[Syllabus: Interference: Young’s Experiment, biprism, colour of thin film, Newton’s ring, Michelson and Fabry-

Perot interferometer]

Interference of light:
Interference of light is the redistribution of energies of the light waves when light from two different sources pass
over each other. The redistribution of energies causes the formation of regions of maximum amplitudes and
regions of minimum amplitudes.
The regions of maximum amplitudes (bright) and minimum amplitude (dark) alternate and may take the form of
straight bands, or circular rings or any other complex shape. The alternate bright and dark bands are called
interference fringes.

Theory of interference:
Let us assume that the electric field components of the two waves arriving at point P vary with time as
E A = E1 sin t .............................(i)
and E B = E2 sin(t +  ) .......................(ii )

where  is the phase difference between them. According to principle of superposition, the resultant electric field
at the point P due to the simultaneous action of the two waves is given by
ER = E A + EB
= E1 sin t + E 2 sin(t +  )
= E1 sin t + E 2 (sin t cos  + cos t sin  )
= ( E1 + E 2 cos  ) sin t + E 2 sin  cos t ................................(iii)
Let
E1 + E2 cos = E cos .........................(iv)
and E2 sin  = E sin  .......... .........................(v)

E R = E cos sin t + E sin  cos t

or , E R = E sin(t +  ) ………….(vi)
Equation (vi) shows that the superposition of two sinusoidal waves produces a sinusoidal wave of same frequency
having a phase difference  and a different amplitude E.
Squaring and adding eq.-(iv) and (v) we get,
( E1 + E 2 cos  ) 2 + ( E 2 sin  ) 2 = E 2 (cos 2  + sin 2  )
E12 + 2 E1 E 2 cos  + E 22 (cos 2  + sin 2  ) = E 2 ..............................(vii)
Thus it is seen that the square of the amplitude of the resultant wave is not a simple sum of the square of the
amplitudes of the super imposing waves, there is an additional term which is known as the interference term.
The intensity of a light wave is given the square of the amplitude, i.e.,
I  E2
From equation (vii)
I = I 1 + I 2 + 2 I 1 I 2 cos  .......................(viii )
We see that the resultant intensity at P on the screen is not just the sum of the two intensities due to the separate
waves. The term 2 I1 I 2 cos  is known as the interference term.
When the phase difference between two waves is zero, i.e., =0 we have the maximum amount of light. Thus,
I max = I1 + I 2 + 2 I1 I 2 . If I1 = I2 = I0, we get, Imax= 4I0
When the phase difference is =180, we have the minimum amount of light, I min = I1 + I 2 − 2 I1 I 2 .
For I1=I2=I0, Imin=0
It means that the resultant intensity I will be less than the sum of the intensities due to the two sources.

At points that lie between the maxima and minima, when, I1 =I2 =I0
I = I 0 + I 0 + 2 I 0 I 0 cos = 2 I 0 (1 + cos )
We get,
I = 4 I 0 cos 2  / 2 ................................................(viii)
Equation (viii) shows that the intensity varies along the screen in accordance with the law of cosine square.
Following figure shows the variation of intensity as a function of phase angle .

Coherent sources: Two sources are said to be Coherent if they emit light waves of the same frequency, nearly
the same amplitude and always in phase with each other or there is a constant phase difference between them. It
means that two sources must emit radiation of the same color. In actual practice, it not possible to have two
independent sources which are coherent. But for experimental purposes, two virtual sources formed from a single
source can act as coherent sources.
Example of production of coherent sources:

Conditions for Interference:

(i) The light source must be coherent and in the same state of polarisation.
(ii) The amplitudes and intensities of the light sources should be equal. This will ensure there is sufficient contrast
between the maxima and the minima of the interference pattern.
(iii) The light source must be monochromatic and a point source. E.g., lasers.
(iv) The sources of light must be near enough to produce clear and sharp fringes.
(v) The source and screen should be far enough to produce wide but sharp fringes.

Young two slit experiment:

Thomas Young performed the double-slit experiment in 1801 that helped understand the interference of light.
This experiment helped prove the wave nature of light.
By proving that light also moved like waves, Young successfully established, that like all electromagnetic waves,
light also underwent interference of light when two light waves passed over each other.
He used one source of light, the sun, but passed it through two small openings or holes, small enough that they
could be called slits. This ensured that clear and sharp fringes of light were produced and that the source of light
would now be coherent. A single source of light also ensured that it was monochromatic. This way, he observed
an interference pattern on the screen, depending on whether the light waves were crest to crest or crest to trough,
i.e., same phase or opposite phase, respectively.

The above diagram demonstrates how the plane light waves arrive at a barrier that contains two parallel slits,
S1 and S2. These two slots serve as a pair of coherent light sources, because waves emerging from them originate
from the same wave front and therefore maintain the same phase relationship. The light from S1 and S2 produces
a visible pattern of bright and dark parallel bands called fringes on the screen.
When the light from S1 and S2 arrive in the same phase, constructive interference takes place, making a bright
fringe appear. When the light from S1 and S2 arrive in different phases, destructive interference takes place,
making a dark fringe appear.

Theory:

Lens

S1 R
S d  O
Source A
S2 T

Light from a monochromatic source passes through a lens and is focused on to a single slit S. It then falls on a
double slit, S1 and S2 separated by a distance d. This produces two wave trains that interfere with each other in the
region on the right of the diagram. The interference pattern at any distance from the double slit may be observed
with a micrometer eyepiece or by placing a screen in the path of the waves at a distance D from the slits.
Let us consider the effects at a point P a distance X from the axis of the apparatus. The path difference between
the two rays at P is S2P-S1P.
For a bright fringe the path difference must be a whole number of wavelengths and for a dark fringe it must be an
odd of half wavelengths.
Consider the triangles S1 PR and S2 PT
𝑑 2
𝑆1 𝑃2 = (𝑋 − ) + 𝐷 2
2
𝑑 2
𝑆2 𝑃2 = (𝑋 + ) + 𝐷2
2
Therefore,
𝑆2 𝑃2 − 𝑆1 𝑃2 = 2𝑥𝑑
(𝑆2 𝑃 + 𝑆1 𝑃)(𝑆2 𝑃 − 𝑆1 𝑃) = 2𝑥𝑑
But, 𝑆2 𝑃 + 𝑆1 𝑃 = 2𝑑
Therefore, the path difference
𝑥𝑑
𝑆2 𝑃 − 𝑆1 𝑃 = 𝐷
𝑥𝑑
For a bright fringe: Condition for bright fringe, m =
𝐷
  𝑥𝑑
For a dark fringe: Condition for dark fringe, (2m + 1) =
2 𝐷

mD
The mth order bright fringe occurs when xm =
d
(m + 1)D
And the (m+1)th order bright fringe occurs when xm+1 =
d
D
The fringe separation,  is given by  = xm+1 − xm =
d
From this equation it is observed that,
(i) the fringe width  is independent of the order of fringe. It is directly proportional to the wavelength of light,
i.e.    .
(ii) the width of the fringe is directly proportional to the distance of the screen from the two slits.
(iii) The width of the fringe is inversely proportional to the distance between the two slits. The closer the slits, the
wider will be the fringes.
Problem:
Young’s Experiment: Calculate the fringe width for light of wave length 550 nm in a Young’s slit experiment
where the double slits are separated by 0.75 mm and the screen is placed 0.80 m from them.
𝐷
10 c know fringe width, 𝑥𝑚 =
𝑑
550×10−9 ×0.80
=
0.75×10−3
= 3.3 × 10−4 𝑚
=0.33mm
Newton’s rings:
Newton’s rings are formed when a plano-convex lens of large radius of curvature placed on a sheet of plane glass,
is illuminated from the top with monochromatic light. The combination forms a thin circular air film of variable
thickness in all directions around the point of contact of the lens and the glass plate. The locus of all points
corresponding to specific thickness of air film falls on a circle whose center is at M. At first Newton observed
these concentric circular fringes and hence they are called Newton’s rings.
The experimental set up for observing Newton’s rings is shown in the following figure. Monochromatic light from
an extended source S is rendered parallel by a lens L. It is incident on a glass plate inclined at 45 to the horizontal
and is reflected normally down on to a plano-convex lens placed on a flat glass plate. Part of the light incident on
the system is reflected from the glass-to-air boundary. The remainder of the light is transmitted through the air
film. It is again reflected from the air-to-glass boundary, say, from point. The two rays reflected from the top and
bottom of the air film are divided through division of amplitude from the same incident rays and therefore
coherent. The condition for brightness or darkness depends on the path difference between the two reflected light
rays.

R R
N P
t
M Q
r
Fig. (a) Fig. (b)

Condition for Bright and Dark Rings:

The optical path difference between the rays is given by  = 2t cos r −  / 2. Since  = 1 for air and cos r =1
for normal incidence of light,
 = 2t cos r −  / 2.
Intensity maxima occur when the optical path difference  = m. . If the difference in the optical path between the
two rays is equal to an integral number of full waves, then the rays meet each other in phase. The crests of one
wave falls on the crests of the other and the waves interfere constructively. Thus, if

2t −  / 2 = m

bright fringe is obtained.

Intensity minima occur when the optical path difference is  = (2m + 1)  . If the difference in the optical path between
2
the two rays is equal to an odd integral number of half waves, then the rays meet each other in opposite phase.
The crests of one wave fall on the troughs of the other and the waves interfere destructively.
Hence, if 2t −  / 2 = (2m + 1) / 2 dark fringe is produced.
Circular fringes:
In Newton’s ring arrangement, a thin air film is enclosed between a Plano-convex lens and a glass plate. The
thickness of the air film at the point of contact is zero and gradually increases as we move outward. The locus of
points where the air film has the same thickness then fall on a circle whose Centre is the point of contact. Thus,
the thickness of air film is constant at points on any circle having the point of lens-glass plate contact as the centre.
The fringes are therefore circular.
Radii of Dark Fringes:
Let R be the radius of curvature of the lens (Fig-b). Let a dark fringe be located at Q. Let the thickness of the air
film at Q be PQ = t. Let the radius of the circular fringe at Q be MQ = rm. By the Pythagoras theorem from ONP

OP 2 = NP 2 + ON 2
or , R 2 = rm2 + ( R − t ) 2
or , rm2 = 2 Rt − t 2
As condition for darkness at Q is that, 2t = m
 rm2 = mR
rm = mR
The radii of dark fringes can be found by inserting values 1, 2, 3,……form. Thus
diameter of m th dark ring

Dm = 2rm
………………(1)
Dm = 2 mR
Similarly we obtain the diameter of the m th bright fringe
Dm = 2 mR
The diameter of the n th dark ring

Dn = 2 nR ………………(2)
Application of Newton’s ring:
Determination of wave-length of monochromatic light and radius of curvature of a plano-convex lens

Squaring and subtracting equ(1) and (2)

D m − D n = 4( m − n )  R
2 2

Dm − Dn
2 2

Or, R =
4(m − n)
Dm − Dn
2 2

R= ………(3)
4 (m − n)
 Dm − Dn
2 2

= ………(4)
4 R ( m − n)
Using equation (3) radius of curvature of a plano-convex lens can be determined.
Using equation (4) wave-length of light can be determined.
Spacing Between Fringes:
It is seen that the diameter of dark rings is given by Dm = 2 mR , m = 1, 2, 3, …..
The diameters of dark rings are proportional to the square root of the natural numbers. Therefore, the diameter of
the ring does not increase in the same proportion as the order of the ring, for example, if m increases as 1,2,3,4......
the diameters are
D1 = 2 R
D2 = 2(1.4) R
D3 = 2(1.7) R
D4 = 2(2) R and so on.
Therefore, the rings get closer and closer, as m increases. This is why the rings are not evenly spaced.

Problem: A series of rings formed in Newton’s rings experiment with sodium light was viewed by reflection.
The diameter of the nth dark ring was found to be 0.28 cm and that of the (n+10) th 0.68 cm. If the wavelength of
sodium light is 589 nm, calculate the radius of curvature of the lens.
𝑟2
We know, = 𝑚
𝑅
(0.14×10−2 )2
Solution: For n th ring, = 𝑛 × 589 × 10−9 ........................................(1)
𝑅
(0.34×10−2 )2
and for (𝑛 + 10)𝑡ℎ 𝑟𝑖𝑛𝑔. = (𝑛 + 10)589 × 10−9 ................. (2)
𝑅
Where R is the radius of curvature of the true lens.
from (2)  (1)
(0.34)2 𝑛 + 10
 = 𝑔𝑖𝑣𝑖𝑛𝑔 𝑛 = 2
(0.14)2 𝑛
Thus from (i), R = 1.66m

Fresnel’s biprism:
A Fresnel Biprism is a thin double prism placed base to base and have very small refracting angle (0.5). This is
equivalent to a single prism with one of its angle nearly 179° and other two of 0.5 each. The interference is
observed by the division of wave front. Monochromatic light through a narrow slit S falls on biprism, which
divides it into two components. One of these component is refracted from upper portion of biprism and appears
to come from S1 where the other one refracted through lower portion and appears to come from S2. Thus S1 and
S2 act as two virtual coherent sources formed from the original source. Light waves arising from S1and S2 interfere
in the shaded region and interference fringes are formed which can be observed on the screen.

Applications of Fresnel's Biprism

Fresnel biprism can be used to determine the wavelength of a light source (monochromatic), thickness of a thin
transparent sheet/ thin film, refractive index of medium etc.
A. Determination of wave length of light
D
As expression for fringe width is  =
d
Biprism can be used to determine the wavelength of given monochromatic light using the expression.
 d
=
D
Experimental Arrangement.
Light from monochromatic source is made to fall on a thin slit mounted vertically on a rigid optical bench fitted
with a scale. The biprism and the screen (in this case an eye piece) are also mounted vertically. The eye piece can
be moved in the plane perpendicular to the axis of bench using a micrometer based translation stage.

Experimental Set-Up: Fresnel Biprism Experiment

(i) Measurement of fringe width: To get β, fringes are first observed in the field of view of the microscope. The
vertical wire of the eyepiece is made to coincide with one of the fringes and screw of micrometer is moved
sideways and number of fringes is counted.
β =Distance moved / number of fringes passed
(ii) Measurement of D: This distance between source and eyepiece is directly measured on the optical bench scale.
(iii) Determination of d: To determine the separation between the two virtual sources ( d ), a convex lens of short
focal length is introduced between the biprism and the eye piece, keeping the distance between the slit and
eyepiece to be more than four times the focal length of lens. The lens is moved along the length of bench to a
position where two images of slits are seen in the plane of cross wires of eye piece. The distance between these
two images of slit is measured by setting the vertical cross wire successively on each of images and recording the
two positions of cross wire using micrometer. Let this separation be d1 . Now the lens is moved such that for
another position of lens, again two images of slit are seen on eye piece. Let d2 be the separation between these
two images.

d1 v
=
d u
d2 v '
=
d u'

Since these two positions of lens are conjugate, the separation between the virtual source ‘d ' is given by using
equations 1 and 2 as d = d1d 2
where d1 and d2 are the distance between S1 and S2 for two positions of lens.
B. Determination of thickness of a thin film:
To determine the thickness of transparent thin sheet (mica), the monochromatic source is replaced by white light
source. The position of this central white fringe is recorded by focusing the cross wire of eye piece on it and taking
this reading of micrometer scale. Now mica sheet is introduced in the path of one wave. (such that it blocks the
one half of biprism). By doing it the one wave traverse an extra optical path and the path difference between the
two waves is not same and entire fringe pattern shifts. The central white fringe is now shifted to another position
of cross wire. If ‘S' is the shift in position of white fringe and μ be the refractive index of mica sheet, thickness ‘t'
of mica sheet is given by
S d
t= cm
D  −1
Time required by light to reach from S1 to point P
S1 P − t t
T= +
c v
c
where v =

S1 P + t (  − 1)
T=
c
Hence equivalent path that is covered by light in air is S1P + t ( − 1) Optical path difference at P

Therefore, nth fringe shift is given by

D(  − 1)t
x =
d
D
As  =
d
 (  − 1)t
x =

where λ is the wavelength of the wave; is displacement of fringes and is fringe width.

Interferometer: An interferometer is an instrument in which the phenomenon of interference is used to make

precise measurements of wavelengths or distances.
Michelson Interferometer:
Principle: In Michelson interferometer, a beam of light from an extended
source is divided into two parts of equal intensities by partial reflection and
refraction. These beams travel in two mutually perpendicular directions and
M
come together after reflection from plane mirrors. The beams overlap on
each other and produce interference fringes.
Construction: The schematic of a simple Michelson interferometer is
shown the following figure. It consists of a beam splitter G1, a
compensating glass plate G2 and two plane mirrors M1 and M2. The beam
splitter, G1, is a partially silvered plane parallel plate. The compensating
plate G2 is a simple plane parallel glass plate having the same thickness as M
G1. The two plates G1 and G2 are held parallel to each other and are inclined
at an angle of 45 with respect to the mirror M2. The mirror M1 is mounted
on a carriage and can be moved exactly parallel to itself with the help of a
micrometer screw. The distance through which the mirror M1 is moved can
be read with the help of a graduated drum attached to the screw.
Displacements of the order of 0.1 m can be easily read. The plane mirrors
Michelson Interferometer.
M1 and M2 can be make perfectly perpendicular with the help of the fine screws attached to them. The interference
bands are observed in the field of view of the telescope T.
Workings: Monochromatic light from an extended source S is rendered parallel by means of a controlling lens L
and is made incident on the beam splitter G1. It is partly reflected at the back surface of G1 along AC and partly
transmitted along AB. The beam AC travels normally towards the plane mirror M1 and is reflected back along the
same path and comes out along AT. The transmitted beam travels towards the mirror M2 and is reflected along
the same path. It is reflected at the back surface of G1 and proceeds along AT. The two beams received along AT
are produced from a single source through division of amplitude and are hence coherent. The superposition of
these two beams leads to interference and produces interference fringes.
From the above figure, it is clearly seen that a light ray starting from the source S and undergoing reflection at the
mirror M1 passes through the glass plate G1 three times. On the other hand, in the absence of G2, the ray reflected
at M2 travels through the glass plate G1 only once. For compensating this path difference, a compensating glass
plate G2 of the same thickness is inserted onto the path AB and is held exactly parallel to G1.
If we look into the instrument from T, we see mirror M 1 and in addition we see a virtual image, M2 of mirror M2.
Depending on the position of the mirrors, image M2 may be in front or behind, or exactly coincident with mirror
M1 .
(a) Circular fringes: Circular fringes are produced with monochromatic light when the mirrors M1 and M2 are
exactly perpendicular to each other.
(b) Localized films: When the two mirrors are tilted, they are
not exactly perpendicular to each other and therefore, the
mirrors M1 and the virtual mirror M2 are not parallel. In this
case the air path difference between them is wedge-shaped and
the fringes appear to be straight line. If one of the mirror is
moved, the fringes move across the field. The position of any
particular bright fringe is taken up by the one next to it. The
fringes can be counted as they pass a reference mark. If m is
the fringes move across the field of view when M1 moves
through a distance d, then
m
d=
2
2d
or ,  =
m

Multiple beam interference: Multiple beam interference occurs if light falls on a glass with parallel sides. A
Fabry-perot interferometer uses this type of glass. This is also known an etalon.

E0

ttE0
tt(r2e-i)2E0 tt(r2e-i)4E0

ttr2e-iE0 tt(r2e-i)3E0

The transmitted wave is an infinite series of multiply reflected spatially overlapped beams. If r and t are the
reflection and transmission coefficients from glass to air, respectively and coefficients from air to glass.
𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑤𝑎𝑣𝑒
𝑡=
𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑤𝑎𝑣𝑒
 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑤𝑎𝑣𝑒
and 𝑟=
𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑤𝑎𝑣𝑒
i t
Let us consider the incident wave as E = E0 e and  is the phase change due to round-trip reflection.
Thus considering Fig.1 the transmitted wave field can be written as

𝐸0𝑡 = 𝑡𝑡′𝐸0 + 𝑡𝑡 ′ 𝑟 2 𝑒 −𝑖 𝐸0 + 𝑡𝑡′(𝑟 2 𝑒 −𝑖 )2 𝐸0 + 𝑡𝑡′(𝑟 2 𝑒 −𝑖 )3 𝐸0 +............................

= 𝑡𝑡′𝐸0 [1+(𝑟 2 𝑒 −𝑖 ) + (𝑟 2 𝑒 −𝑖 )2 +(𝑟 2 𝑒 −𝑖 )3 +........................................................]
𝑡𝑡′𝐸0
=
1−𝑟 2 𝑒 −𝑖

𝐸 2
The transmittance is defined as, 𝑇 = | 𝐸𝑜𝑡|
0
2
𝑡𝑡′
𝑇=| |
1 − 𝑟 2 𝑒 −𝑖
(1 − 𝑅)2
=
(1 + 𝑅𝑒 −𝑖 )(1 − 𝑅𝑒 +𝑖 )
(1 − 𝑅)2
=
(1 + 𝑅 2 − 2𝑅 𝑐𝑜𝑠𝛿)
(1 − 𝑅)2
=
𝛿
(1 + 𝑅 2 − 2𝑅[1 − 2𝑠𝑖𝑛2 ]
2
(1 − 𝑅)2
=
𝛿
(1 − 2𝑅 + 𝑅 2 + 4𝑅𝑠𝑖𝑛2 ]
2
(1 − 𝑅)2
𝑇=
𝛿
(1 − 𝑅)2 + 4𝑅𝑠𝑖𝑛2
2
𝛿
The intensity will be maximum when 𝑠𝑖𝑛2 = 0 or, 𝛿 = 2𝑚𝜋, where, 𝑚 = 0, 1, 2, 3 … … … … ….
2
𝛿
The intensity will be minimum when, 𝑠𝑖𝑛 2 = 1 or,  = (2m+1)  𝑚 = 0, 1, 2, 3 … … … … ….
2
4𝑅
The term 𝐹 = is called the coefficient of fineness.
(1−𝑅)2
FABRY-PEROT INTERFEROMETER
The Fabry-Perot interferometer uses the phenomenon of multiple beam interference that arises when light shines
through a cavity bounded by two reflective parallel surfaces. Each time the light encounters one of the surfaces, a
portion of it is transmitted out, and the remaining part is reflected back. The net effect is to break a single beam
into multiple beams which interfere with each other. If the additional optical path length of the reflected beam
(due to multiple reflections) is an integral multiple of the light's wavelength, then the reflected beams will interfere
constructively. More is the number of reflection inside the cavity, sharper is the interference maximum. Using
Fabry-Perot (FP) interferometer as a spectroscopic tool, concepts of finesse and free spectral range can be
understood.
Principle of Working: The basic principle of working of the Fabry-Perot interferometer is schematically
explained in the adjacent figure.

Fig 1: Schematics of a Fabry-Perot Interferometer

Two partial mirrors G1 and G2 are aligned parallel to one another at a distance d, forming refracting cavity. When
radiated by a monochromatic light (a laser here) of wave length λ at an angle of incidence , multiple reflections
takes place inside the cavity. Part of the light is transmitted each time the light reaches the second reflecting
surface. All such transmitted light rays interfere with each other to give rise to a maxima or minima depending on
the path difference between them. Let n be the refractive index of the medium in the cavity (in this case it is air).
Then the optical path difference between two neighbouring rays is
 = 2nd cos
Then the phase difference is given by
 2
= 


 = 2nd cos 

In the figure below calculation of path difference for a general cavity is shown
shown where α and β are the angles of incidence and refraction, respectively.

Thus the resultant transmitted light intensity IT is

1
IT = I 0  (3)
4R 2 
1+ sin
(1 − R)2 2
where, I0 is the incident intensity, R is the reflectivity of the mirrors. It can be noticed that IT
varies with δ.
IT is maximum when  = m (m = 0,1, 2) or  = 2m (4)

IT is minimum when  = (2m + 1) (m = 0,1, 2) or  = (2m + 1) (5)
2
The complete interference pattern appears as a set of concentric rings. The sharpness of the rings depends on a
parameter called coefficient of fineness, F, defined as
4R
F=
(1 − R) 2
Determination of wavelength λ: Using the relations 1 and 4 (or 5) wavelength of the incident light can be
determined accurately. Let the initial separation between the mirrors is d1. If one counts the number of fringes
(say maxima) appearing or disappearing at the centre (θ ≈0) by varying the distance between the mirrors to d 2,
then λ can be determined as follows:
2d1 = m1 2d2 = m2 (m2 − m1 ) = number of maxima counted.

2(d 2 − d1 )
= (6)
m2 − m1
Apparatus:
1. Optical Rail (1 meter)
2. Fabry-Perot setup (Fixed Mirror mount with Two
Etalon)
3 Movable Mirror with Kinematic and fine linear
Micrometer (0-10 mm)
4 Diode Laser mount with
Kinematic (5 V) Power Supply
5 Achromatic Lens mount
6 Frosted Glass viewing Screen and mount with
Micrometer
Fig 3. The Fabry-Perot set up
Procedure:
I. Alignment of Fabry-Perot Interferometer to
observe concentric circular fringes
1) Mount and lock the diode laser on the optical rail
towards one end.
2) Mount the Fabry-Perot interferometer towards the
middle of the optical rail (about
40 cm from laser). It has one fixed and one movable mirror
as shown in Fig. 4.
3) Adjust the three screws behind the
movable mirror to make sure that the two
mirrors are visibly parallel to each other
approximately and the distance between
them is about 2mm.
MAKE SURE THAT THE TWO MIRRORS
NEVER TOUCH EACH OTHER’S
SURFACES)
4) Mount the ground glass screen at the other extreme end.
5) Switch on the diode laser and adjust it such that the beam
passes through the centre of the two mirrors. Adjust the two black screws (for movement in x and y directions)
behind the movable mirror to let the multiple reflected beams coincide on the screen. It means both the mirrors
are now nearly parallel.
6) Place a lens (f= 100mm) in front of the laser to expand the beam to create a broad source. Adjust the position
of the lens so that the entire reflection cavity is illuminated. With all the components perfectly set, the observer
can find a series of very intense, concentric circular interference rings on the ground glass screen.
II. Measurement of the Wavelength of a diode Laser
1) Setup the F-P interferometer as described above to observe clear circular fringes at the centre of the ground
glass screen.
2) Determine the least count of the micrometer screw attached to the movable mirror. Record the reading d1 of the
fine micrometer.
3) Turn the micrometer slowly and count the number of fringes that appear (or
disappear) at the centre of the ground glass screen. Keep on recording the
micrometer reading d2 after every count 50 fringes.
CAUTION: The micrometer screw is extremely sensitive. So move it very
slowly to avoid collapse of many fringes while counting, which will lead to
error.
4) Acquire enough data and fill up the observation table. Plot a suitable graph to get
a straight line. Find slope of the graph and use Eq. 6 to determine λ.
Precautions:
1. Do not touch or contact in any way either the front or back surfaces of the
mirror pieces. Doing so will permanently damage the mirror coatings.
2. Avoid eye exposure to the direct laser beam.
3. Move the micrometer screw very slowly.