1. Fresnel’s Biprism :

Some objected Young’s Experiment result and said that: “the bright fringes

were probably due to some complicated modification of the light by the

edges of the slits and not interference” Fresnel did new experiments to prove

that they were interference. S is the real source, and the other S’s are virtual.

Figure (4) Diagram of Fresnel's biprism experiment

To find d: the distance between sources or virtual images sources

When a: distance between the slit and biprism

( ) (21)

( )

In fig 4 the central point between be region fringe of maximum intensity

occurs there.

The fringes width of dark or bright fringe is

(22)

Where D: distance between source and screen

The wavelength is given by:

A

a

D

B

D= a+c

Example 2 :a) calculate the separation between the coherent sources formed by biprism whose inclined faces make angles of 2 with its base ,the slit being 0.1 m away from the biprism (n=1.5)?

( )

Example 3: In a Fresnel’s bi-prism experiment, the refracting angle of prism

were 1.5 o and refractive index of the glass was 1.5.with the angle slit 5 cm

from the bi-prism and using light of wavelength 580nm,fringes were formed

on a screen 1m from the single slit ,calculate the fringes width?

Solution

For a thin prism, a deviation ( ) ( ) ( ) (

) in this

case, therefore S1S =[ (5+0.75)x

] cm

S1 S2 =2S1S=7.5x

= 0.131 cm

The fringe width is

3. Fresnel’s Mirrors:

Here, Fresnel used two mirrors. Light from a slit S is reflected (not

refracted) from two plane mirrors slightly inclined to each other. The

mirrors produce two virtual images of S. They act exactly like two

sources formed by the biprism.

Figure (5): Fresnel's mirrors.

L

a b

Light from a slit is reflected in two plane mirrors slightly inclined to each

other. The mirrors produce two virtual images of the slit, a narrow slit S is

illuminated by a monochromatic light wavelength

It is placed parallel to the line of intersection of mirror surface, one portion

of cylinder wavelength is reflected from the first mirror and another portion

of the wave front is reflected from the second mirror. The virtual sources S1

and S2 are coherence sources which produced from the source S. the fringes

from (ae) region are equal width. The distance between virtual sources is d

Fringes width

( )

(23)

Then the distance between two virtual slits Then the eq.23 becomes:

( )

Compared between the fringes by Fresnel’s biprism and Mirrors: both cases are

similar ,but the double mirror are narrower than Fresnel’s biprism fringes.

4.Lloyd’s Mirrors :

This method is used to single long mirror about 30 cm. The

interference is produced by the light reflected from the mirror and the

light coming directly from the source (the slit 1)

Figure (6): Set up Lloyed’s Mirrors

O

A cylindrical wave front coming from a narrow slit S1 falls on the

mirror which reflects a portion of incident wave front, giving rise to a

virtual image of the slit s’.

The slit s1 and s2 represent at two coherent sources. The point O

equidistant from s and s’

An arrangement for producing an interference pattern with a single light

source

•Waves reach point o either by a direct path or by reflection.

•The reflected ray can be treated as a ray from the source S’ behind the

mirror.

o

O

O is central (zero-order) fringes. With white light the central point o is

expected white, but in practice it is dark. The phase change leads to a

path difference of λ/2 and here destructive interference occurs.

To determine of wavelength , the fringes width is:

Comparison between the fringes produced by biprism and

Lloyd’s mirror:

1. In biprism the complete set of fringes is obtained. In Lloyd’s

mirror a few fringes on one side of the central fringe are noted, the

central fringe which being invisible

2. In biprism the zero-order (central fringe) is bright, but the central

fringe of Lloyd’s mirror is dark.

3. The central fringe is less sharp in biprism than in Lloyd’s mirror.

Example (3):

In young’s experiment, interference bands are produced on screen placed 1.5

m from two slits 0.15 mm apart and illuminated by the light of wavelength

6000Å. Find

a) fringe width

b) Change in fringe width if the screen is taken away from the slits by 50

cm?

Solution

a)

b)

= 2mm

Example (4):

A light falls on two slits ( 2 mm) apart and produces on a screen 1 m away

from the fourth –order bright line 1 mm from the center of the pattern .what

is the wavelength of the light used?

Solution:

N=4 order , d=2mm

D=1 m

(

)(

)

( )( (

)

(

)

(

Example (5): yellow light pass through two slits and interference

pattern is observed on a screen

1) The bright fringes will increase in width if the yellow light is replaced

blue

2) The bright fringes will increase in width if the distance between slit

minimized

3) The intensity of light decreases if it is far from the central fringes

4) The intensity of light is constant if it is far from the central fringe

which is the correct statement?

Solution:

d:distance between slit is smaller than the distance between slit and

screen, so the angle is very small

(

)(

)

n: no. of bright lines is proportional inversely to the wavelength

Yellow wavelength (small frequency) is larger than blue light.

Blue light is changed; the wavelength λ is decrease, so the number of

bright lines is increase.

This statement is correct

2) The bright fringes will increase in width if the distance between slit

minimized

If The bright fringes will increase in width if the distance between slit

minimized, the number of bright lines (n) decreses.

(

)

The bright fringes will increase in width if the distance between slit

minimized

This statement is incorrect

3) The intensity of light decreases if it is far from the central fringes

This statement is correct

Intensity related to the light level. Intensity is proportional inversely

to the distance. If the distance is increased the intensity is decrease(the

light dimmer)

4)The intensity of light is constant if it is far from the central fringe

which is the correct statement?

This statement is incorrect

Example (6):

Suppose in double slit arrangement, d=0.150 mm, L=120 cm, the

wavelength =833 nm, and the width fringe y=2 cm

a) What is the path difference δ for the rays from the two slits arriving at

point P

b) Express this path difference in terms of λ?

c) Does point P corresponds to a maximum or minimum, is intermediate

condition?

Solution:

a) δ = d sin θ bright = d

b) from a) we get

Path difference

c) at point P the path difference δ = the multiple

wavelength , so the intensity y at point p is maximum

COHERENT SOURCES

The result is that the difference in phase between any pair of points in the

two sources always remains constant, and so the interference fringes are

stationary. Such as Young's experiment and in Fresnel's mirrors and biprism,

the two sources Sl and S2 always have a point-to-point correspondence of

phase, since they are both derived from the same source any interference

experiment with light that the sources must have this point-to-point phase

relation, and sources that have this relation are called coherent sources.

While special arrangements are necessary for producing coherent sources of

light, the same is not true of microwaves, which are radio waves of a few

centimeters wavelength. These are produced by an oscillator which emits a

continuous wave, the phase of which remains constant over a time long

compared with the duration of an observation. Two independent microwave

sources of the same frequency are therefore coherent and can be used to

demonstrate interference. Because of the convenient magnitude of their

wavelength, microwaves are used to illustrate many common optical

interference and diffraction effects. If in Young's experiment the source slit S (Fig. 6) is made too wide or the

angle between the rays which leave it too large, the double slit no longer

represents two coherent sources and the interference fringes disappear

Figure (6): diagram of young’s Experiment

Interference fringes with white light and monochromatic

If the slit illuminated by white light in the in Young's experiment the source

slit S is used and Fresnel’s biprism. The interference pattern consist of a

central white light around it on both sides a few coloured fringes. The

central fringe will be dark and others will be colors. With white light, fringes

are observed only when the path difference is small.

(8-3-2 ). INTERFERENCE BY DVSION OF AMPLITUDE

1-MICHELSONt INTERFEROMETER

Principle: two beams are interference formed by division of amplitude.

Construction:

The main optical parts consist of two highly polished plane mirrors M1 and

M2and two plane-parallel plates of glass Gland G2• Sometimes the rear side

of the plate G1 is lightly silvered , so that the light coming from the source S

is divided into (1) a reflected and (2) a transmitted beam of equal intensity.

The light reflected normally from mirror M1 passes through G1 a third time

and reaches the eye as shown. The light reflected from the mirror M 2 passes

back through G2 for the second time, is reflected from the surface of G1 and

into the Even when the above adjustments have been made, , will not

produce the desired system of fringes in this case. The reason

for this will appear when we consider the origin of the fringes. Second, the

light must in general be monochromatic, or nearly so. Especially is this true

if the distances of M1 and M2 from G1 are appreciably different

Construction of Michelson

Interference condition for the two rays is determined by their path length

Differences. Light, but it is indispensable when white light is used.

The effective thickness of the air film is varied by moving mirror M1

parallel to itself with a finely threaded screw adjustment. Under these

conditions. As M1 is moved, the fringe pattern shifts. For example, if a dark

fringe appears in the field of view (corresponding to destructive

interference) and M1 is then moved a distance λ/4 toward M, the path

difference changes by λ/2 (twice the separation between M1 and What was a

dark fringe now becomes a bright fringe. As M1 is moved an additional

distance λ /4 toward M, the bright fringe becomes a dark fringe. Thus, the

fringe pattern shifts by one-half fringe each time M1 is moved a distance λ

/4. The wavelength of light is then measured by counting the number of

fringe shifts for a given displacement of M1. If the wavelength is accurately

Known (as with a laser beam), mirror displacements can be measured to

Within a fraction of the wavelength.

An extended source suitable for use with a Michelson interferometer may be

obtained in anyone of several ways. A sodium flame or a mercury are, if

large enough, may be used without the screen L shown in Fig. 9. If the

source is small, a ground-glass screen or a lens at L will extend the field of

view. Looking at the mirror M 1 through the plate G1 one then sees the

whole mirror filled with light.

In order to obtain the fringes, the next step is to measure the distances of M1

and M 2 to the back surface of G1 roughly with a millimeter scale and to

move M 1 until they are the same to within a few millimeters. The mirror M2

is now adjusted to be perpendicular to M1 by observing the images of a

common pin, or any sharp point, placed between the source and G1 Two

pairs of images will be seen, one coming from reflection at the front surface

of G1 and the other from reflection at its back surface. When the tilting

screws on M2 are turned until one pair of images falls exactly on the other,

the interference fringes should appear. When they first appear, the fringes

will not be clear unless the eye is focused on or near the back mirror M1 so

the observer should look constantly at this mirror while searching for the

fringes.

The purpose of the plate G2, called the compensating plate, is to

render the path in glass of the two rays equal. This is not essential for

producing fringes in monochromatic

Types of fringes

1) Circular fringes

Circular fringes are produced with monochromatic light when the

mirrors M1 and M2 are e xactly perpendicular to each other.

The ones used in most kinds of measurement with the interferometer. Their

origin can be understood by reference to the diagram of Fig. 10. Here the

real mirror M2 has been replaced by its virtual image M2’ formed by

reflection in G1•

M2’ is then parallel to M1• Owing to the several reflections in the real

interferometer, we may now think of the extended source as being at L, behind the observer, and as forming two virtual images L1 and L2 in M1 and

M;. These virtual sources are coherent in that the phases of corresponding

points in the two are exactly the same at all instants. If d is the separation

M1M2’ the virtual sources will be separated by 2d.

Figure (8) Schematic of the Michelson interferometer.

When d is exactly an integral number of half wavelengths, i.e., the path

difference 2d equal to an integral number of whole wavelengths, all rays of

light reflected normal to the mirrors will be in phase. Rays of light reflected

at an angle, however, will in general not be in phase. The path difference

between the two rays coming to the eye from corresponding points P' and P"

is 2d cos θ, as shown in the figure. The angle θ is necessarily the same for

the two rays when M1 is parallel to M2’ so that the rays are parallel. Hence

when the eye is focused to receive parallel rays (a small telescope is more

satisfactory here, especially for large values of d) the rays will reinforce

each other to produce maxima for those angles e satisfying the relation

Figure (10)

Since for a given m, A., and d the angle θ is constant, the maxima

will lie in the form of circles about the foot of the perpendicular

from the eye to the mirrors. By expanding the cosine, it can be

shown from Eq. (l3g) that the radii of the rings are proportional

To the square roots of integers, as in the case of Newton's rings.

The intensity

Distribution across the rings follows Eq. (13b), in which the phase

difference is given by

Fringes of this kind, where parallel beams are brought to interference with a

Phase difference determined by the angle of inclination θ, are often referred

to as fringes of equal inclination.

This type may remain visible over very large path differences

The upper part of Fig. 10 shows how the circular fringes look under different

conditions. Starting with M1 a few centimeters beyond M2, the fringe

system will have the general appearance shown in (a) with the rings very

closely spaced. If M1 is now moved slowly toward M2 so that d is

decreased, its radius because the product 2d cos θ must remain constant. The

rings therefore shrink and vanish at the center, a ring disappearing each time

2d decreases by λ, or d by λ/2. This follows from the fact that at the center

cos θ = 1, so that mλ=2d

To change m by unity, d must change by λ/2. Now as M1 approaches M2 the rings become more widely spaced, as indicated in Fig. 13P(b), until

finally we reach a critical position where the central fringe has spread out to

cover the whole field

Appearance of the various types of fringes observed in the

Michelson interferometer.

Upper row, circular fringes. Lower row, localized fringes. Path

difference increases outward, in both directions, from the center.

2) Straight fringes: If M1 and M2 are not exactly perpendicular figure (11

b) , a wedge shaped air film is formed between M1 &M2’.the fringes become

practicaly straight (see fig)

When the two rays are viewed as shown, the image of M2 produced by

the mirror M is at which is nearly parallel to M1. (Because M1 and M2

are not exactly perpendicular to each other, the image is at a slight angle

to M1.)Hence, the space between and M1 is the equivalent of a wedge-

shaped air film. When M1 actually intersects M2’ in the middle (11 b)

The fringes are fringes of equal thickness. The fringes located in air film

it.

a b

3) White light fringes: if white light is used, the central fringe will be

dark and other will be coloured.fringes are observed only when the path

difference is small. These fringes are important because they are used to

locate the position of zero path difference.

Figure 11:Micheslon’ interferometer using white light

a)wit compensating plate b) with beam splitter

VISIBILITY OF THE FRINGES

There are three principal types of measurement that can be made with the

interferometer:

(I) width and fine structure of spectrum lines, (2) lengths or displacements

in terms of wavelengths of light, and (3) refractive indices

In case of Michelson interferometer, the intensity is given by:

(13)

Michelson tested the lines from various sources and concluded that a certain

red line in the spectrum of

cadmium was the most satisfactory. He measured the visibility, defined as

where Imax and Imin are the intensities at the maxima and minima of the fringe

pattern. The more slowly V decreases with increasing path difference, the

sharper the line.

Example: which case will the visibility be 1? What does a visibility of one

mean?

Imax= 4I, I min = 0

Uses of Michelson's Interferometer

1. Determination of wavelength of monochromatic light

2d1= m1 λ (1)

2d2=m2 λ

( )

𝛌 ( )

d2 -d1: no. of fringes counting the center of the field of view.

(Conversely, if x is increased, the fringe pattern will expand.)

N: no. of fringes

2- Determination of different in wavelength between two neighbor lines

or two waves

Let the source of light emit close wavelength 𝛌 and ,condition λ1>λ2.

The apparatus is adjusted to form circular rings. the arrangement of the

position of mirror M1 is moved and reached when a bright fringes of one set

falls on the bright fringe of the other an fringes are again distinct. So the

small difference λ is given by:

3-thickness of a thin transparent sheet

4-determination of the refractive index of gases

The path difference introduce between the two interfering beam is 2(n-1) L

Where n: Refractive index of gas

L: length of the tube. If m fringes cross the center of the field of view thus:

2(n-1)L=mλ

Example (6):

In an experiment for determine the refractive index of gas using Michelson

interferometer a shift of 140 fringes is observed. when all the gas is removed

from the tube. If the wavelength of light used is 5460Å and the length of the

tube is 20 cm, calculate the refractive index of the gas?

Solution