1. The distance of the nth bright fringe from the centre
with monochromatic light
Where

2. c A E
Fringes of equal width S C d a F B Z1 Z2
Fringes of large width b D

3. d
d/2 s1 s2  z1 

4. For green light,
For red light,

6. A Transparent material G of thickness t and refractive index  is placed in the path of one beam.
WHAT WILL HAPPEN?

7. G

8. Fringe pattern will remain sameor
Interference fringe pattern will shift

9. Light wave from S2 will go directly to P but light wave from S1 will go to P partly through air and partly through Transparent material G.
Time required for the light to reach from S1 to the point
P is
C0 is the velocity of light in air and C its velocity in the medium
By introducing thin plate the effective optical path
changes.

10. The time needed for both the path will be same.
…….(1)

11. Time required for the light to reach from S1 to the point
P is
Clearly effective path in air from S1 to P is S1P +(-1)t

12. So the air path S1P has been increased by S1P +(-1)t
as a result of the introduction of the plate.
…….(1)

13. So the path difference between the beams reaching
P, from S1 and S2 ()
If there is no transparent plate then we know the path diff

14. So the path difference will be
If P is the centre of the nth bright fringe, then

15. It means that the introduction of the plate in the path
At n = 0 the shift y0 of central bright fringe is
It means that the introduction of the plate in the path
Of one of the interfering beams displaces the entire
Fringe system through a distance
This displacement is towards the beam in the path of which
The plate is introduced.

16. So central fringe at C is shifted from C to point P on The screen on insertion of transparent plate material Of refractive index 
Knowing the distance through which the central fringe is shifted, D, d and  the thickness of the material t can be calculated.

17. We have to use white light to determine the thickness of the material.
For monochromatic light central fringe will similar to
other bright bright fringe. For white light central fringe
is white.

18. THE LLOYD’S MIRROR ARRANGEMENT

19. Light directly coming from the slit S1interferes with the light reflected from the mirror forming an Interference pattern in the region BC of the screen.
For two sources one is real and the other one is virtual.

20. The central fringe will be dark.
Reflected beam undergoes a sudden phase change of  On reflection.SO at
S2P – S1P = n
We can get minima (destructive interference)
S2P – S1P = (2n+1)/2
and
We get maxima (constructive interference)

21. Using the principle of optical reversibility we
Can say that there will be an abrupt phase change
Of  when light gets reflected by the denser medium
No such phase change occurs when reflection takes
Place at rarer medium.

22. Phase change on Reflection,
Refraction

23. Principle of optical reversibility
In the absence of any absorption, a light ray that
is reflected or refracted will retrace its original
path if its direction is reversed.

24. n1 o n2 n2>n1 fig1 a  amplitude of incident ray
r1  reflection coefficient
t1,  transmission coefficient n1 o n2
n2>n1
fig1

25. r1, r2  reflection coefficients
t1, t2  transmission coefficients
n1, n2  refractive index of two media
n2>n1
a  amplitude of incident ray
ar  amplitude of reflected ray
at  amplitude of refracted ray

26. ar12
ar1
at1t2 I n1
at1r2 II n2
at1
at1r1
fig2

27. According to principle of optical reversibility
the two rays of amplitudes ar12 and at1t2 must
combine to give the incident ray of fig 1.
So,
ar12 + at1t2 = a
Stokes’ relation
 t1t2 = 1- r12
………(1)
The two rays of amplitudes at1r1 and at1r2 must
cancel each other.
So, at1r1 + at1r2 = 0
……………(2)
 r2 = -r1
Stokes’ relation

28. Relation (2) represents that the coefficient of reflection from the surface of a denser medium is equal in magnitude to the coefficient of reflection from the surface of a rarer medium but opposite in sign.
Intensity of reflected light is the same for a ray incident from either side of the boundary.
Negative sign in amplitude indicates a phase change of  occurring due to reflection at medium I and II. There are two possibilities:
If there is no phase change on reflection at medium I, there must be a phase change of  on reflection from medium II.
If there is no phase change on reflection from medium II, there must be a phase change of  on reflection from medium I.
Equation (2) does not give any information as to which of the two reflection gives the phase change.

29. Phase change  occurs when light gets reflected from denser medium.
 LLOYD’S MIRROR
From equation (2) we may say that
no phase change will occur when light gets
Reflected by a rarer medium. or
There is no phase change on reflection from medium II, there must be a phase change of  on reflection from medium I.

30. Interference in thin films due
to reflection

31. Colours of oil film on water
Colours of soap bubble
Interference of thin film
Interference by division of amplitude

32. II I d a c b
DIVISION OF AMPLITUDE

33. If plane wave falls on a thin film then the wave
reflected from the upper surface interferes with
the wave reflected from the lower surface.
Thin films are material layers of about 1 µm
thickness. For thin-film optics, the thickness of the layers of material must be on the order of the wavelengths of visible light. Layers at this scale can have remarkable reflective properties due to light wave interference.

34. Q H L  A B K E i S
 > 1 i i P N r O F L’

35. X = (PF +FE) – PK X = (PN+NF+FE) - PK
The optical path difference between the rays PQ and EH is
X = (PF +FE) – PK
X = (PN+NF+FE) - PK
Here <SPL = <LPK = i
In Δ EKP, <KPE = <90-i
<EKP = <90
so, <KEP = i
Similarly in Δ PNE, < PEN = r

36. Now, PK =  PN
 X =  (PN+NF+FE)-PN
X= (NF+FE)

37. Q H  A B K E i r S i
 > 1 i P J N r
r+ C O F  R
r+ L

38. EC = CL=t and FE = FL X =  (NF+FL) =  NL …………..(i)
EC is normal to OA.triangles ECF and FCL are congruent.
EC = CL=t and FE = FL
X =  (NF+FL)
=  NL …………..(i)
Angle between the inclined surfaces is the same as the angle
Between the normals at P and F.
SO, <PRF = 
Again the exterior angle <PEJ of Δ PRF is equal to the sum
of the interior angles,
<PEJ = r + 

39. From equ (i), x = 2t COS (r + )
Now JR and EL are parallel and PEL cuts these parallel lines
Such that <FLC = <NFJ= r + 
In right angled triangle ENL ,
COS (r + ) = NL/EL
NL = EL COS (r + )
NL = 2t COS (r + )
From equ (i),
x = 2t COS (r + )

40. 2 tcos(r+)-/2 Effective path difference between the interfering
Since PQ is the reflected wave train from a denser medium
Therefore there occurs a phase change of  or a path
Difference of /2.
Effective path difference between the interfering
waves PQ and EH is
Δ =
2 tcos(r+)-/2

41. 2 tcos(r+) = (2n+1) /2 ……(1)
Condition for constructive interference
2 tcos(r+) - /2 = n
2 tcos(r+) = (2n+1) /2 ……(1)
Condition for destructive interference
2 tcos(r+)=n ……(2)

42. 2 tcosr=(2n+1) /2 2 tcosr=n From equ (1) and (2)
So bright and dark fringes of different orders
will be observed at different thickness of the film.
Practically  is very small , therefore
Cos(r+ )cosr and so the condition will be
2 tcosr=(2n+1) /2
and
2 tcosr=n

43. For monochromatic light beam incident on a wedge shaped film ,  are constant. So change in path difference is only due to varying thickness of the film. At a particular point thickness is constant. So we get a bright or dark fringe at that point due to constant path difference.
Thickness of the film continuously changes. So equidistant interference fringes are observed parallel to the line of intersection of the two surfaces means parallel to the edge of the wedge .

44. Q H  A B K E i r S i
 > 1 i P J N r
r+ C O F  R
r+ L

45. 2 tn cos(r+)=(2n+1) /2 B Pn+m Pn+1 Pn  A Qn Qn+1 Qn+m C x
Suppose nth bright fringe at Pn.
Thickness of airfilm will be at Pn = PnQn = tn
Relation for bright film will be
2 tn cos(r+)=(2n+1) /2

46. 2 tn=(2n+1) /2 = 2  PnQn ……(3)
For nearly normal incidence cosr = 1
2 tn=(2n+1) /2 = 2  PnQn ……(3)
Next bright fringe will appear at Pn+1 for n+1th fringe
2Pn+1Qn+1 = [2(n+1)+1] /2…..(4)
2tn+1= [2(n+1)+1] /2
Subtracting (3) from (4)
2Pn+1Qn  PnQn = 
Pn+1Qn+1 – PnQn = /2
tn+1 – tn = /2
For air film Pn+1Qn+1 – PnQn = /2

47. Pn+1Qn+1 – PnQn = /2 Pn+mQn+m – PnQn = m/2 tn+m – tn = m/2
So next bright fringr will appear where air thickness
will increase by /2.
For (n+m) th bright fringe
Pn+mQn+m – PnQn = m/2
tn+m – tn = m/2
Therefore let at x distance from Qn m th bright fringe appears then

48. L x For small  Fringe width Pn+m Pn+1 Pn   A Qn Qn+1 Qn+m C
Pn+mQn+m – PnQn Pn L  x  A Qn
Qn+1
Qn+m C
For small 
Fringe width

49. eye B
Air film  A O C O’
The interfering rays do not enter the eye parallel to each
other but they appear to diverge from a point near the
film.