Lecture 23: Diffraction• In addition to interference, another property that’s unique

to waves is diffraction– this is the ability of a wave to spread out after passing

through an opening• We’ve already seen this behavior: it’s the basis of the

double-slit apparatus– each slit acts as a “point source” of light due to the

diffraction of the incoming light• Diffraction can be understood as an example of Huygens’

Principle, which says that each point on a wave front canbe considered as a point source for an outgoing wave

Poisson’s Spot• One of those unwilling to abandon Newton’s particle

theory for light was Simeon Poisson• He pointed out the following:

– let’s say light hits a perfectly circular object– if the wave theory is right, every point on the edge of the

object should act as a wave source– the distance from the each point on the edge to the center of

the object’s shadow is the same– therefore all the waves would add up constructively, forming

a bright spot in the center of the shadow• Poisson noted that this was so ridiculous that the wave

theory of light had to be wrong!• But not long afterwards, the bright spot (now called

“Poisson’s spot”) was in fact observed experimentally– Poisson thereby unintentionally helped to prove the theory

he was trying to discredit!

Diffraction by a Slit• So far, we’ve treated a slit as a point source of light

– but that’s not really true– a slit has some width:

a

• So there are really an infinite number ofpoint sources– one for each point on the slit

• These waves can interfere with one anotherjust like waves from different slits can

• The result is a diffraction pattern on thescreen behind the slits– we’ll consider the case where the screen is

far away, so that all the rays leaving the slitcan be considered parallel

• How can we analyze interference of an infinite number ofwaves?

• Start by dividing the slit into two halves:

• In fact, for every ray from the top half, there’s a ray fromthe bottom half with

a/2

a/2

θ

123456

• The difference in pathlength between rays 1and 3 is

• That’s also the pathlength differencebetween rays 2 and 4,and 3 and 6

! =a

2sin"

! =a

2sin"

• So if it happens that rays 1 and 3 differ in path length by ,there will be a dark spot on the screen– because every ray from the top half will be cancelled by a ray

from the bottom half of the slit– so the condition for a minimum is

• The fact that we chose to divide the slit in halves was purelyarbitrary– we could have split it into quarters, and found that the condition

for a minimum was:

±!

2

a

2sin! = ±

"

2

sin! = ±"

a

a

4sin! = ±

"

2

sin! = ±2"

a

• By considering any even number of “sub-slits”, we arriveat the general rule for all the minima that appear on thescreen:

• Note that m = 0 does not give a minimum– this corresponds to the point at the center of the screen,

which is a maximum

sin! =m"

a; m = ±1,±2,±3…

Intensity of the Diffraction Pattern• Now we know where the intensity is zero, but we don’t

know that value at any other point• We can use phasors to find out!• In this case, there are an infinite number of waves adding

at each point on the screen– if we consider every small portion of this slit as a source of

waves, the phasor diagram for a given point on the screenwould look like:

• The total phase shift β is related to the path lengthdifference from the top and bottom of the slit:

β

! =2"

#asin$

ER

Phasors lie on arcof a circle

• Define the total length of the phasor chain as Eo

• We can use geometry to find ER:

• The arc length is given by Eo = βR• From the triangle, we see that:

R

β

sin!

2=ER/ 2

R=ER

2"!

Eo

• Which means the total electric field is:

so the intensity varies as:

ER=2E

o

!sin

!

2

I = Imax

2Eo

!sin

!2

"

#$

%

&'

2

= Imax

(Eo

)asin*sin

)asin*(

+,-

./0

"

#$%

&'

2

• Note what happens when the slit becomes much biggerthan the wavelength:

• The angular spread due to diffraction becomes smaller– that’s why the ray approximation works well in many

cases

a = 25λa = 100λ

Back to the Double Slit• When we studied double-slit interference earlier, the

equations said we should see alternating dark and brightregions, with all the bright regions having the sameintensity– Experimentally, we did see the bright and dark regions, but

the intensity clearly decreased as one moves away from thecenter of the pattern

• That’s because both interference and diffraction arehappening– the total intensity is the product of the two effects

I = Imaxcos

2 !d sin"#

$%&

'()

#Eo

!asin"sin

!asin"#

$%&

'()

*

+,-

./

2

• This pattern looks like: