Interference and Diffraction of Light

Lessons

Senior 5 · Physics

Interference and Diffraction of Light

4 lessons

Lesson 3 of 4

The Intensity Distribution of a Fringe Pattern

25 min

How bright is a fringe?

Young's formula tells you where the bright and dark fringes are, but not how the brightness changes in between. A light meter moved slowly across the pattern shows that the brightness rises and falls smoothly, and that the bright fringes are much brighter than you might expect: twice as bright as the two slits' light simply added together. The calculation that explains this is the same one engineers use to combine radio signals from two aerials and to design the anti-reflection coatings on spectacles and camera lenses.

Adding the two waves

At a point P on the screen, let the electric field of the wave from each slit have the same amplitude $E_{0}$ , and let the wave from slit 2 lag by a phase $\phi$ , caused by the path difference:

E_1 = E_0\sin(\omega t), \qquad E_2 = E_0\sin(\omega t + \phi), \qquad \phi = \frac{2\pi}{\lambda}d\sin\theta

By superposition, and using $\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2}$ ,

E = E_1 + E_2 = 2E_0\cos\frac{\phi}{2}\,\sin\left(\omega t + \frac{\phi}{2}\right)

The resultant is a wave of the same frequency, with amplitude $2E_0\cos\frac{\phi}{2}$ . The same result comes from adding the two waves as rotating arrows, phasors, of length $E_{0}$ with an angle $\phi$ between them:

Intensity is proportional to the square of the amplitude. If one slit alone gives intensity $I_0 \propto E_0^2$ , then

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

For small angles, $\sin\theta \approx \frac{y}{D}$ , and using $w = \frac{\lambda D}{d}$ , this becomes

I = 4I_0\cos^2\left(\frac{\pi y}{w}\right)

At the bright fringes, $\phi = 0, 2\pi, 4\pi, \ldots$ and $I = 4 I_{0}$ : four times one slit, because the amplitudes add and intensity goes as amplitude squared.
At the dark fringes, $\phi = \pi, 3\pi, \ldots$ and $I = 0$ .
The average of $\cos^2$ over a cycle is $\frac{1}{2}$ , so the average intensity is $2 I_{0}$ : exactly the sum of the two slits. Interference redistributes energy across the screen; it neither creates nor destroys it.

Definition

The intensity distribution of a double-slit pattern is $I = 4I_0\cos^2\left(\frac{\phi}{2}\right)$ , where $I_{0}$ is the intensity from one slit alone and $\phi = \frac{2\pi d\sin\theta}{\lambda}$ is the phase difference. A phasor is a rotating arrow whose length is the amplitude of a wave and whose angle is its phase; waves are added by adding their phasors tip to tail. Example: a quarter of a fringe from the centre, $\phi = \frac{\pi}{2}$ and $I = 4I_0\cos^2 45^\circ = 2I_0$ .

Worked example 1: intensity between the fringes

In the pattern of Lesson 2 (fringe spacing $w = 3.6$ mm), each slit alone gives intensity $I_{0}$ at the screen. Find the intensity (a) at the centre, (b) 1.0 mm from the centre, (c) 1.8 mm from the centre.

Solution

(a) $y = 0$ : $I = 4 I_{0}$ .

(b) $I = 4I_0\cos^2\left(\frac{\pi(1.0)}{3.6}\right) = 4I_0\cos^2(50^\circ) = 4I_0(0.413) = 1.7I_0$ .

Worked example 2: where is the intensity a quarter of the maximum?

Find the smallest path difference, in wavelengths, at which the intensity falls to $I_{0}$ , a quarter of its maximum.

Solution

$4I_0\cos^2\frac{\phi}{2} = I_0$ gives $\cos\frac{\phi}{2} = \frac{1}{2}$ , so $\frac{\phi}{2} = 60^\circ$ and $\phi = 120^\circ = \frac{2\pi}{3}$ .

Path difference $= \frac{\phi}{2\pi}\lambda = \frac{\lambda}{3}$ .

Unequal slits and real patterns

If the two slits give amplitudes $E_{1}$ and $E_{2}$ that are not equal, the phasors no longer cancel completely. The bright fringes have intensity proportional to $(E_{1} + E_{2})^{2}$ and the dark fringes $(E_{1} - E_{2})^{2}$ , so the dark fringes are not quite dark and the contrast of the pattern falls.

In a real experiment the bright fringes also do not all have the same height. Each slit has a finite width, so the light from each spreads out by diffraction (Lesson 4), and the cos² pattern is multiplied by the single-slit diffraction pattern: the fringes are brightest at the centre and fade away to the sides.

Worked example 3: one slit partly covered

One slit is partly covered so that the amplitude of its light is half that of the other. Find the ratio of the maximum to the minimum intensity in the pattern.

Solution

Amplitudes in the ratio $2 : 1$ . Maximum $\propto (2 + 1)^2 = 9$ ; minimum $\propto (2 - 1)^2 = 1$ . The ratio is 9 : 1. The dark fringes now have a ninth of the brightness of the bright ones, instead of zero.

Warning

Do not add intensities. For coherent waves you must add the amplitudes (with their phases) and then square. Adding intensities, $I_{0} + I_{0} = 2 I_{0}$ , gives only the average; it is what happens with incoherent sources such as two lamps.

Tip

Phase difference first. Every intensity question becomes easy once you have $\phi$ : convert path difference with $\phi = \frac{2\pi\Delta}{\lambda}$ , or position with $\phi = \frac{2\pi y}{w}$ , then use $4I_0\cos^2\frac{\phi}{2}$ .

Intensity in YDSE (Visual method-phasors), Khan Academy India. Watch how adding two phasors gives the resultant amplitude 2E₀cos(φ ÷ 2) and the intensity 4I₀cos²(φ ÷ 2).

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Practice

Show that the sum of $E_0\sin(\omega t)$ and $E_0\sin(\omega t + \phi)$ has amplitude $2E_0\cos\frac{\phi}{2}$ .
Why is the maximum intensity in a double-slit pattern four times, not twice, the intensity from one slit?
Find the intensity, in terms of $I_{0}$ , where the phase difference is (a) $\frac{\pi}{3}$ , (b) $π \pi$ , (c) $2\pi$ .
Find the intensity at a point where the path difference is $\frac{\lambda}{4}$ .
In a pattern with fringe spacing 2.0 mm, find the intensity 0.50 mm from the central maximum.
The two slits give amplitudes in the ratio 3 : 1. Find the ratio of maximum to minimum intensity.
Show that the average intensity over one fringe is $2 I_{0}$ , and explain what this tells you about energy.
A student uses two separate identical lamps instead of two slits and predicts fringes with $I_{\max} = 4I_0$ . Evaluate this prediction.

Show answersHide answers

$\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2}$ with $A = \omega t + \phi$ , $B = \omega t$ : $2E_0\cos\frac{\phi}{2}\sin\left(\omega t + \frac{\phi}{2}\right)$ .
The amplitudes add, giving $2 E_{0}$ ; intensity is proportional to amplitude squared, $(2 E_{0})^{2} = 4 E_{0}^{2}$ .
(a) $4I_0\cos^2 30^\circ = 3I_0$ ; (b) 0; (c) $4 I_{0}$ .
$\phi = \frac{\pi}{2}$ : $I = 4I_0\cos^2 45^\circ = 2I_0$ .
$I = 4I_0\cos^2\left(\frac{\pi(0.50)}{2.0}\right) = 4I_0\cos^2 45^\circ = 2I_0$ .
$(3 + 1)^{2} : (3 - 1)^{2} = 16 : 4 = 4 : 1$ .
The average of $\cos^2$ over a cycle is $\frac{1}{2}$ , so $\overline{I} = 4I_0 \times \frac{1}{2} = 2I_0$ , the sum of the two slits' intensities. Energy is redistributed from the dark fringes to the bright ones, not created or destroyed.
Wrong. Separate lamps are incoherent: their phase difference changes randomly about $10^{8}$ times a second, so the cos² term averages to $\frac{1}{2}$ everywhere. The screen is uniformly lit at $2 I_{0}$ , with no fringes.

Key points

Coherent waves: add amplitudes with phases. Two equal waves give amplitude $2E_0\cos\frac{\phi}{2}$ .
$I = 4I_0\cos^2\frac{\phi}{2} = 4I_0\cos^2\frac{\pi y}{w}$ : maxima $4 I_{0}$ , minima 0, average $2 I_{0}$ .
Unequal amplitudes: $I_{\max} \propto (E_1 + E_2)^2$ , $I_{\min} \propto (E_1 - E_2)^2$ : lower contrast.
Real fringes fade to the sides because each slit's light is diffracted.

Sources and credits

Video: Intensity in YDSE (Visual method-phasors), Khan Academy India - English, embedded from YouTube.

Written by Ganzaa. The intensity graph (calculated) and the phasor diagram (drawn for $\phi = 60^\circ$ ), the worked examples and the practice exercises are by Ganzaa.

1How bright is a fringe?

2Adding the two waves

3Unequal slits and real patterns

How bright is a fringe?

Adding the two waves

Unequal slits and real patterns