_1$ and $S_2$ dipping up and down identically in a water trough. At each point the phase difference between the two disturbances stays fixed in time, so the nodal and antinodal lines stay put.

Building the pattern from path differences

• Point $P$ on the perpendicular bisector, $S_1P=S_2P$: the two waves take the same time, arrive in phase, so with $y_1=a\cos\omega t$ and $y_2=a\cos\omega t$ we get $y=2a\cos\omega t$ and $I=4I_0$, where $I_0\propto a^2$ is the intensity from one source alone. This is constructive interference.
• Point $Q$ with $S_2Q-S_1Q=2\lambda$: a path difference of $2\lambda$ is a phase difference of $4\pi$, so $y_2=a\cos(\omega t-4\pi)=a\cos\omega t$ — again in phase, again $I=4I_0$.
• Point $R$ with $S_2R-S_1R=-2.5\lambda$: a path difference of $2.5\lambda$ is a phase difference of $5\pi$, so $y_2=a\cos(\omega t+5\pi)=-a\cos\omega t$. The two displacements cancel exactly and the intensity is zero. This is destructive interference.
• Summarising, for two coherent sources vibrating in phase and an arbitrary point $P$:

$$S_1P\sim S_2P=n\lambda\quad(n=0,1,2,\dots)\;\Longrightarrow\;\text{constructive},\;I=4I_0$$

$$S_1P\sim S_2P=\left(n+\dfrac{1}{2}\right)\lambda\quad(n=0,1,2,\dots)\;\Longrightarrow\;\text{destructive},\;I=0$$

• The symbol $\sim$ means the magnitude of the difference. The locus of points with a fixed path difference $0,\pm\lambda,\pm2\lambda,\dots$ is a family of hyperbolae, which is why the fringes on a distant screen look like straight bands.

Derivation of the intensity formula — the 3-mark core

• Let the displacements at a general point be

$$y_1=a\cos\omega t,\qquad y_2=a\cos(\omega t+\phi)$$

• where $\phi$ is the constant phase difference. By superposition,

$$y=y_1+y_2=a\left[\cos\omega t+\cos(\omega t+\phi)\right]$$

• Using $\cos C+\cos D=2\cos\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$,

$$y=2a\cos\left(\dfrac{\phi}{2}\right)\cos\left(\omega t+\dfrac{\phi}{2}\right)$$

• This is a harmonic wave of the same frequency, with resultant amplitude

$$A=2a\cos\left(\dfrac{\phi}{2}\right)$$

• Since intensity is proportional to the square of the amplitude, and writing $I_0=a^2$ for the intensity of each source alone,

$$A^2=4a^2\cos^2\left(\dfrac{\phi}{2}\right)\qquad\Longrightarrow\qquad I=4I_0\cos^2\left(\dfrac{\phi}{2}\right)$$

• Conditions read off this result:
• Constructive: $\cos^2(\phi/2)=1$, i.e. $\phi=0,\pm2\pi,\pm4\pi,\dots=2n\pi$, equivalently path difference $\Delta x=n\lambda$; $I_{\max}=4I_0$.
• Destructive: $\cos^2(\phi/2)=0$, i.e. $\phi=\pm\pi,\pm3\pi,\pm5\pi,\dots=(2n-1)\pi$, equivalently path difference $\Delta x=(2n-1)\dfrac{\lambda}{2}$; $I_{\min}=0$.
• The intensity therefore varies between $0$ and $4I_0$ — four times the intensity due to a single slit. That statement, with this expression, is exactly what a 5-mark "justify the statement" question wants.

Unequal amplitudes — the general result

• If the two waves have different amplitudes, take $y_1=a\sin\omega t$ and $y_2=b\sin(\omega t+\phi)$. Expanding and collecting,

$$y=\sin\omega t\,(a+b\cos\phi)+\cos\omega t\,(b\sin\phi)$$

• Put $a+b\cos\phi=R\cos\theta$ and $b\sin\phi=R\sin\theta$, so that $y=R\sin(\omega t+\theta)$. Squaring and adding,

$$R^2=(a+b\cos\phi)^2+(b\sin\phi)^2=a^2+b^2+2ab\cos\phi$$

$$R=\sqrt{a^2+b^2+2ab\cos\phi}$$

• With $I_1=Ka^2$, $I_2=Kb^2$ and $I_R=KR^2$,

$$I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$$

• The third term is the interference term. Maxima and minima:

$$I_{\max}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2,\qquad I_{\min}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$$

$$\dfrac{I_{\max}}{I_{\min}}=\dfrac{(a_1+a_2)^2}{(a_1-a_2)^2}$$

• Since intensity is proportional to slit width, $\dfrac{w_1}{w_2}=\dfrac{I_1}{I_2}=\dfrac{a_1^2}{a_2^2}$.

Incoherent addition

• If the two sources do not hold a constant phase difference — if $\phi$ changes rapidly and randomly with time — then the positions of maxima and minima shift too fast to be seen and the eye or detector records a time average.
• Since $\langle\cos\phi\rangle=0$ over random phases, the interference term drops out:

$$\langle I_R\rangle=I_1+I_2\qquad\text{and, for equal sources,}\qquad I=2I_0$$

• The intensities simply add, everywhere, and the screen is uniformly illuminated. This is what happens when two separate light sources illuminate a wall.

The comparison worth memorising

Standard results, collected

• Coherent sources: same frequency, constant phase difference.
• $A=2a\cos(\phi/2)$; $I=4I_0\cos^2(\phi/2)$ with $I_0=a^2$.
• Constructive at $\phi=2n\pi$ or $\Delta x=n\lambda$; destructive at $\phi=(2n-1)\pi$ or $\Delta x=(2n-1)\lambda/2$.
• $\Delta\phi=\dfrac{2\pi}{\lambda}\Delta x$ converts between the two languages.
• $I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$ for unequal sources; incoherent limit $I_R=I_1+I_2$.
• $\dfrac{I_{\max}}{I_{\min}}=\left(\dfrac{a_1+a_2}{a_1-a_2}\right)^2$ and $\dfrac{w_1}{w_2}=\dfrac{a_1^2}{a_2^2}$.