1

Chapter 33INTERFERENCE & DIFFRACTION

33.1 Phase Difference & Coherence

A sum of two harmonic waves of the same frequency and wave vector is a harmonic wave whose amplitude depends on the phase difference of two waves (interference).A common source of the phase difference is a difference of path lengths x∆ :

02 360x xλ λδ π ∆ ∆= =

Example 1. What is the minimal path difference sufficient to produce 90o phase difference in 750 nm light?

0

00 0 90 1

436090 360 750 187.5x x nm nmλδ λ∆= = ⇒ ∆ = = =

Another important cause of phase difference is the 180o phase change when light is being reflected by the surface of the medium in which light travels more slowly.

2

The interference of two overlapping waves can be observed only if both sources are coherent.

The length of the wave packets in which the phase is the same is called the coherence length (or, time-wise, the coherence time).

33.2 Interference in Thin FilmsReflected rays 1 and 2 interfere with each other. Depending on the phase / path difference (i. e. film thickness) the interference can be constructive or destructive. Note: ray 1 also undergoes the 180o phase change! If the light is (almost) perpendicular to the surface, then the overall phase difference between rays 1 and 2 is

0 0 2180 360 , /water

twater air waternλδ λ λ= + =

3

Here both reflected (interfering) rays experience the 180o phase change.

If the thickness of the film varies, the monochromatic light produces alternating bright and dark bands (fringes).

The distance between the bright and dark fringes is such that the path difference 2t is '/ 2λ

.

The circular fringes from an air film between plane and spherical glass are called Newton’s rings

4

42.4 10 .radθ −= ⋅

Example 2. An air wedge is made by putting a thin slip of paper between two glass plates;

Describe the fringes for light with 750 nm wavelength.

The m-th dark fringe with position x is observed when the path difference 2t is equal to m wavelengths (do not forget an extra 180o phase shift for ray 2!),

2 't m mλ λ= =The local wedge thickness is related to the angle as and t xθ=

4 7 1

2 / 2 / ;/ 2 / 2 2.4 10 / 7.5 10 6400

m t xm x m m

λ θ λ

θ λ − − −

= =

= = ⋅ ⋅ ⋅ =

where m/x is the density of fringes (# of fringes per meter)

5

33.2 Two-Slit Interference Pattern

One of the ways to observe interference is to look for the interference of two rays from the same source coming through two different slits.

At large distance from the slits the rays from both slits are almost parallel and the path difference is sind θ≈ . Thus, the interference maxima are at

sin , 0,1, 2,3...md m mθ λ= =and minima are at

( )12sin , 0,1, 2,3...md m mθ λ= − =

where m is called the order number.

6

( )2 sin /dδ π θ λ=

( )tan /m my L m L dθ λ= =

/L dλ

In general, the phase shift between two rays is The distance on the screen between the center and the m-th bright point is

and for small angles the fringes are equally spaced at a from each other.

Example 3. The spacing between fringes for 650 nm light coming through two slits separated by 1.2 mm is 1.6 mm. Find the distance to the screen

( )

3 3

9

1

1.2 10 1.6 10650 10

// 1.6

/ 2.95

m

m m

y m L dy y y L d mm

L d y m m

λδ λ

δ λ − −

−

+

⋅ ⋅ ⋅⋅

= ⇒

= − = = ⇒

= = =

7

δ( )1 0 2 0sin , sinE A t E A tω ω δ= = +

( )( ) ( )

( ) ( )

1 2 0 0

0

1 1 1 12 2 2 2

sin sin

2 cos / 2 sin / 2

sin sin 2cos sin

E E E A t A t

A t

ω ω δ

δ ω δ

α β α β α β

= + = + +

= +

+ = − + ( )02 cos / 2A δ

Calculation of intensity

Consider two identical waves with phase shift ,

The resultant wave function is

i.e. the amplitude of the resultant wave is

.

The intensity of the wave is proportional to the square of the amplitude. Therefore,

( )204 cos / 2I I δ=

where I0 is the intensity of the light from one wave.

8

On the screen one sees the intensity pattern.

Another way to produce a two-slit interference pattern is to use Lloyd’s mirror (below):

9

33.4 Diffraction pattern of a single slit

When light goes through a narrow slit, one sees on the screen a pattern like this instead of a sharp image of the slit.

Most of the intensity is in the central diffraction maximum (peak).

10

The first zero in the intensity is at 1 1

1 12 2sin / sina aθ λ θ λ= ⇔ =

At this angle, the waves coming out of each point in the upper half of the slit are exactly 1800 out of phase from the waves coming out of the point a/2 lower down the slit and cancel each other (see the path difference in the figure).

Similar derivation leads to the following general equation for the directions of the diffraction minima (zeroes):

sin , 1,2,3,...ma m mθ λ= =

11

Expressed via the distance to the screen L, the distance from the central maximum to the firstdiffraction minimum is

1 1tany L θ=

Example 4. The distance between the first diffraction minima on the left and right of the central maximum is 4.8 cm. Find the distance to the screen if the wavelength is 600 nm and the slit width is 0.1 mm.

The distance between the first left and right diffraction minima is

1 12 2 tany L θ=

1θ 1sin / aθ λ=where the angle is given by the equation Since

2sin sincos 1 sin

tan α αα α

α−

≡ =the distance

( )

( ) ( )( )

12 21

27 421

7 4

sin 2 /1 1 1 sin 1 /

2 0.048 1 6 10 /102 1 /2 / 2 6 10 /10

2 2 tan 2 ,

7.98

L aa

y aa

y L L

L m m

θ λθ λ

λλ

θ

− −

− −

− −

⋅ − ⋅−

⋅ ⋅

= = =

= = =i

12

Interference-diffraction pattern of two slits

The figure illustrates the interference-diffraction pattern from two slit whoseseparation is ten times the slit width, d = 10a.

Here the tenth interference maximum on each side is missing since it coincides with the first diffraction minimum. The central diffraction maximum contains 19 interference peaks – the central interference maximum + 9 maxima on either side.

In general, the pattern is the set of interference peaks modulated by the diffraction curve. If the ratio m = d/a is integer, the m-th interference maximum will fall exactly at the first diffraction minimum and the m-th fringe will not be seen. There will be m – 1 fringes on each wing of the central diffraction maximum to the total of

2( 1) 1 2 1, /N m m m d a= − + = − =fringes.

13

33.5 Fraunhofer and Fresnel diffraction

Diffraction patterns, like the single-slit pattern, that are observed on a screen which is far away from the obstacle (slit) so that the rays are almost parallel, are called Fraunhofer diffraction patterns.

The diffraction pattern observed near an aperture or an obstacle is called a Fresnel diffraction pattern.

14

Example of Fresnel diffraction ofa straight edge. Below it is a graph of intensity vs.distance fromthe edge

15

33.6 Diffraction and resolution

Diffraction, especially for circular apertures, limits the resolution of the optical devices.

The angle for the first diffraction minimum is related to the diameter of the opening D as

sin 1.22 /Dθ λ=For high resolution, one often deals with small angles,

1.22 /Dθ λ≈

Therefore, the critical angular resolution of two sources is

1.22 /c Dα λ=(the first diffraction minimum from the first source falls on the central maximum of the second).

16

The equation 1.22 /c Dα λ= is called Rayleigh’s criterion for resolution.

cα α> cα α=To increase resolution, one has to either decrease the wavelength or increase the entrance aperture of an instrument.

17

Example 5. Small lights, 1 cm apart, are attached to a wall. What is the eye pupil diameter if the lights merge when one is 25 m away? The wavelength is 800 nm.

The given critical angle is 4/ 0.01/ 25 4 10c d Lα −= = = ⋅

On the other hand, the critical angle is

7 4

1.22 / 1.22 /

1.22 8 10 / 4 10 2.44c cD D

m mmα λ λ α

− −

= ⇒ =

= ⋅ ⋅ ⋅ =

Such small size of the pupil means that the lights are very bright.

18

33.7 Diffraction gratings

The wavelength of light can be measured using the diffraction gratings – sets of equally spaced lines or slits.

θsind θ

At an angle the optical path difference for rays coming from adjacent slits is .

If we have N slits, at the amplitudes from all slits add up and the intensity is

0θ =20N I

At the rays from slits cancel each other and we have a interference minimum.

sin / 2d θ λ=

At the waves again add up and we have a maximum, and so on. Thus, the maxima are at

sind θ λ=

sin , 0,1,2,3,...md m mθ λ= =The larger the number of slits N, the sharper the picture since the width of the interference maximum is min2 2 / Ndθ λ≈wile the intensity of the maximum is proportional to

2N

.

19

Each wavelength emitted by the source produces a separate image of the collimating slit called a spectral line. The set of lines corresponding to m = 1 is called the first order spectrum, m = 2 – the second order spectrum, and so on.

Example 6. Diffraction grating has 14000 lines per cm. At what angles will the two lines of wavelengths 612.2 and 613 nm be seen in the first order?

The distance between the slits is 1/14000 cm = 0.71 x 10-4 cm. For the first order,

sind θ λ=and the angles are

( )

( )

9

6

9

6

1 1 1 0612.2 101 0.7110

1 1 1 0613102 0.7110

sin / sin sin 0.8626 59.61 ;

sin / sin sin 0.8634 59.70

d

d

θ λ

θ λ

−

−

−

−

− − −⋅⋅

− − −⋅⋅

= = = = = = = =

The resolving power of the spectroscope is /R mNλ λ≡ ∆ =where is the smallest difference between two nearby wavelengths that still can be resolved. In the example above, R should be

λ∆613

613 612.2 767R −= =meaning that we need at least 767 or more slits to resolve these two lines.

20

Review of Chapter 33Interference

x∆02 360x x

λ λδ π ∆ ∆= =Phase difference due to a path difference :

The interference is constructive if the phase difference is zero or an integer times 360o. The interference is destructive if the phase difference is an odd integer times 180o

An additional 180o phase difference is added if a light wave is reflected from a medium with a slower wave speed

The interference of light reflected from the outer and inner surfaces of thin film produces interference fringes; the path difference for normal reflection is twice the thickness of the film

Interference maxima and minima for two slits separated by a distance d are

sin , 0,1,2,3...md m mθ λ= =

( )12sin , 0,1,2,3...md m mθ λ= − =

21

Diffraction occurs when a portion of a wavefront is limited by an obstacle or an aperture.Fraunhofer patterns are observed at large distances (almost parallel rays), Frensel ones – at small.

The zeroes in the single-slit diffraction pattern occur at angles

sin , 1,2,3,...ma m mθ λ= =

The interference-diffraction pattern of two identical slits is the two-slit interference pattern modulated by the single-slit diffraction pattern.Rayleigh’s criterion for optical resolution by circular apertures is 1.22 /c Dα λ=

Interference maxima for diffraction gratings are atsin , 0,1,2,3,...md m mθ λ= =

The resolving power of a grating (spectroscope) is

/R mNλ λ≡ ∆ =