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

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

d d/2 s1 s2  z1 

For green light, For red light,

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

G

Fringe pattern will remain same or Interference fringe pattern will shift

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.

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

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

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

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

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

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.

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.

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.

THE LLOYD’S MIRROR ARRANGEMENT

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.

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)

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.

Phase change on Reflection, Refraction

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.

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

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

ar12 ar1 at1t2 I n1 at1r2 II n2 at1 at1r1 fig2

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

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.

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.

Interference in thin films due to reflection

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

II I d a c b DIVISION OF AMPLITUDE

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.

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

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

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

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

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 + 

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 + )

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

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)

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

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 .

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

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

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+1 - 2  PnQn =  Pn+1Qn+1 – PnQn = /2 tn+1 – tn = /2 For air film Pn+1Qn+1 – PnQn = /2

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

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

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.