1

Interference and Diffraction

From previous lecture:

•Short review on waves

•Superposition of waves and interference

•Interference maxima and minima for 2 slit experiment

This lecture:

•Complete Interference: thin films

•Diffraction

Interference max ad min

Interference max:

!r = d sin" =m#

Interf. min:

!r = d sin" =(m+1/2)#

Relation between Path and Phase difference:

!

"r = # $ "% = 2& rad

!

"# =2$

%"r Eg.:

Position of fringes on screen:

!

"r = d sin# $ d tan# =dy

L

Question: interference

The distance between the slits in a double-slit experiment is increased by a factor of 4. If the distance between the fringes is small compared with the distance from the slits to the screen, the distance between adjacent fringes near the center of the interference pattern

A) increases by a factor of 4.B) depends on the width of the slits. C) decreases by a factor of 2.D) decreases by a factor of 4.

A math explanation

Superposition of 2 waves with equal frequency and amplitude:

D = D1+D2 = A sin(kx1 - "t +#10) + A sin(kx2 - "t +#20) =>

!

D(x, t) = 2Acos"#

2

$

% & '

( ) sin[kxavg *+t + (,0)avg ]

!

"# =#2$#

1

!

xavg =x1+ x

2

2

!

("0)avg =

"01

+ "02

2

From trigonometry:

!

sin" + sin# = 2cos1

2(" $#)

%

& ' (

) * sin

1

2(" + #)

%

& ' (

) *

Amplitude maximum for:

!

"# = 2m$ or!r = d sin" =m#

Intensity

Intensity of 2 slits if there were no interference:

With interference:

!

I" 2Acos#$

2

%

& ' (

) *

2

Min.: I = 0, Max: I ∝4A2

Average I: ∝2A2

Notice: intensity decreases as y increases

For N slits (diffraction grating):

!

I = 2I1"2A

2

!

I = N2I1

Interference in Thin Films

There is no phase change when the wave is reflected from a boundary leading to a medium of lower index of refraction (analogous to a pulse on a string reflecting from a free support)

Light undergoes a phase change of 180° upon reflection from a medium of higher index of refraction medium(analogous to a pulse on a string reflected from a rigid support)

Remember: the wavelength of light n in a medium with index of refraction n is !n = !/n with ! wavelength in vacuum

More on interference in thin films

Ray 1 has 180!phase shift.

Ray 2 travels 2t (for normal incidence) more

than ray 1

Phase difference:

For constructive interference2nt = (m + 1/2)# m = 0, 1, 2 …

This takes into account both the difference in optical path length for the two rays and the 180° phase change

For destructive interference2nt = m! m = 0, 1, 2 …

!

"# + 2#$r

%n

= 2m# & "# + 2#2nt

%= 2m#

Soap Bubble and butterflies

The iridescent colours of soap bubbles are caused

by interfering light waves and are determined by the

thickness of the film. Black is seen when

t<<wavelength hence no phase difference

Interference occurs when light hitting the wing combines

with light reflected off the wing.

Question: CD playerHow deep (t) are the pits on a CD, given that it is read by interference?

•120 nm

•240 nm

•480 nm

•960 nm

Huygen’s principle (1678)

All points on a wave front is the source of spherical wavelets that with speed, frequency equal to initial wave

When a plane wave meets a small aperture, the result depends on the size of the aperture a respect to !

!<<a

!>>a

The transmitted wave is concentrated in the forward direction, and at near distances wave fronts have the shape of the aperture

Similar to wave from a point sourceresult of of interference of wave fronts

Wavefrontat t=0

Wavefrontat time t

Single slit diffraction

•Each portion of the slit acts as a source of waves (Huygens)

•These waves interfere according to path length difference

•Fraunhofer diffraction: screen very far so that rays leave the slit in parallel directions

Ray 1 travels

farther than ray 3Interf. min:

In general:

!

"r13

=a

2sin#

!

a

2sin" =

#

2

m = ±1, ±2, ±3, …

!

asin" = m#

More on Single SLITThe diffraction pattern is actually an interference pattern.The central max is twice as large as secondary ones.The narrower the slit, the wider the diffr. pattern. If diffraction pattern to small to be seen

m = ±1, ±2, ±3, …

!

asin" = m#

!

" /a <<1

Resolution

Sources far enough => maxima do not overlap =>their images can be

resolved

Not resolved

Since #<<a

To be resolved the angle between 2 sources must begreater than

!

sin" # "

!

"min

= # /a

Circular apertures

well resolved just resolved not resolved

The limiting angle of resolution of the circular aperture is

D diameter of aperture

Resolution of Eye

Suppose the optics of your eye are diffraction limited, how far away can you resolve headlight separation on a car?

Pupil diameter D~ 0.2cm = 2x10-3 m

Light wavelength #~ 500nm = 5 x 10-7 m

"min =1.22#/D= 3 x 10-4 radHeadlight separation ~ 2m

X-ray diffraction by crystals

Laue pattern for Be

Bragg’s law gives constructive

interference:

2dsin$=m! m=1,2,3,...

Diffraction on DNA