1. UNIT-III
CHAPTER-I
INTERFERENCE

2. STRUCTURE:
3.1.1. INTRODUCTION
OBJECTIVES
3.1.2. LIGHT AS A WAVE
3.1.3. PHASE DIFFERENCE
3.1.4. PATH DIFFERENCE
3.1.5. RELATION BETWEEN PHASE DIFFERENCE AND PATH DIFFERENCE
3.1.6. COHERENCE
3.1.7. INTERFERENCE (DEFINITION)
1. PRINCIPLE OF SUPERPOSITION
2. TYPES OF INTERFERENCE(CONSTRUCTIVE & DESTRUCTIVE)
3. WHEN THE LIGHT WAVES HAVING DIFFERENT FREQUENCIES AND VARYING
PHASE DIFFERENCE.
4. WHEN THE LIGHT WAVES HAVING EQUAL FREQUENCIES AND CONSTANT PHASE
DIFFEENCE
5.CONDITIONS FOR SUSTAINED INTERFERENCE
3.1.8. INTERFERENCE IN THINFILMS
1. STOKE’SPRINCIPLE
2. OPTICAL PATH OF LIGHT
3.1.9. INTERFERENCE IN THINFILMS OF UNIFORM THICKNESS DUE TO REFLECTED
AND TRANSMITTED LIGHTS
3.1.10. INTERFERENCE IN THINFILMS OF NON UNIFORM THICKNESS
1. WEDGE METHOD
(i) DETERMINATION OF THICKNESS OF PAPER
(ii) DETERMINATIO OF FLATNESS OF GIVEN GLASS PLATES
2. NEWTON’S RINGS
(i) DETERMINATION OF WAVELENGTH OF MONOCHROMATIC LIGHT
(ii) DETERMINATION OF REFRACTIVE INDEX OF A GIVEN LIQUID
3.1.11. MICHELSON INTERFERROMETER
3.1.12. QUIZ
3.1.13. SOLVED EXAMPLES
3.1.14. PROBLEMS FOR PRACTICE
3.1.15. SUMMARY

3. 3.1.1. INTRODUCTION
Sunlight, as the rainbow shows us, is a composite of all colors of visible
spectrum. The colors reveal themselves in the rainbow because the incident wave lengths
are bent through different angles as they pass through raindrops that produce a bow.
However, soap bubbles and oil slicks can also show striking colors produced not by
refraction but by constructive and destructive interference of light. The interfering waves
combine either to enhance or to suppress certain colors in the spectrum of sun light.
Interference of light waves is thus a superposition phenomenon.
This selective superposition of wavelengths has many applications. When light
encounters an ordinary glass surface, for example about 4% of incident light energy is
reflected, thus weakening the transmitted beam by the amount. This unwanted loss of light
can be a real problem in optical systems with many components. A thin transparent
“interference film” deposited on glass surface, Can reduce the amount of reflected light by
destructive interference. The bluish cast of camera lens reveals the presence of such a
coating. Interference coatings can also be used to enhance –rather than reduce- the ability
of a surface to reflect light.
To understand interference, we must go beyond the restrictions of geometrical
optics and employ the full power of wave optics. In fact, as you will see, the existence of
interference phenomena is perhaps our most convincing evidence that light is a wave-
because interference cannot be explained other than with waves.
OBJECTIVES:
After going through this chapter we should able to: know
1) About light wave.
2) The relation between path difference and phase difference.
3) About interference.
4) About the interference phenomenon takes place in thin films.
5) Determination of thickness of a paper using wedge method.
6) Determination of wavelength of monochromatic light using Newton’s rings.
3.1.2) LIGHT AS A WAVE:
Light has dual nature. It has both particle nature and wave nature but it cannot act as a
wave and particle simultaneously. Light is also having the properties such as reflection,
refraction, interference, diffraction, polarization. Light is also having two most famous
effects like photoelectric effect and Compton Effect. To explain the properties like
interference diffraction and polarization we need to adopt wave nature for light.
Light is a transverse wave.
The first person to advance a convincing wave theory for light was DUTCH PHYSICIST
CHRISTIAN HUYGENS, in 1678.
Light wave should be represented as follows.
Figure1

4. Whose displacement is given b
Where ' a' is amplitude of wave.
'ωt' is Phase, Which gives the position and direction of wave at a time 't'
3.1.3) PHASE DIFFERENCES:
The angular separation between two points of wave is called Phase difference.
It is measured in radians or degrees.
3.1.4) PATH DIFFERENCE:
The linear separation between two points of wave is called Path Difference
It is measured in mm or cm.
Fig.
3.1.5) RELATION BETWEEN PHASE DIFFERENCE AND PATH DIFFERENCE:
Figure
2 Phase Difference = λ Path Difference
Phase Difference = Path Difference
Phase Difference = Path Difference

5. Therefore for
2 Phase Difference = λ (Path Difference)
Phase Difference (δ)= What is the Path Difference(X)
=
δ = ( ) x path difference
Phase Difference= ( ) x Path Difference
Path Difference = ( ) x Phase Difference
3.1.6) COHERENCE:
Two light waves are said to be coherent if they are having
1) Same Frequency
2) Almost Same Amplitude
3) Moving in a medium with either zero or constant Phase Difference
4) S
3.1.7) INTERFERENCE:
Interference is the optical phenomenon in which brightness and darkness are
produced by the combination of two similar light waves.
DEFINITION:
When two light waves of same frequency having constant phase difference coincide in
space and time. There is modification in the intensity of light.
The resultant intensity at any point depends upon amplitudes and phase relationships
between two waves.
This modification in the intensity is due to superposition of two light waves are called
Interference.
And the pattern dark and bright fringes produced are called Interference pattern.

6. (i)PRINCIPLES OF SUPERPOSITION:
When two or more waves reach a point simultaneously, the resultant displacement
at that point is the algebraic sum of the displacements produced by the individual waves in
absence of others.
Explanation: Let us consider the two waves of same frequency “ ” is
If y1 is the displacement produced by one wave and
y2 is the displacement produced by second wave
δ is the phase difference between these two waves
The resultant displacement produced by the superposition of these waves is
(1)
Let (2)
(3)
Where a1 and a2 are the amplitudes of these two waves
Substitute equation (2) and (3) in (1), we get

7. (4)
Let (5)
(6)
Sub eq(5) and eq(6) in (4) then
(7)
This is an equation for resultant displacement.
Where A is resultant amplitude which can be calculated as follows.
By squaring and Adding equation (5) and equation (6)
=
Square of the amplitude (A²) = Intensity of light (I)
(2)TYPES OF INTERFERENCE:
Interference is of two types.
1. Constructive interference / constructive superposition
2. Destructive interference / destructive superposition

8. Figure: Constructive and Destructive Interference
(A)CONSTRUCTIVE INTERFERENCE:
When crust of one light wave falls on the crust of another wave then the
resultant intensity increases and this type of interference is called constructive
interference.
Here we get maximum intensity.
The intensity becomes maximum ) when

9. , n=0, 1, 2, 3...............
For δ=0, 2π, 4π.............we get bright band
Path Difference = ) X Phase Difference
Path Difference= ) x2nπ
Path Difference=
This is the condition for constructive interference.
`
For δ=0, 2π, 4 ………
s
(B)DESTRUCTIVE INTERFERENCE:
When crust of one light wave falls on the trough of another wave then the
resultant intensity decreases. This type of interference is called destructive interference.
Here we get minimum intensity.
The intensity becomes minimum ) when
, n=0, 1, 2, 3...............
For .............we get dark band.
Path Difference = )X Phase Difference
Path Difference=
Path Difference=

10. This is the condition for constructive interference.
`
For δ= , 3π, 5 ………
Variation of Intensity with phase:-
3) When two light waves having different frequencies and varying phase difference:
If two light waves have different frequencies and the phase difference between
light waves is not constant then we get unsustained interference pattern i.e., Uniform
illumination is observed i.e., bright and dark bands cannot be seen.

11. 4)When two light waves having equal frequencies and constant phase difference:
If two light waves having equal frequencies and constant phase difference then we
get sustained interference pattern. i.e we get alternative bright and dark fringes. Here the
bright fringe becomes very bright and dark fringe becomes very dark.
To get sustained interference light waves has to be satisfying the following conditions.
5)Conditions for sustained or Good interference pattern :
1) We require two monochromatic light sources.
2) These two sources must be coherent i.e., they constant phase difference.
3) The frequency must be the same.
4) The amplitude must be the same.
5) They must travel in the same directions.
6) The two sources must be the same
7) These two sources must be as near as possible and the screen must be as far from
them as possible.
3.1.9) INTERFERENCE IN THIN FILMS:
To understand interference in thin films first of all we have to understand some basic
concepts like stokes principle and optical path.
1) STOKE'S PRINCIPLE:
STATEMENT:-
According to stoke principle, a light beam which is initially passing through a rarer
medium reflects back into the same rare medium from denser medium suffering sudden
phase change of ∏ or path change of λ/2.Such a phase change cannot be observed for a
light beam light beam which is initially passing through denser medium and reflects back
into the same medium.

12. 2) The second basic point is Optical Path:
The optical path travel led by a light beam in a medium of refractive index 'μ' is not equal
to actual path travel led by the light beam.
Optical path travelled by light beam =µ X Actual path travelled by light
Thin films are of two types
1) Thin films which have uniform thickness
2) Thin films which have non uniform thickness
Striking colors observed on thin films due to interference

13. 3.1.10) INTERFERENCE IN THINFILMS OF UNIFORM THICKNESS DUE TO REFLECTED AND
TRANSMITTED LIGHTS:
a) Interference due to reflected light :
Let us consider a transparent thin film of uniform thickness t and refractive index as shown
in figure.
AB- is the incident ray on thin film.
A part of light ray reflected along BR and transmitted along BC.
BC reflected from lower surface of thin film along CD and DR1 is transmitted ray .Now the
interference takes place between reflected rays BR and DR1.
These are reflected from upper and lower surfaces of thin film.
The actual path difference between BR and DR1 is = (BC+CD) in film - BE in air
The optical path difference between BR and DR1 = (BC+CD) µ -BE air---------- (1)
For BE
From Snell's law
From ∆BED,
∆BFD,
Therefore

14. BE= µBF--------------- (2)
Sub eq’n (2) in eq’n (1)
Path difference ∆= (BC+CD) µ - µBF
= (BC+CD-BF) µ
= (BC+CF) µ ------------ (3)
∆le BCQ, ∆le PCQ are congruent triangles.
Therefore
BC=PC
BQ=PQ=t
QC is common side and all angles are equal.
Therefore ∆= (PC+CF) µ
∆=µ (PF) ------------- (4)
From ∆le PBF
PF= (PB)
= (BQ+QP)
=
= 2t ---------- (5)
Therefore eq’n (4) becomes
∆= µ (2t)
∆=2µt --------- (6)
But according to stokes principles the light ray BR is undergoing additional path change of
λ /2.
Therefore the total path difference between BR and DR is equal to 2μtcosr± λ /2

15. Condition for bright band/bright fringes
We get bright fringes when path difference =n λ
Condition for dark band/dark fringes
we get dark fringes when the path difference
3.4) NOTE:
In thin film, due to transmitted light
The conditions for bright is
The conditions for dark is
We can say the interference pattern due to reflected and transmitted rays are
complementary each other.
3.1.11) INTERFERENCE IN THIN FILMS OF NON UNIFORM THICKNESS:
1) Wedge method
2) Newton’s Rings

16. (A)WEDGE METHOD:
Let us consider two plane surfaces GH; GH1which is inclined at an angle gives wedge shape
which encloses an air film
AB is the incident ray on GH. The interference takes place between two reflected rays BR
and DR1.One is reflected from upper surface and other one is from lower surface of air
film.
The optical path difference between BR and DR1 is given by
Δ = µ (BC+CD)-BE----------------> (1)
For BE
From Snell's law
From ∆BED,
∆BFD,
Therefore
BE= µBF--------------- (2)

17. Substitute equation (2) in equation (1)
Δ =µ (BC+CD)-µBF
Δ =µ (BC+CD-BF)
Δ =µ (BC- BF+CD)
Δ =µ (FC+CD) -----------------------------> (3)
ΔCDQ, ΔCPQ are congruent triangles
CD=CP
Therefore Eq’n (3) becomes
Δ =µ (FC+CP)
Δ =µ (FP) -----------------------> (4)
From ΔFPD
FP=2t ------------------>(5) (from figure DP=2t)
Sub (5) in (4)
Δ=2µt --------------------------------- (6)
From stoke principle
The ray BR undergoes reflection from denser medium; it suffers on additional path change
λ/2.
Therefore the total path difference=
Condition for bright fringe:
We get bright fringes when path difference =nλ

18. Condition for dark fringe:
We get dark fringe when path difference=
3.4.2)TYPES OF FRINGES IN WEDGE SHAPED FILMS:
In wedge shaped films ,at the edge of wedge the thickness of air film will be same .If we
draw a line at this edge all along this length we observed straight parallel alternative bright
and dark fringes with equal Spacing.
Fig
3.4.3)Applications of Wedge method:
1) Determination of thickness of a paper or diameter of a wire/hair.
2) Verification of flatness of the given transparent surface.
3.4.4)Determination of thickness of paper or thin film
First we measure the spacing between two consecutive dark/bright fringes which is known
as fringe width to determine the thickness of paper.
Spacing between two consecutive dark fringes
Let us assume that nth dark fringes is formed at a distance Xn and (n+1)th dark fringe at a
distance Xn+1 from wedge.
Now Xn+1 – Xn gives the values of fringe width ''.

19. Condition for dark fringe is -------------- (a)
For normal incidence i=0
r=0
For air µ =1
Eq--(a) becomes
------------------------->(b)
For nth dark fringe
--------------------------(c)
From figure (ii)
---------------- (d)
Sub eq (d) in eq(c)

20. For (n+1)th dark fringe
-------------- (f)
From fig(ii)
------------------------>(g)
sub eq's(g) in (f)
------------------ (h)
Now fringe width
-----------------(i)
As α is very small ,
-------------------- (j)
Thickness of paper:
If the angle between two wedges is very small then AB=AC=l
From ΔABC,

21. This is equation (k)
Sub eq’n (k) in eq’n(i), fringe width becomes
This is an expression for thickness of paper.
Where
λ = wavelength incident light
l = length of air film
β = Fringe width
(B)NEWTON'S RINGS:
Plano-convex lens on flat black surface
When a Plano convex lens with its convex surfaces is placed on glass plate air film is
formed between two whose thickness increases gradually. The thickness of air film at the
point of contact is zero. When monochromatic light is allowed to fall normally then we get
fringes which are circular. These fringes are concentric circle, uniform in thickness and with
the point of contact as the center. This phenomenon was first described by Newton that
why they are known as Newton’s rings.

22. Experimental arrangement
L is a planoconvex lens with large radius of curvature .
G is a plane glass plate
L is placed with its convex surface on G.
monochromatic light is incident on L normally by using 45° arrangement of glass
plate.
A part of light is reflected from curved surface of lens and another one is reflected
from plane surface of glass plate.
Now interference take place between these two rays
i.e one is reflected from upper surface of air film and another one reflected from
lower surface of air film.
Thus we get alternative bright and dark fringes .
Here we get central ring as dark ring , the reason is as follows :
The path difference between two light rays due to reflected light equal to
Due to large radius of curvature of lens, α is so small and it is neglected.
For normal incidence i=0, r=0.
For air film μ=1
Path difference from equation 1 becomes
Path difference =2t+λ/2------------ (2)
At point of contact of lens and glass plate, t=0 then eq--( 2) becomes

23. The path difference = λ/2---------- (3)
This is a condition for minimum intensity.
So central ring is dark due to reflected light.
Conditions for bright and dark rings:
Conditions for bright rings:
We observe maximum intensity/ bright ring when path difference=n λ
2t+λ/2=n λ (from eq---2)
2t= (2n-1) λ/2----------- (4)
Conditions for dark rings:
We observe minimum intensity/ dark ring
When Path difference = (2n+1) λ/2
(from 2)
2t=n λ------------------ (5)
Calculation of Diameter of bright and Dark rings formed due to reflection of light :
fig

24. (Here d=t)
From Figure
t = thickness of air film
D=Diameter of Newton's rings
r=Radius of Newton’s ring =D/2
R = Radius of curvature of convex lens

25. From properties of chords in circle
ON * OD = OP *OQ
t(2R - t) = r * r
2Rt-t ² = r ²
2Rt-t ²= (D/2)² =D²/4
D²= 8Rt-4t²
As the thickness of air film is very small,t² is neglected.
D²= 8Rt
D²= 4R(2t)
Condition for Diameter of bright ring
From (4)
Diameter of bright rings is proportional to square root of odd natural numbers.
Diameter of Dark Ring:
From eq-5
The diameter of the dark ring is proportional to square root of natural numbers.

26. Note:
The distance between rings decreases when the order of rings increases.
Reason:
Condition for dark ring is
D16= 8√Rλ
D9= 6√Rλ
D4= 4√Rλ
D1= 2√Rλ
D16 -D9=2√Rλ ------------->7 Fringes
D9-D4 = 2√Rλ -------------->5 fringes
D4-D1= 2√Rλ ---------------->3 Fringes
Here the distance( 2√Rλ) remains constant but number of rings increases .
From this we can say that width decreases as order increases.
APPLICATIONS:
1) Determination of wavelength of monochromatic source
2) Determination of refractive index of liquid
1) Determination of wavelength of monochromatic source
First Newton's Rings are formed by respective experimental arrangement.
The condition for diameter of dark ring is given by
If we consider mth dark ring

27. The diameter for mth dark ring is given by
------------------ (a)
If we consider nth dark ring
The diameter for nth dark ring is given by
---------------------- (b)
Equation (a) - equation(b)
This is equation(c)
R is radius of curvature of lens is measured by spherometer.
R=l²/6h+h/2
Where
l is the distance between two legs of spherometer.
h is height of convex lens.
(ii) DETERMINATION OF REFRACTIVE INDEX OF A LIQUID
First the experiment is performed when there is an air film is formed between glass plate
and Plano convex lens.
The diameter of nth and mth rings are measured.
For air film,
This is equation (d)
Now the liquid whose refractive index is to be measured is powered in the container
without disturbing the whole arrangement.
Again we measured the diameter of mth and nth rings whose diameters are D'²m and D'²n
For liquid medium

28. This is equation (e)
From eq (d) /eq (e) then we get
APPLICATIONS
1) Determination of wavelength of monochromatic light.
2) Determination of difference of nearly equal wavelengths
1)DETERMINATION OF WAVELENGTH OF MONOCHROMATIC LIGHT:
Michelson interferometer is set for circular fringes set bright at center.
Condition for bright spot
The mirror M1 is set at a distance x1 for one fringe.
Let it be zeroth fringe.
The mirror M1 is moved from x1 to x2 N fringes are crossed
2(X2~X1) =Nλ
λ=2/N(X2~X1)
2)DETERMINATION OF DIFFERENCE IN WAVELENGTH:
There are two spectral lines D1 and D2 of sodium light.
Their wavelengths are nearly equal.
The difference in their wavelengths of D1 and D2 lines.
So they form two separate fringe n .
As thickness between M2 and M2¹ is very small, two fringe patterns coincide practically.
By moving M1 we separate these two fringe pattern s.
We move M 1 so that nth bright fringe of λ1 coincides with (n+1)th dark fringes of λ2.
Note Down this distance , At this position we observe indistinctness.
We move M2 again another indistinctness is observed .Let the fringe distance be x.
For nth fringe

29. 2x=nλ1
This is equation (1)
For (n+1)th fringe
2x= (n+1)λ2
This is equation (2)
Equation (2) – equation (1)
This is difference between wavelengths.
3.1.13)QUIZ
WHY RAINBOW FORMS:

30. WHY MORPHO’S BUTTERFLY CHANGES IT’S COLOR FROM BLUE TO BROWN
.
.
3.1.14) SOLVED EXAMPLES:
(1)Two coherent sources of intensity 10 w and 25 w interfere to from fringes. Find
the ratio of maximum intensity to minimum intensity.
Solution:
Given that =
Hence =
Now =
=
=19.724
(2)Two sinusoidal waves of equal amplitudes are wavelength out of phase. What is the
amplitude of the resultant?
Solution:
When two waves of equal amplitude but different phase superpose, the resultant intensity
is
I=4
Or resultant amplitude is
A=2acos

31. Where Ø is the phase difference.
Given path difference =
Hence phase difference Ø= Χ =
Hence A=2a cos
=2a =
(3) A soap film of refractive index 4/3 and of thickness 1.5Χ cm is illuminated by white
light incident at an angle of 60º. The light reflected by it is examined by a spectrometer in
which is found a dark band corresponding to a wavelength of 5Χ Calculate the order
of interference of the dark band.
Solution:
For dark band 2µtcosr= λ
Given =60º
Hence µ=
Or sin = =
=0.6511
i.e.,
3.1.15) PROBLEMS FOR PRACTICE
1) Two coherent source of intensity ratio interfere. Prove that in the interference
pattern
2) Two coherent sources whose intensity ratio is 81:1 produce interference fringes
.Deduce the ratio of maximum intensity to minimum Intensity.
3) A parallel beam of light is incident on a thin glass plate such that the angle of
refraction into the plate is 60°. Calculate the smallest thickness of the glass plate
which will appear dark by reflection.
4) Fringes of equal thickness are observed in a thin glass wedge of refractive index, the
fringe spacing is 1mm and wavelength of light is 5893 A°. Calculate the angle of
wedge.
5) In Newton's Rings experiment, the diameter of the 4th and 12th dark rings is 0.400 cm
and 0.700cm respectively. Find the diameter of 20th dark ring.
6) In Newton's Rings experiment, the diameter of the 10th rings from 1.40 cm to
1.27cm when a liquid is introduced between the lens and the plate. Calculate the

32. refractive index of the liquid.
7) Light containing two wavelengths λ1 and λ2 falls normally on a Plano convex lens
of radius of curvature R resting on a glass plate .If nth dark ring due to λ1 coincides
with (n+1)th dark ring due to λ2 prove that radius of nth dark ring of λ1 is
8) In a Michelson interferometer 200 fringes cross the field of view when the movable
mirror is displaced through 0.0589 mm .Calculate the monochromatic light
wavelength.
9) The movable mirror of Michelson's interferometer is moved through a distance of
0.02603 mm. find the number of fringes shifted across the cross wire of eye piece of
the telescope ,if a wavelength of 5200 A° is used.
10) In an experiment with Michelson's interferometer scale readings for a pair of
maximum indistinctness were found to be 0.6939 mm and 0.9884 mm .If the mean
wavelength of the two components of D line are 5893 A°. Deduce the difference
between wavelengths.
3.1.17) SUMMARY:
We have seen that two independent sources of light cannot acts as coherent sources. Even
two different parts of the same lamp cannot acts as coherent sources of light. Hence we
choose a ray which splits into two different parts such as reflected and transmitted to get
two coherent waves. Due to the path difference between such split rays interference
occurs. The phenomena of interference are very useful in many areas such as holography
optical switching calibration of instruments, displacements and measurements.
 The displacement of wave light is
y=a sin (wt+ δ)
 Relation between path difference and phase difference
 Path Difference= λ /2∏ * Phase Difference
From principle of superposition
 The intensity of resultant wave is given by
I=a1²+a2²+2a1a2cosδ
 Intensity I= A²
When a1 =a2 =a
 I=4a²cos²δ/2
 A²=4a²cos²δ/2
 A=2acosδ/2
 Condition for maximum intensity /Bright fringe
Path difference=nλ
Phase difference =2nπ for n=0, 1,2.......
I max= (a1+a2)²
 Condition for minimum Intensity/dark fringe
Path difference= (2n±1) λ / 2
Phase difference= (2n±1) π for n=0, 1, 2.....
I min= (a1-a2)²

33. Interference in thin film for uniform thickness thin films:
Condition for bright and dark fringes due to reflected light
 For Bright 2μtcosr= (2n ± 1) λ /2
 For dark 2μtcosr=n λ
Conditions for bright and dark fringes due to transmitted light
 For dark 2μtcosr=(2n ± 1) λ /2
 For bright 2μtcosr=n λ
Interference due to reflected and transmitted lights are complimentary
to each other.
B)FOR NONUNIFORM THICKNESS THIN FILMS:
Wedge method:
 Bright fringe condition 2μtcos(r+α )=(2n ± 1) λ /2
 Dark fringe condition 2μtcos(r+α )=nλ
 Fringe width β =λ /2sin α or λ /2α
 Thickness of thin film t= λl/2β
Newton's Rings:
 Condition for bright ring 2t=(2n -1) λ /2
 Condition for dark ring 2t=nλ
 Diameter of bright ring Db= √2R λ √ (2n-1
 Diameter of dark ring Dd= √2R λ √n
 Wavelength of monochromatic light λ=D²m - D²n/4R(m - n) where R=l²/6h+h/2
 Refractive index of a liquid µ liquid= D²m - D²n / D'²m - D'²n