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10A SUPERPOSITION NO DEFINITIONS. JUST PHYSICS.

10.1.1 Constructive and Destructive Interference .............................................................................. 2

10.1.2 Phase Difference ................................................................................................................ 4

10.1.3 Path Difference ................................................................................................................... 6

10.2.1 Ripple Tank Interference Pattern ........................................................................................... 9

10.2.2 Interference Pattern Spacing ............................................................................................ 14

10.2.3 Necessary Conditions for Interference .............................................................................. 16

10.3.1 Double Slit Interference Pattern ........................................................................................... 18

10.3.2 Derivation of the dsinθ formula ......................................................................................... 21

10.3.3 Fringe Separation ............................................................................................................. 23

10.3.4 Young’s Double Slit Experimental Set-up ......................................................................... 25

10.4.1 Diffraction Grating Interference Pattern ................................................................................ 27

10.4.2 White Light Interference Pattern ....................................................................................... 33

10.5.1 Single-Slit Interference Pattern ............................................................................................ 35

10.5.2 Diffraction ......................................................................................................................... 40

10.5.3 Rayleigh’s Criterion .......................................................................................................... 41

10.5.4 Diffraction Envelope ......................................................................................................... 44

10.A N-slit Interference ................................................................................................................... 46

10.B Single Slit D.I. ......................................................................................................................... 51

Online resources are provided at https://xmphysics.com/super

https://xmphysics.com/super


10.1.1 Constructive and Destructive Interference

When two waves “collide”, they don’t rebound off each other. They don’t even slow down or change

direction. They just pass right through each other, fully intact, as if nothing ever happened.

see video at xmphysics.com

But something did happen when they were overlapping. When waves overlap, they superpose to

become one resultant wave. Superposition is closely related to the phenomenon called interference.

When the superposition results in a larger wave, (e.g. when two positive pulses superpose), the

interference is said to be constructive. When the superposition results in a smaller wave, (e.g. when

a positive pulse superpose with a negative pulse), the interference is said to be destructive.


So what determines the resultant wave? Well, the outcome of superposition is governed by the

Principle of Superposition, which states that when waves overlap, the resultant wave’s

displacement is equal to the summation of all the overlapping waves’ displacement. The example on

the following page should help you confirm your understanding.


http://www.xmphysics.com/10.1.1





resultant two waves overlap

0

0

0

0

0

0

0

0

0

0

0

0

0

0


10.1.2 Phase Difference

In the previous section, we saw how pulses can superpose. Actually, we are more interested in the

superposition of continuous sinusoidal waves (drawn in red and blue below).

It turns out that when two sinusoidal waves (of the same period) superpose, the resultant wave is yet

another sinusoidal wave (with the same period). If the amplitude of each superposing waves is A,

then the amplitude of the resultant wave can range from 0 to 2A, depending on the phase difference

(between the two sinusoidal waves).

If the phase difference between the two sinusoidal waves is 0, 2, 4, 6 and so on, the two waves

will be exactly aligned to each other. The two waves are said to superpose in-phase. The outcome

of the constructive interference is a resultant wave of amplitude 2A (shown in magenta).

constructive interference

If the phase difference between the two sinusoidal waves is , 3, 5, 7 and so on, the two waves

will line up exactly opposite to each other. The technical jargon is that they superpose completely out

of phase, a.k.a in antiphase. The outcome of the destructive interference is a resultant wave of

amplitude 0.

resultant



destructive interference

What if the phase difference is some in-between values? Well, the outcome will still be a sinusoidal

wave, but the amplitude will be between 0 and 2A. For example, when the phase difference is 0.5,

the resultant amplitude is 2A . Math buffs will eagerly remind us that

sin sin( 0.5 ) 2 sin( 0.25 )x x x . The sin( )R formula, they will say.

“In-between” interference see slides at xmphysics.com

Do realize that when summing (superposing) waves, we sum up the displacement, not the amplitude

nor the intensity. When two waves are in-phase or in antiphase, the summation or difference of the

original waves’ amplitudes happen to be equal to the resultant wave’s amplitude. But these are the

exceptions rather than the norm. In general, you have to sum up the displacement point by point,

instant by instant.

resultant

resultant



10.1.3 Path Difference

Imagine you make a speaker play a monotone. You walk around the speaker and hear the same

boring monotone everywhere. Now you find another speaker and make it play the exact same

monotone. Is the sound twice as loud now as before? You are not quite sure, but as you walk around,

you distinctly hear the tone alternates between louds and softs. There are locations where having two

speakers actually produce a quieter sound, or even silence.

What on earth is happening? Let’s think through the situation. Let’s assume that the sound waves

from speaker A and speaker B are in phase with each other at the time when they leave the speakers.

However, these two sound waves take two different paths, and therefore travel two different distances,

before they arrive at your position P. The so called path difference | AP BP | , will introduce a

phase difference between the two waves when they arrive at P. This means that we can have

constructive interferences at some locations but destructive interferences at other locations.

Let’s use a few examples.

Let’s have A, B and P positioned such that AP and

BP correspond to 1λ and 3λ respectively. This

means that the wave from B travels a longer

distance of | 3 | 2 to arrive at P (compared

to A).

So two waves are arriving at P, but the one from

B is lagging the one from A by two complete

cycles, which makes the two waves exactly

aligned (upon arrival at P). So as far as P is

concerned, there is a superposition of two waves

which are in-phase with each other. Constructive

interference will occur, resulting in a loud sound being heard at P.

A

B

= 2.0

P

P

A

B

1

3

P



How about the above scenario? Wave A travels | 2 0.5 | 1.5 longer than wave B before arriving

at P. So wave A will arrive 1.5 cycles behind wave B, which makes them line up exactly opposite to

each other at P. As far as P is concerned, there is a superposition of two waves which are in antiphase

with each other. Destructive interference between the two waves results in silence at P.

One more example. Wave A travels | 2.25 1.75 | 0.5 longer than wave B before arriving at P.

The path difference of 0.5 introduces a phase difference of π radians between the waves arriving at

P. Destructive interference between the two waves results in silence at P.

In a nutshell, the path difference determines the phase difference of the waves arriving at the

destination. It is thus the quantity to evaluate if we want to know whether the interference at a location

is constructive interference (C.I.) or destructive interference (D.I.). For two in-phase wave sources,

C.I. occurs wherever n , 0,1,2...n

D.I. occurs wherever 1

( )2

n , 0,1,2...n

Do you want to experience first hand such an interference phenomenon? Just get two handphones

to emit the same tone (4000 Hz is good). Then move your head from side to side. You should be able

A

B

= 1.5

P

A

B 0.5

2

P

P

A

B

= 0.5λ A

B

1.75

2.25

P

P

P


to loud and soft sounds at different spots! The quiet spots are where the two sound waves undergo

destructive interference!




10.2.1 Ripple Tank Interference Pattern

If you can have a swimming pool all to yourself and the water is completely calm, you should try this.

Dip one of your feet into the water at constant intervals. That will send out ripples and cause the water

surface to oscillate everywhere. Nice. But why not double the fun by dipping both feet? With two wave

sources instead of one, would the water surface oscillate twice as vigorously as before?


Traditionally, the two-source interference pattern is demonstrated by using a ripple tank, which

comprises a shallow tank of water and two dippers motorized to produce two wave sources of chosen

frequencies.


Besides the expanding circles, there is an unmistakable pattern formed on the water surface. Most

strikingly, there are lines (on the surface) along which the water is now completely calm (despite being

hit by two waves). These lines are called the nodal lines. If you look very carefully, the locations

where the maximum amplitude of oscillation occur also form lines (between each pair of nodal lines).

These lines are called the antinodal lines.





The easiest way to understand how this pattern is formed is to try to construct the pattern ourselves.

Let’s start by drawing

concentric circles centred at

two dippers A and B. These

circles are meant to represent

the wave crests (of the

outward propagating waves).

So the distance between two

circular crests corresponds to

one wavelength.

Next we mark out the points

where the crest lines

intersect, and join those

points up with smooth curves.


These are obviously the

antinodal lines. Because the

path difference can only be

n at locations where “crest-

meets-crest”.

More specifically, the one at

the middle is called the 0th

order antinodal line (aka

0n antinodal line), since

these are locations with path

difference of 0λ.

On either side of the 0n

antinodal line, we have the

1n , or 1st order antinodal

lines, where the path

difference is exactly 1.


Now let’s add in the dashed

circles that represent the

wave troughs. This time, we

look for points where a crest

line intersects with a trough

line. Join these points up with

a smooth curve.

These are the nodal lines.

Because the path difference

must be 1

( )2

n at locations

where “crest-meets-trough”.

There is always be one nodal

line in between each pair of

antinodal lines.


Voila. We have mapped out

the antinodal and nodal lines,

and they tally exactly with the

pattern that is observed in the

ripple tank. How satisfying.


10.2.2 Interference Pattern Spacing

Before we start to perform calculations, it is useful to develop a “gut feel” of how the wavelength

and the distance between the wave sources d affect the appearance of the two-source interference

pattern.

As the pictures below illustrate, for the same d, a longer results in a more “spaced-out” pattern. Not

only is the spacing between the lines (both nodal and antinodal) larger, there are also fewer lines

formed.

d = 5.0 units

= 1.0, 1.5, 2.0, 3.0, 4.0 units

Increasing wavelength

On the other hand, for the same , a shorter d results in larger spacing between the lines (both nodal

and antinodal), and fewer lines formed.

= 1.0 unit d = 5.0, 4.0, 3.0, 2.0, 1.0 units

Decreasing source separation

d



Think about it. The highest order antinodal line formed is related to d . This is because the

maximum path difference is actually equal to d, which occurs on either sides of the sources. This

explains why when is large or d is small, less orders of lines are formed.

The effect of changing d or on the pattern can be simulated using transparencies. Cute.

see video at xmphysics .com

increasing

max=d max=d



10.2.3 Necessary Conditions for Interference

Many people use the terms superposition and interference interchangeably. But strictly speaking, they

are not the same thing. When two waves of the same kind overlap, superposition always occurs. But

there are 2 scenarios when two waves superpose but do not interfere with each other:

1. The waves are incoherent.

Two waves are said to be coherent if they maintain a constant phase difference1 between them.

Conversely, if the phase relationship changes continuously, they are said to be incoherent.

Let’s consider two waves with intensity I. If they superpose in-phase, the resultant intensity will be

4I due to C.I. If they superpose in antiphase, the resultant intensity would be 0 due to D.I. Now,

what if the phase difference between these two waves is changing randomly from time to time?2

artist’s impression of two incoherent waves

This will cause the interference to alternate between constructive and destructive interference. If

this random phase changing occurs very frequently, then all we can detect is the time-averaged

intensity of 4 0

22

II . Now think about it. Isn’t 2 I I I the same outcome as if there is no

interference? For this reason, we say that incoherent waves do not interfere.

1 Take note. Coherence does not require the phase difference to be zero. It only requires that the phase difference to be unchanging. 2 This happens if the waves transmit in short bursts. So the phase difference between the waves is only maintained for a short period of time before the next burst changes the phase difference again.

2I 0I 2I 4I



2. The waves are polarised perpendicularly to each other3.

Now let’s say wave X, with amplitude Ax and intensity Ix, is polarised horizontally. Wave Y, with

amplitude Ay and intensity Iy, is polarised vertically.

When they superpose, the resultant horizontal amplitude and vertical amplitude will always remain

as Ax and Ay, regardless of the phase difference between X and Y. And the intensity of the

resultant wave is always x yI I , regardless of the phase difference between X and Y. Since the

outcome is regardless of the phase difference between the waves, we say that there is no

interference between two perpendicularly polarised waves.

3 Obviously, this is applicable only to transverse waves since only transverse waves can be polarised.


10.3.1 Double Slit Interference Pattern

If two sound waves in anti-phase superpose destructively to result in silence, then two light waves in

anti-phase should also superpose destructively to result in darkness.

In the past, it was quite challenging to set up light sources and narrow slits suitable to demonstrate

the interference of light. It is much easier today: laser pointers and slit slides can be bought on the

cheap. Look and behold, when we shine a laser beam through two very narrow slits (called a double

slit), we obtain a beautiful pattern of bright and dark spots on a screen.


So what’s producing those bright and dot spots? The slits, being lit up by the laser beam, act like two

in-phase light sources, emitting two coherent light waves propagating towards the screen.

n=

0

n=

1

n=

2

n=

3

n=

3

n=

2

n=

1

bright fringes

n=

0.5

n=

0.5

n=

1.5

n=

2.5

n=

2.5

n=

1.5

dark fringes

slit slit laser beam

screen




The two light waves travel different distances to arrive at each position on the screen. The path

difference translates into a phase difference. At locations where the two light waves arrive in-phase,

bright spots are formed. At locations where they arrive in antiphase, dark spots are formed.

Basically, the double-slit interference pattern is the ripple tank interference pattern performed with

light instead of water waves. Some key differences are:

1. The air is basically transparent to light. For this reason, we can only see the fringes on the screen,

and not in the space between the slits and the screen.

2. Visible light is an EM wave with frequency of ~1014 Hz. The human eye cannot track the oscillation

of the electric field strength. We definitely cannot “see” the instantaneous displacement. All we

can “see” is the amplitude and intensity of the light wave. That’s why C.I. and D.I. of light waves

manifest to us as bright and dark fringes respectively.

3. The distance between the slits is extremely short compared to the distance to the screen (so the

diagram above is not reflective of the true scale). This remarkable geometry results in many

interesting properties which we can exploit to simplify the analysis. Read on.

slit slit

laser beam

screen

bright bright bright bright bright dark dark dark dark dark dark


10.3.2 Derivation of the dsinθ formula

For the double-slit, the path difference can be calculated using this sind formula. The detailed

derivation of this formula is arguably not required by the H2 syllabus. But the geometrical trick involved

is so fascinating I have to share it with you.

(a) Consider the circular sector PAB with sector angle β.

(b) If we make β really small, the sector would be squeezed until it looks like a thin line.

(c) If we were to zoom into the arc AB, we would see that the radii AP and AB appear as parallel lines,

and the arc AB a straight line. Remember that in O-level optics, you were taught that rays

originating from a point at infinity can be treated as parallel rays when they arrive at a lens. It is

the same situation here.

(a) Consider light waves departing from slits A and B, arriving at destination P on the screen. Note

that PAB’ is a circular section, which means PA and PB’ are radii and AB’ is a circular arc. This

diagram is drawn totally not to scale. The distance between the slits, called the slit separation d,

(a) (b) (c)

(a) (b) (c)

L

d d



is usually less than 1 mm. The distance between the slits and the screen, called the screen

distance L, is seldom less than 30 cm.

(b) If drawn to scale, AOB would look like a single point. And OP makes an angle of θ with the centre

line.

(c) If we zoom into the double slits, we would see that the two radii PA and PB’ are practically parallel

lines, and the arc AB’ is practically a straight line at right angle to PA and PB’. From the right

angled triangle ABB’, we obtain BB' sind . But BB' AP BP is actually the path difference!

Hence sind .

In general, to calculate the path difference | AP BP | , we must first calculate AP and BP.

However, for double-slits, the path difference can be calculated directly using sind . As long as

L d , the paths can be treated as parallel lines, and the path difference at each angle θ is constant.

Hence

Bright fringes are formed at angles θ where sind n , 0,1,2...n

Dark fringes are formed at angles θ where 1

sin ( )2

d n , 0,1,2...n


10.3.3 Fringe Separation

Very often, you are asked to calculate the distance between adjacent fringes, called the fringe

separation ∆y. But between which two? You might ask. Well, any two. Because the double slit pattern

usually has a constant fringe spacing. A little mathematics would tell us why.

n=

0

equal spacing

y1

L

double

slit

screen

intensity

θ1

y

n=

0

n=

1

n=

2

n=

3

n=

1

n=

2

n=

3

This diagram is not

drawn to scale.

θ is typically only a

few degrees for

double slits

∆y



Let’s use θn to denote the angle at which the nth order bright fringe is formed. Let’s also use yn to

denote the distance from the central axis.

Let’s start from the fact that

sin nd n , 0,1,2...n

Rearranging the equation, we get

sin n nd

, 0,1,2...n

For double slits, d is of the order of 0.1 mm, and is between 400 nm and 750 nm, making d

very

small. This means that the low order fringes ( 10n ) are all formed at very small θ.

For small values of θ, we can replace sin with y

L. This is because if θ is small, sin tan

y

L .

So

nyn

L d

, 0,1,2...n

In other words, the bright fringes are formed at

n

Ly n

d

, 0,1,2...n

The fringe separation is thus

1

1( ) ( )

n ny y y

L Ln n

d d

L

d

The mathematics actually shows us that the fringe separation is constant only for the fringes which

are formed at small θ. In fact, from the shape of the inverse sine function, we can deduce that the

fringes formed at larger values of θ ought to show larger fringe separations. However, the higher order

fringes of a double slit pattern are usually too dim to be seen anyway (due to the diffraction envelope

effect, which will be discussed later). All we get to see are the lower order fringes, which are formed

at small values of θ, which thus display a constant fringe separation.


10.3.4 Young’s Double Slit Experimental Set-up

Historically, the double-slit interference pattern provided the first evidence of interference of light,

which proved beyond doubt that light is indeed a wave4. But what took us so long to notice the

interference of light?

It turns out that light waves can be quite “erratic”. An ordinary light source consists of a very large

number of randomly oriented atomic emitters. Each excited atom radiates a polarized wavetrain for

roughly 10-8 s. All these wavetrains superpose to form a resultant wave whose phase (and polarization)

changes continuously in a completely unpredictable fashion5. This means that with two independent

light sources (e.g. two light bulbs), we get two incoherent light waves, since both waves are randomly

changing their phases with time. This is why we do not see bright and dark fringes when we switch

on two light bulbs!

So how did Thomas Young obtain two coherent light sources? Well, he shone the light from one single

light source through two slits. The light wave leaving the light source is changing phase continuously

with time. The light waves leaving the two slits are also changing phase continuously with time. But

since they originated from the same light source, their phases are changing in a coordinated manner.

This means that the light waves leaving the two slits are actually coherent with each other. Genius.

4 Until the photoelectric effect proved beyond doubt that light is a stream of particles called photons. Well, this is how science works: we modify our understanding as we take in new evidences. You will learn about the photoelectric effect in the topic of quantum physics. 5 Mathematically, natural light is modelled as two arbitrary, incoherent, perpendicularly polarized waves of equal amplitude.

double slit screen

light

source

☼

not drawn to scale

two

co

he

ren

t ligh

t wa

ve

s

(randomly

changing

phase)



Traditionally, an additional narrow slit is inserted before the double slit. The purpose of the single slit

is simply to reduce the width of the light source6. In other words, we are trying to obtain a line or point

source. This is necessary because the fringe separation of a double slit pattern is typically only a few

mm. If the light source itself is a few mm thick, every dark fringe (formed by light wave from one

particular point of the light source) will be filled up by the bright fringes (formed by light waves from

other points of the light source). The pattern of alternating bright and dark fringes will then be

concealed from us.

Worked Example 1

The main features of the apparatus for a double slit interference demonstration, and some of the

dimensions are illustrated in the diagram below.

Calculate the separation of the bright fringes on the screen if

a) light of wavelength 600 nm is used.

b) light of wavelength 400 nm is used.

6 It is not to diffract (i.e. spread) the light, as taught in many schools.

double slit screen narrow slit

light

source

☼

not drawn to scale


Solution

a)

9

3

( )

1.50(600 10 )

3.0 10

0.30 mm

Ly

d

y

b) Since y ,

400 600

400 20.30 0.20 mm

600 3y y

10.4.1 Diffraction Grating Interference Pattern

A diffraction7 grating is basically a set of regularly spaced slits. With today’s technology, it is easy to

fabricate a grating with hundreds, if not thousands of slits per millimetre. When we shine laser through

a grating, each ray leaving each slit acts as a wave source. While the double-slit pattern is formed by

the superposition of 2 light waves, the diffraction grating pattern is formed by the superposition of

hundreds if not thousands of light waves.


Did you notice that the bright fringes produced by a grating are much brighter and narrower than those

produced by a double slit? To discuss why this is so, let’s consider a grating with only 10 (regularly

spaced) slits.

7 We will study diffraction in detail in later sections. For the time being, just take it to mean spreading.

n=

0

n=

1

n=

1

bright fringes

n=

2

n=

2




Recall that for the double slit, C.I. occurs when 0 2sin , 1 , d and so on? This is because when

the path difference is an integer number of wavelengths, the two rays arrive in phase at P, superpose

constructively to form a resultant wave of amplitude 2A (assuming the amplitude of each individual

ray to be A).

Now that we have 10 slits, does C.I. also occur when 0 2sin , 1 , d and so on? The answer is

yes. This is because sind now represent the path difference between every pair of adjacent

rays. So when sind n , we get not just ray 1 and ray 2, but all 10 rays arriving in phase at P,

superposing constructively to form a resultant wave of amplitude 10A!

resultant wave

2 waves in-phase

2A

A

resultant wave

10 waves in-phase

10A

A


This explains why, whether 2 slits or N slits, bright fringes occur at the same values of (provided d

is the same). However, the bright fringes produced by N slits are much brighter since they are formed

by a super C.I. of N number of rays (compared to just 2 rays for the double slit).

Ok. Now let’s think about the dark fringes. Recall that for the double slit, D.I. occurs when

0 5 1 5 2 5sin . , . , . d and so on? This is because when the path differences is an odd integer

number of half wavelengths, the two rays arrive in antiphase at P and superpose destructively to

cancel out each other.

When we have 10 slits, however, the first D.I. requires a path difference of only 0.1 (instead 0.5 ).

Why? Because we have 10 rays now. What 0 1sin . d means is that the 10 rays will arrive at P

with each successive ray lagging by 1/10 of a cycle. Guess what is the outcome of superposing 10

sinusoidal functions which are “spaced out equally”? Zero!

In fact, these 10 rays also undergo complete D.I. when the path difference is 0.1, 0.2, 0.3, 0.4,

0.5, 0.6, 0.7, 0.8 and 0.9. If you want a complete analysis, do read Appendix A. For the H2

syllabus, however, it is sufficient just to have the idea that when we have N slits, there are actually

(N-1) dark fringes between each pair of bright fringes. A grating provides hundreds if not thousands

of slits. So for a grating, D.I. occurs pretty much at every value of where 0 2sin , 1 , ...d . In

other words, the interference pattern is dark everywhere except at precise values of where

0 2sin , 1 , ...d where (very) bright fringes are formed. This explains why gratings produce such

narrow bright fringes.

resultant wave

2 waves out of

phase by 0.1 cycle

A

0 A

resultant wave 10 waves regularly out of

phase by 0.1 cycle

0 A

A


To sum things up, refer to the graph below, which shows the intensity profile of a typical diffraction

grating interference pattern.

Things to note:

Positions of bright fringes

Similar to a double slit, a grating produces bright fringes at angles where

sind n , 0,1,2...n

However, because the slit separation d for a grating is usually much closer, the fringes are formed

at much large values of θ. (A typical 200-lines-per-mm grating has 1 mm

5 um200

d vs

0.1 mmd for a typical double slit).

grating

screen

intensity

θ

n=

0

n=

1

n=

2

n=

3

n=

1

n=

2

n=

3

θ


Fringe spacing

Unlike the double slit, the fringe spacing for the grating is usually not constant. The angular

separation between the higher order bright fringes are larger.

1sin sin ( )n

nd n

d

It has to do with the non-linearity of the inverse sine function. As n increases, θ increases at

increasing rate. That’s why the fringe spacing keeps increasing. (It should be clear to you that the

Ly

d

formula is not applicable to grating).

Fringe brightness

The bright fringes produced by a grating are much brighter compared to those produced by a

double slit. Why? Because the bright fringes of a grating are formed by the superposition of

hundreds of waves from hundreds of slits, whereas a double-slit has, well, 2 waves only. Even for

a grating of 100 slits, we are talking about a bright fringe with amplitude 100A and intensity 10,000,

compared to the double slit’s 2A and 4. 8

Fringe width

The bright fringes produced by a grating are much narrower. Why? When there are hundreds of

waves, the slightest misalignment between adjacent rays result in a lot of destructive interference

among them, such that the amplitude of the resultant wave is always negligibly small if not zero.

In this sense, the grating is a lot more finely tuned than the double slit. All the energy is

concentrated very precisely at angles where sind n , 0,1,2...n , and little is left elsewhere.

Contrast and resolution

Since the grating produces very narrow and bright fringes, interference pattern produced by a

grating offers very good contrast and spatial resolution. This explains why the grating is widely

used for spectral analysis.

8 In practice, the slit width is also usually narrower for a grating compared to a double slit. So the amplitude of an individual ray for the grating is usually lower. Nevertheless, this factor is not enough to change the outcome of the comparison.


Worked Example 2

A parallel beam of light of wavelength 600 nm is incident normally on a diffraction grating. The grating

has 500 lines per millimeter.

a) Determine the total number of bright fringes formed.

b) Determine the angles at which the fringes are formed.

Solution

a) 3

61.0 102.0 10 m

500d

6 9

( sin )

(2.0 10 )sin90 (700 10 )

2.857

d n

n

n

The highest order fringe formed is the 2n order. Therefore, 5 fringes are formed.

b) 1st order fringe: 6 9

1

1

( sin )

(2.0 10 )sin (1)(700 10 )

20.5

d n

2nd order fringe: 6 9

2

2

( sin )

(2.0 10 )sin (2)(700 10 )

44.4

d n


10.4.2 White Light Interference Pattern

So far, we have only discussed interference patterns produced by monochromatic light. What if we

are using a beam of white light instead?

Well, white light contains light of all colours in the visible spectrum. In other words, white light consists

of light waves of wavelength between 400 nm to 700 nm. Since coherence is required for interference,

each wavelength (colour) can only interfere with itself. So an interference pattern is produced for each

colour. The resultant interference pattern is simply all the different coloured interference patterns

overlaid on one another.

For example, when white light passes through a double-slit, colourful fringes are formed on the other

side, as shown below. (For reference, the interference pattern for monochromatic green light

( 500 nm ) is shown in the top row)

The key to understanding the pattern is to remember that the fringes formed by monochromatic red

light ( 700 nm ) are broader and more spaced out, and those formed by monochromatic violet light

( 400 nm ) are narrower and closer together.

This explains why the central white fringe has a reddish tint at the edge. This is because even though

the 0th order bright fringe of every colour is centred at the middle, red with the longest wavelengths

has the widest fringe. On the either side of the central white fringe, we see more colourful fringes.

You should understand the appearance if you remember that the higher order fringes of different

colours are formed at different positions. That’s how the colours get separated.

n=0 n=1 n=2 n=3 n=1 n=2 n=3



We know that if we have more slits, we can obtain narrower bright fringes. If each colour forms very

thin fringes, the different colours will be more separated. Shown below is the pattern formed by

passing white light through a 15-slit grating.

Here, we see a narrow white line at the middle, formed by the recombination of the (very narrow) 0th

order bright fringe of every colour. On either side, we see a distinct 1st order spectrum, formed by the

1st order fringe of all the colours, beginning with violet, ending with red. Even further out, we have the

2nd order spectrum, where the colours are even more separated. Notice also that the violet end of the

3rd order spectrum actually starts before the red end of the 2nd order spectrum, resulting in overlapping

spectrums.

So you see, beside refraction, interference can also cause the separation of colours. To many people,

when they see rainbow colours around them, they assume that refraction is involved. Actually, nature

has in store for us many natural double slits and gratings, producing rainbows everywhere.

Those shimmering colours of thin oil layers, seashells, butterflies and compact discs are all interference patterns.

n=0 n=1 n=2 n=3 n=1 n=2 n=3


10.5.1 Single-Slit Interference Pattern

If we pass light through one single slit, will we see an interference pattern?


There is only one slit. One wave. Who is interfering with who? Good that you’re thinking. Now here’s

my confession. In the past I treated light through a slit as a point source. I intentionally kept things

simple so that I would not lose you. But now I have to tell you the truth: a slit is not a point source. As

long as it has a width, it is actually a continuous line of point sources. Rays from each point in the slit

interferes with rays from every other point in the slit. So what do we get? Interference pattern!

It is easy to understand why C.I. occur at 0 . Each point in the slit propagates a light ray towards

the screen. If the screen is placed far enough, all the rays that converge at a faraway destination point

n=

1

n=

1

equal spacing

n=

2

n=

2

n=

3

n=

3

A

B

C

A, B and C




P on the screen become parallel rays at the slit. At 0 , all the rays travel (practically) the same

distance to the centre point on the screen. The path difference is zero. All the rays arrive in-phase,

resulting in the maximum possible intensity.

What happens on either sides of 0 ? Since the rays must now travel different distances to arrive

at the destination P, the rays no longer arrive in-phase. Obviously, the rays can only superpose to a

lower intensity compared to when 0 . In fact, as increases, the interference among the rays

become more and more destructive, resulting in the intensity decreasing progressively.

Does complete destructive interference among the rays ever occur? The answer is yes. Note that the

path difference between the two rays from the edge of the slit (labelled as A and C in the diagrams)

is sinb , b being the slit width. Let me tell you that the first time complete destructive interference

occurs is when sinb .

You will have to read Appendix B for the proof, if not, you can take it from me that complete D.I. occurs

whenever sinb n , 0,1,2...n

A

B

C

A B C

A B C


As a result, the single-slit interference pattern consists of a broad central bright band, called the central

maximum, flanked by far dimmer bright bands on both sides. The intensity of the central maximum

peaks at the centre, and tapers off towards both sides, hitting zero at the so-called first minimum.

n=

1

n=

1

equal spacing

n=

2

n=

2

n=

3

n=

3

central

maximum

first

minimum

first

minimum

L

single slit

screen

intensity

θ1

y

n=

1

n=

2

n=

3

n=

1

n=

2

n=

3

This diagram is not

drawn to scale.

θ1 is typically only a

few degrees for

single slits

w

θ1


The central maximum contains the vast majority of the total light energy. In practice, the light beyond

the first minimum is often too dim to be seen. For this reason perhaps, the H2 syllabus also ends at

the first minimum. You are not required to know of the higher order bright and dark bands. You only

need to know that the first minimum occur at

1sin

b

where b is the slit width.

As you can interpret from the formula, by narrowing the slit that the light passes through, the emergent

light beam actually spreads wider. This may be counter-intuitive at first, but makes total sense once

you remind yourself that light is a wave.

θ

b=5

15° 15° θ

15° 15° θ

15° 15°

b= b=10


Worked Example 3

Parallel light of wavelength 590 nm passing through a rectangular slit is incident on a screen.

Calculate the width of the central fringe w as observed on the screen.

Solution

First we calculate the first minima angle 1.

1

9

1 3

4

(sin )

590 10

0.60 10

9.833 10 rad

b

Since 1 is very small, w can be approximated by the arc length subtended by 21.

1

4

( )

(2 )

2.4(2 9.833 10 )

4.72 mm

s r

w L


10.5.2 Diffraction

Actually, the spreading of waves is a well-known phenomenon long before the single slit interference

pattern was observed. For example, when sound passes through an opening, it spreads outward

(after entering the opening). When a water ripple meets an obstacle, it spreads inward (after passing

around the obstacle). The spreading of waves is called diffraction.

Light waves have very short wavelength (<750 um for visible light). So even a 1 mm slit would be too

wide to cause any appreciable spreading in the emergent beam. (Think 1sin

b

). This is why the

diffraction of light usually requires careful and deliberate setup before they are observable.

Sound waves and water waves have wavelengths ranging from centimeters to meters. So the

diffraction of sounds waves and water waves as they interact with common objects are much more

readily observable.

The following are typical illustrations of diffraction found in many textbooks. Strictly speaking, they are

what I call artists’ impressions of the diffraction phenomena because the actual picture is a lot

messier9.

see animation at xmphysics.com

Nevertheless, the main idea is that the degree of spreading depends on the width of the slit (or

obstacle) relative to the wavelength. The narrower the slit and the longer the wavelength, the more

the spreading. In the extreme case where the slit has negligible width compared to the wavelength,

the wave energy is spread (almost) uniformly in all directions.

9 The picture for diffraction of light is actually much neater because of the simplification that is offered by the parallel ray approximation.




10.5.3 Rayleigh’s Criterion

I believe you have heard of a pin-hole camera before. It is found in every lower secondary science

text book. Let me ask you, for the sharpest image, should the pin-hole be tiny, or large? The tinier the

better, right? If the pin-hole is large, then many rays from one point on the object can arrive at multiple

points on the film, resulting in a blurred image. The sharpest image is formed when the pin-hole is so

tiny that only one ray from each point on the object arrives at one point in the image. Right? Nope.

Plot twist. It turns out that, yes, pin holes should be small, but not too small. When it’s too small, the

image actually starts to blur again! Why?

One word. Diffraction! When the pin-hole is too small, diffraction of light sets in. So each point object

does not form a point image, but rather a smeared blob. Think of the width of central maximum of the

single-slit interference pattern. Based on 1b

, the smaller the pin-hole, the larger the smear.

This smearing is of grave concern to microscopy or telescopy because it limits an optical instrument’s

resolving power, which is the ability to distinguish two very close objects.



Say we have two point objects A and B. In another universe where light does not diffract, two point

images A’ and B’ would have been formed. If A and B have an angular separation of , A’ and B’ will

also be separated angularly by angle . In our universe light does diffract. So instead of two point

images, we get two smeared blobs. The intensity profile of each blob follows that of a single-slit

diffraction pattern: peaking at 0 (at A’ and B’), tapering off on either sides, and reaching zero at

the first minimum angle10 of 1b

. If the angular separation is comparable to the first minimum

angle 1, then we have a concern, because the two smeared blobs will start to overlap and it becomes

impossible for us to tell whether there is one or two objects.

10 From the single slit formula, the first minima occurs at 1sin

b

. Since we are dealing with very small

values value of 1, 1 1sinb

. Also because the light emitted by the two point objects are incoherent, they

do not interfere with each other. We can simply sum up their intensities.

A B A B A B

A’ B’

A’ B’ A’ B’

No diffraction With diffraction Just resolved

1

1


The generally accepted criterion for whether two images are distinguishable is the Rayleigh criterion,

which states that two images are just resolved when the central maximum of one image coincides

with the first minimum of the other image. Hence according to the Rayleigh criterion, two images are

just resolved when

1b

In any optical system, the area that is receiving the light is acting as the “slit”. For animals, it is the

size of the pupil of the eyes that decides the amount of diffraction occurring. That’s why eagles, with

their large pupils, have very good eye sight. For satellite dishes, having a larger dish will reduce the

amount of diffraction. To have high resolving power to distinguish astronomical objects with very small

angular separation, gigantic dishes are built.

1 1 1

angular separation between the objects

central maximum central maximum

first minimum first minimum


10.5.4 Diffraction Envelope

In this section, we will do the double-single-slit grand slam.

When you first saw the double slit interference pattern, did you notice that the bright fringes are not

equally bright? Now that you have learnt about diffraction, we can finally discuss the so-called

diffraction envelope.

The double slit is actually also two single slits. So each light ray will interfere with (1) rays from the

same slit (single slit interference) and (2) rays from the other slit (double slit interference). As such,

the outcome of the superposition at any angle depends on both sind and sinb .

The grand outcome is a double slit interference pattern inscribed by a single-slit diffraction envelope.

The intensity peaks at the 0th order bright fridge, while the higher order fringes on either sides

decrease in brightness due to single-slit interference. In fact, very often, only the (double slit) fringes

formed within the (single slit’s) central maximum are bright enough to be visible. The higher order

(double slit) fringes formed beyond the (single slit’s) first minimum angle 1 are too dim to be seen.

n=

0

equal spacing

double slit fringes in the

single slit’s central maximum



intensity

intensity

intensity

single slit interference pattern slit width b

single slit diffraction envelope

double slit fringes

“ideal” double slit interference pattern slit width 0

slit separation d

double slit interference pattern slit width b

slit separation d

1=/b

∆y = L/d

∆y = L/d

1=/b 1=/b


10.A N-slit Interference

(Beyond H2 syllabus)

Before we discuss N-slit interference, let’s first study what happens when we superpose N sinusoidal

waves.

Let’s consider the superposition of 3 sinusoidal waves which (1) have the same amplitude A and (2)

have the same phase difference θ between them. Look at the diagram below to see what I mean.

To get total constructive interference, we need the 3 sinusoids to be exactly aligned. This happens

whenever 2n , 0,1,2...n

θ θ

resultant

http://www.xmphysics.com/10.A


To get total destructive interference, we need to “space out” the 3 sinusoids evenly. This occurs when

120 and when 240 .

Even if we add more sinusoidal waves, we always get C.I. when the waves are exactly aligned, and

D.I. when the waves are “spaced out evenly”. E.g. with 4 sinusoids, total D.I. occur when 90 ,

180 and 270 . With 5 sinusoids, it will be 72 , 144 , 216 and 288 .

In other words, for N waves,

Total C.I. occurs at 2n , 0,1,2...n

Total D.I. occurs at ( )2m

nN

, 0,1,2...n and 1..( 1)m N .

120° 120°

resultant

240° 240°

resultant


Ok. We are now ready to discuss N-slit interference. Let’s start by adding a 3rd slit to the double slit

experiment. Note the common slit separation d between slits A and B, and between slits B and C.

If the screen is placed far enough from the slits, the 3 rays that converge at a faraway destination

point P on the screen become parallel rays at the slits. Ray C travels a longer distance of CC’’

compared to ray B, which in turn travels a longer distance of BB’ compared to ray A. Both path

differences are equal to sind . Applying what we learnt from Section 10.A, we realize that the first

few total C.I. and total D.I. will occur at the following angles:

sind

C.I. 0

D.I. 1

3 ,

2

3

C.I.

D.I. 1

(1 )3 ,

2(1 )

3

C.I. 2

D.I. 1

(2 )3 ,

2(2 )

3

d

d

d

d


In short, for three regularly spaced slits,


Dark fringes are formed at angles θ where 3

sin ( )m

d n , 0,1,2...n , 1 or 2m

This resulting interference pattern has an intensity profile shown in the graph below.

By extension, for N slits (with a common slit separation d)


Dark fringes are formed at angles θ where sin ( )m

d nN

, 0,1,2...n and 1 .. 1m N

L

3 slits

screen

intensity

θ

n=

0

n=

1

n=

2

n=

3

n=

1

n=

2

n=

3

θ

m=

1

m=

2

m=

1

m=

2

m=

1

m=

2


Note that as N increases

1. the positions of the bright fringes do not change.

2. there are more dark fringes. In fact, there are 1N dark fringes between every two bright fringes.

3. the intensity of the bright fringes should increase. Assuming that each ray arrives at the

destination point on the screen with amplitude A and intensity 0, the intensity of the resulting

bright fringes should have amplitude NA and intensity 2

0N I . In the graphs above, the intensity

have been “equalised” across the different N values because we have chosen to plot the

normalized intensity max

I

I, where maxI is 2

0N I .

θ

θ

θ

θ

θ

θ

N=2

N=3

N=4

N=5

N=10

N=20


10.B Single Slit D.I.

(beyond H2 syllabus)

The single slit pattern is the result of superposition of an infinite number of rays propagating from an

infinite number of points along the slit width. The outcome obviously depends on the phase difference

among the rays.

Notice that the path difference between the two rays from the edge of the slit (labelled as A and C in

the diagrams) is sinb , b being the slit width. Consider the case when sinb . You may be

thinking of C.I. occurring between A and C since their path difference is . Or you may be thinking of

D.I. occurring between A and B since their path difference is 2

. And what about all the other rays?

You may be thinking, “what a mess”.

Surprisingly, there is a simple way to sort out this mess. In our minds, we can split the slit into two

halves. The top half sends out rays A1 to AN while the second half send out rays B1 to BN. If A1 and B1

are in anti-phase with each other, so are A2 and B2, A3 and B3, and every pair of rays until AN and BN.

Pairing up the rays this way makes it clear that all the rays will superpose to zero when A and B are

in antiphase, which occurs when A and C are in-phase. So we conclude that a complete D.I. occurs

when sinb .

A

B

C

A B C

http://www.xmphysics.com/10.B


If we now mentally divide the single slit into 2 slits of width 2

b each, it is easy to realize that the next

complete D.I. occurs when sin2

b . We can also mentally divide the single slit into 3 slits of width

3

b each, and realize that complete D.I. also occurs when sin

3

b . In fact, there is nothing to stop

us from mentally dividing the single slit of width b into n number of slits of width b

n, and conclude that

complete D.I. occurs when sinb

n , 1,2...n

A1 to

AN

B1 to

BN

l 10a superposition - wordpress.com

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