35-3 Interference and Double-Slit Intensity

35-3 Interference and Double-Slit Intensity#

Prompts

  • What is coherent light? Why must the two waves in Young’s experiment be coherent for an interference pattern to appear? What happens with incoherent sources?

  • For two waves with electric fields \(E_1 = E_0\sin(\omega t)\) and \(E_2 = E_0\sin(\omega t + \phi)\) arriving at a point, how does the intensity \(I\) depend on the phase difference \(\phi\)?

  • Sketch a phasor diagram for two waves of equal amplitude with phase difference \(\phi\). How do you find the amplitude of the resultant wave? Why does \(I = 4I_0\cos^2(\phi/2)\)?

  • Relate the phase difference \(\phi\) to the angle \(\theta\) and path length difference \(\Delta L\) in the double-slit setup. At what \(\phi\) do maxima and minima occur?

  • Why is the average intensity on the screen \(2I_0\) whether the sources are coherent or incoherent? What does interference do to the energy?

Lecture Notes#

Overview#

  • For an interference pattern to be visible, the two waves must be coherent—their phase difference must stay constant in time.

  • The intensity at a point on the screen is \(I = 4I_0\cos^2(\phi/2)\), where \(\phi\) is the phase difference and \(I_0\) is the intensity from one slit alone.

  • Phasors (section 16-6) provide a geometric way to add the electric fields and find the resultant intensity.

  • Interference redistributes energy; it does not create or destroy it. The average intensity is \(2I_0\) in both coherent and incoherent cases.

Coherence#

Coherent light: the phase difference between the two waves at any point remains constant in time.

  • In Young’s experiment, both slits are illuminated by the same wave from the single slit \(S_0\). The light at \(S_1\) and \(S_2\) is therefore coherent.

  • Incoherent light: phase difference varies randomly (e.g., two independent incandescent bulbs). The interference fluctuates rapidly between constructive and destructive; the eye averages over time and sees uniform illumination \(2I_0\)—no fringes.

Why a single slit?

The single slit in Young’s setup serves two roles: (1) it creates coherent light (waves from one small source); (2) diffraction spreads that light to illuminate both slits. Direct sunlight at the two slits would be incoherent—the slits are too far apart.

Lasers emit coherent light (atoms emit cooperatively). Ordinary sources (incandescent, fluorescent) are incoherent over typical slit separations.

Intensity in double-slit interference#

The electric fields at point \(P\) from the two slits are

(281)#\[ E_1 = E_0\sin(\omega t), \qquad E_2 = E_0\sin(\omega t + \phi) \]

where \(\phi\) is the phase difference (from path length difference). The intensity at \(P\) is

(282)#\[ I = 4I_0\cos^2\frac{\phi}{2} \]

Here \(I_0\) is the intensity that would appear on the screen from one slit when the other is covered (assumed uniform over the fringe region).

Phase difference and angle:

(283)#\[ \phi = \frac{2\pi}{\lambda}\,d\sin\theta = \frac{2\pi}{\lambda}\,\Delta L \]

So we can also write

(284)#\[ I = 4I_0\cos^2\left(\frac{\pi d\sin\theta}{\lambda}\right) \]

Maxima and minima from the intensity formula#

From \(I = 4I_0\cos^2(\phi/2)\):

Condition

\(\phi\)

\(d\sin\theta\)

Maximum (\(I = 4I_0\))

\(\phi = 2m\pi\)

\(m\lambda\)

Minimum (\(I = 0\))

\(\phi = (2m+1)\pi\)

\((m + \frac{1}{2})\lambda\)

These match the conditions from section 35-2. The intensity formula encodes both the locations and the gradual variation between bright and dark fringes.

Phasor derivation#

Two waves of amplitude \(E_0\) with phase difference \(\phi\) are represented by phasors of length \(E_0\) at angle \(\phi\) apart. Their vector sum has magnitude

(285)#\[ E = 2E_0\cos\frac{\phi}{2} \]

(From the isosceles triangle: the resultant bisects the angle \(\phi\); each half is \(E_0\cos(\phi/2)\), so \(E = 2E_0\cos(\phi/2)\).)

Since intensity \(I \propto E^2\) and \(I_0 \propto E_0^2\):

(286)#\[ \frac{I}{I_0} = \frac{E^2}{E_0^2} = 4\cos^2\frac{\phi}{2} \]

which gives Eq. (282).

Phasors for light

Same phasor method as section 16-6 (mechanical waves). For light, we add electric field phasors; intensity is proportional to \(E^2\).

Energy and average intensity#

Interference redistributes energy over the screen—it does not create or destroy it.

  • Coherent sources: \(I\) varies from \(0\) to \(4I_0\); average \(I_{\text{avg}} = 2I_0\) (from \(\langle\cos^2(\phi/2)\rangle = \frac{1}{2}\)).

  • Incoherent sources: no fringes; uniform \(I = 2I_0\) everywhere.

In both cases, the total power on the screen is the same. Coherent light concentrates it into bright fringes; incoherent light spreads it uniformly.

Combining more than two waves#

For three or more waves arriving at a point:

  1. Draw phasors for each wave, end to end, preserving the phase relations.

  2. The vector sum gives the resultant amplitude; its length squared (times a constant) gives the intensity.

This generalizes the two-phasor construction and is used for multiple slits (Chapter 36) and other interference configurations.

Summary#

  • Coherent light: constant phase difference; required for visible interference. Incoherent: no fringes, uniform \(2I_0\).

  • Intensity: \(I = 4I_0\cos^2(\phi/2)\) with \(\phi = (2\pi/\lambda)d\sin\theta\).

  • Phasors: add electric field phasors; resultant amplitude \(E = 2E_0\cos(\phi/2)\).

  • Energy: average intensity \(2I_0\); interference redistributes, does not create energy.