Post on 19-Nov-2020
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