1

• For two beams to interfere and produce a stable pattern, they– Must have nearly the same frequency;

– Must have equal or nearly equal amplitudes• the pattern will be very clear.

– Need not be in-phase; but phase difference must be constant coherent.

Conditions for Interference

2

Wavefront-splitting Interferometers:Young’s Experiment• Young was able to

produce interference even before laser was invented.

• In his experiment, a single wavefront was selected and split into two coherent sources, which were then used to produce interference.

3

• Consider a plane wave illuminating a slit. – Light will diffract out in all

directions from this primary slit.

• This diffracted ray then encounters two slits, S1 and S2, on another barrier. – The wavefront arriving at the

two slits will be exactly in-phase.

– The slits constitute two coherent secondary sources.

– Interference occurs wherever these two waves overlap.

4

5

• Consider the OPD traverse by rays from S1 and S2

to point P, on the screen:OPD = l2 – l1

• It can be approximated by taking a perpendicular line from S1 to l2:

OPD = d sin

6

• For a bright fringe to exist, the OPD must be an integral number of the wavelength:

OPD = m, m = 0, 1, 2, …d sin = m

• Consider sin ≈ :

It gives the angular position of the mth order fringe.

d

mm

7

• IN order to determine the vertical position of the mth order bright fringe, on the viewing screen.

• Let y = 0 be the horizontal position midpoint between the two slits and ym be the vertical position of the mth order fringe along the viewing screen.

• The position of ym can be determined from the geometry of the experiment set-up.

Ld

my

d

m

L

y

m

sin

8

• For a dark fringe to exist, OPD = (m + ½), m = 0, 1, 2, …d sin = (m + ½)

• Consider sin ≈ :

• The position of the mth order dark fringe:

• The separation between two consecutive maxima/minima can be shown as

d

mm

21

L

d

mym

21

Ld

y

9

Dark fringes bright

fringesm 3

22

1

1

0

-1

-1

-2

-2

10

• The phase difference between the two rays = k(l2 – l1)

• The total irradiance at an arbitrary point on the viewing screen is

I = I1 + I2 + 2(I1I2)1/2 cos

• In this experiment, the irradiance contribution from both sources are equal i.e. I1 = I2 = Io

I = 2Io (1 + cos )

L

ydkI

llkI

I

o

o

o

2cos4

2cos4

2cos4

2

122

2

11

• The irradiance has a maximum value when

• Consecutive maxima are separated by y.

d

Lny

nnL

yd

L

ydk

n

,...2,1,0,2

...20d

L

d

L

12

Example

A viewing screen is separated from a double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe is 4.5 cm from the center of the line.

(a) Determine the wavelength of the light.(b) Calculate the distance between adjacent bright

fringes.

13

Solution

L = 1.2 m, d = 0.030 mm, m = 2, ybright = 4.5 cm

(a) ybright = m

= l 560 nm

(b) Dy = ym+1 – ym =

= 2.2 cm

d

L

d

L

14

Amplitude-splitting Interferometers:Thin Film• Beamsplitter is an optical device that splits a

beam of light into two – reflected beam and transmitted beam.

• Examples: thin stretched plastic films, uncoated glass plate, half-silvered mirror, soap bubbles.

• Both the transmitted and the reflected waves would have lower amplitudes than the incident wave.

• These two waves, when brought together again at a detector, would produce interference pattern as long as the original coherence between the two had not been destroyed.

15

16

• Consider a film of uniform thickness t and index of refraction n.

• Assume :i. Light rays travelling in air

are nearly normal to the two surfaces of the film.

ii. The film is non-absorbent.iii. The amplitude-reflection

coefficients at the interface are so low that only the first two reflected beams need to be considered.

17

• In practice, the amplitudes of the higher-order reflected beams generally decreases very rapidly.

• Part of the wave is reflected, at surface A, as ray 1 with a 180o phase change and another part is refracted into the film without phase change.

• These two rays can be consider as arising from two coherent; they are parallel on leaving the film.

18

• The OPD between these two rays isOPD = nf[AB + BC] – n1(AD)

• From the diagram, AB = BC = d/cos t,

AD = AC (nf sin f/n1)

AC = 2d tan t

then OPD = (2nfd/cos t) – n1(AD)

= 2nfd(1 – sin2t)/cos t

= 2nfd cos t

t

t

A

B

C

D

d

n1

nf

n2

Snell’s Law at point A

• Case 1: – Film is immersed in 1 medium (n1 = n2 = n)

– n < nf (soap film in air)

• The 1st reflected ray experience phase change by 180o.

• The 2nd ray, when it exits the upper surface, did not undergo phase change.– There exist a relative phase difference of between these

two rays.

• If OPD = m where m = 0, 1, 2,… then a central dark fringe is observed.

194)2(

4)2(cos

2)2(cos2

n

ft

tf

mn

md

mdn

20

• If OPD = where m = 0, 1, 2,… then a central

bright fringe is observed.

• A monochromatic light beam incident on a thin film can have the reflected beam either reinforced or diminished depending on the phase difference and hence angle of incidence and film thickness.

2

m

4)12(cos

2)12(cos2

nt

tf

md

mdn

21

• Case 2.1: Film is immersed in 2 mediums.– n1 < nf < n2

• Both reflected rays experience phase change by 180o. There is no relative phase difference between these two rays.

• Case 2.2: Film is immersed in 2 mediums.– n1 > nf > n2

• Both reflected rays are in-phase with the incident ray. There is no relative phase difference between these two rays.

• The above equations would have to be modified accordingly.

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22

Fringes of Equal Thickness

• Fringes of equal thickness exist for which optical thickness, nfd, is the dominant parameter rather than i.– Under the white-light illumination, the iridescence of thin

film (soap bubbles, oil slicks, even oxidized metal surfaces) is the result of variations in film thickness.

• Each fringe is the locus of all point in the film for which the optical thickness is a constant.– In general nf does not vary.

• This is useful in determining the surface features of optical elements.

23

• Consider the surface of an optical element (to be examined) in contact with an optical flat.– Optical flat – A flat surface that

deviates by not more than about /4 from a perfect plane.

– Individual flats of /200 can be achieved.

• If the two surfaces are plane, so that the film is wedge-shaped, the fringes will be practically straight, following the lines of equal thickness.– Fizeau’s fringes

24

• For a thin wedge of small angle , the optical path length difference between two reflected rays may be approximated from the thin film case i.e.

OPD = 2nfd cos t ≈ 2nfdm

where d is the thickness at a particular point that is d = x

a

x

d

25

• The condition for bright fringes becomes2nfdm = (m + ½)

2nf (xm) = (m + ½)

• From one fringe to the next, m increases by 1, the optical thickness of the film nd change by /2.

• If the test surface is not perfectly flat, the fringes will be irregular in shape.

• Calculating the order of the fringes, and as the film is air (n = 1.0), the position of the imperfect surface can be located for further polishing.

26

• A long-focus lens placed on a test plate produces a set of circular fringes, known as Newton’s rings.

• Observation are usually made at normal incidence according to the arrangement shown. – The beamsplitter reflects

light from source, S down to the plates.

– After reflection it is then transmitted by the beamsplitter to the low-power microscope above it.

S

Newton’s Rings

Circular fringes is called Newton’s rings.

The radius of the nth dark fringes

Important in the testing of optical lenses.

A circular pattern is obtained only when the lens is ground to a perfectly symmetric curvature.

n

Rmr

27

• The position of the maxima can be determined according to that of Fizeau’s fringes.

• Assuming the lenses are just touching at the centre, and let the radius of curvature of the lens be R, the thickness of the air film can be calculated as

d = r2/2R

• Substituting the above expression yields

(Minima)

Fig 37-18a, p.1191

n

Rmr

d

28

• The amount of uniformity in the concentric circular pattern is a measure of the degree of perfection in the shape of the lens.

• As the diameter of the fringes, r, depends on wavelength, white light will produce only a few coloured rings near the point of contact.

• Monochromatic light gives an extensive fringe system.

• When the contact is perfect the central spot is black due to the relative phase change by .

Newton’s Rings

Circular fringes is called Newton’s rings.

The radius of the nth dark fringes

Important in the testing of optical lenses.

A circular pattern is obtained only when the lens is ground to a perfectly symmetric curvature.

n

Rmr

29

Non-reflecting Films• A simple and very important application of the

principles of interference in thin films is the production of coated surfaces.

• A film of transparent substance of refractive index n’ (= n½) is deposited on glass of a larger index n to a thickness of one-quarter of the wavelength of light in the film so that

• Light reflected at normal incidence is almost completely suppressed by (destructive – rare-to-dense reflection) interference, this corresponds to the condition m = 0.

'4nd

30

• Light is NOT destroyed by a non-reflecting film, it is merely redistributed such that a decrease of reflection carries with it a corresponding increase of transmission.

• They can be use in optical system which require greatly reduced loss of light.

31

Mirrored Interferometers• Mirrored interferometers utilize arrangements of

mirrors and beamsplitters.

• Most important of these is Michelson interferometer.

, M2

, M1

, O, C

32

• An extended source (as oppose to point source) emits a wave part of which is directed at the beamsplitter, O.

• The wave is divided into two, one segment travels to movable mirror, M2, and the other travels to the fixed mirror, M1.

• These are then reflected by the mirrors and return to the beamsplitter.– The ray from M2 passes through the beamsplitter going

downward.– The ray from M1 deflected by the beamsplitter toward

the detector.

• They unite to produce interference pattern.

33

• One ray passes through the beamsplitter trice (reflect) while the other (transmit) only go through it once.

• The compensator plate is used to even out the path difference due to the beamsplitter.

• It is an exact duplicate of the beamsplitter, with the exception of any possible silvering of thin film coating on the beamsplitter.

• It is position at 45o so that it is parallel to the beamsplitter.

• Any optical path difference that exist now is from the actual path difference

34

• Fringes will not be observe by simply following the arrangement shown.– Mirrors M1 and M2 must be made exactly perpendicular

to each other. In physical arrangement, there are screws at M1 used to control the tilt of the mirror.

– Light in general must be monochromatic or nearly so, especially if the distances of OM1 and OM2 are appreciable.

– The distances OM1 and OM2 should be roughly within a few millimetres.

• The fringes are circular when they appear.

35

• Consider the figure below:

• The fixed mirror, M1 is replaced by M1’, its virtual reflection by the beamsplitter, O.

• These two mirrors, M1’ and M2, will have virtual images of the extended source at L, labelled L1 and L2.– These virtual sources are coherent – the phases of

corresponding points in the two are exactly the same at all instants.

P’L2

P’’L1

2d

PL

M2M1'

d

• If d is the separation between M1’ and M2, the virtual sources will be separated by 2d.

• When d is exactly an integral number of half-wavelength, 2d equals to an integral number of whole wavelength.– All rays of light reflected normal to the mirror will be in

phase.

• Reflected rays at an angle will in general not be in phase.– Its path difference is 2dcosq.– The rays will reinforce each other to

produce a maxima for 2dcosq = m

– For a given m, , and d, the angle q is a constant.• the maxima will lie in the form of circles about the foot of the

perpendicular from the eye to the mirrors.36

P’ P’’

2d

2dcos

37

• Notice that the reflected rays are parallel; they do not meet. The observer’s eyes, focus at infinity, serve as the condensing lens for the image to form.

• As M2 moves towards M1’, d decreases. From the expression earlier, – cos qm increases;

– qm decreases;– The mth order ring shrinks towards the centre.– Each remaining ring broadens as more and more

fringes vanish at the centre.• http://www.optics.arizona.edu/jcwyant/JoseDiaz/MichelsonInterferometerFri

nges.htm

• When d = 0, the central fringe would have spread out, filling the entire field of view.

38

• As M2 moves away from M1’ d increases, widely spaced fringes reappear from the centre.

• These will gradually become more closely spaced as the path difference increases.

• When M1’ and M2 are not parallel, inclined wrt each other making small angle, it results in wedge-shaped air film between then.

• Fizeau fringes, a pattern of straight parallel fringes, are observed.

• The interference fringes appear to diverge from a point behind the mirrors. The eye would have to focus on this point to observe the localized fringes.

39

P’

L2 L1

PL

M2M1'

40

d reducing d = 0 d increasing

Fringes and the relative position of the movable mirror and fixed mirror.

• The fringes are not exactly straight when d has an appreciable value.– They are in general curved and are always convex towards the

thin edge of the wedge. – If d is decreased, the fringes will move to the left across the

field, a new fringe crossing the centre each time d changes by /2.

– d ≈ 0; the fringes become straighter.– When d increases, the fringes curve in the opposite direction.