Interference due to transmitted light in thin films

s

A

B

R

P

C

M

D

Q

AIR

AIR

i

r

r r

r

i

r

i

N

t

S

A

B

C

D

R

Q

i

r

r r

r

i

r

i

M

AIR

AIR

t

N

P

C’

Here at B reflection takes place at the rarerMedium .so no phase change change occurs.

BM normal to CD and DN normal to BR.

So optical path difference between DQ and BR is

x = (BC+CD) -BN

n1sini = n2sinrSinr = sini=sini/sinr

In BND, sini=BN/BDIn BMD, sinr=MD/BD

=sini/sinr =BN/MD

BN = MD

Since <CPB=<CBP=rand BC=CP=CD

x = (BC+CD) –BN = (DP)- (MD) = (DP-MD) = (MP)

In BPM cosr = MP/BPSo MP = BP cosr = 2t cos r since BP=2t

SO, x = PM = 2 t cos r

Conditions for constructive interference will be

2 t cos r = n

And conditions for destructive interference will

2

)1n2(rcost 2

One of the important applications of the thin film interference is reducing the reflectivity of lens surface.

ai ar

at

n1

n2

n1> n2

taann

n2a

raann

nna

ii21

1t

ii21

21r

ai, ar, at are amplitudes of incident,reflected andtransmitted waves.

n2>n1 , ar is negative showing that reflection occurs at a denser medium a phase change comes

21

1

21

21

nn

n2t,

nn

nnr

r and t are reflection and transmissioncoefficients

n2

n1

a r’a

t’a

21

2

12

12

nn

n2't,r

nn

nn'r

2

2

21

21

221

21

rnn

nn

)nn(

nn41'tt1

These are the Stokes’ relations.

Non reflecting films• Reflectivity is the fraction of incident light reflected

by a surface for normal incidence.

• Reflectivity depends upon the refractive index of the material. It is given by

• For glass = 1.5.

• Reflectivity = 0.04

• 4% of incident light is reflected for normal incidence. Remaining 96% is transmitted.

2

1

1

• The loss of energy due to reflection is one major reason of clarity reduction. There is also a reduction in the intensity of the images since less light is transmitted through the lenses.

• When films are coated on lens of prism surface the reflectivity of these surfaces is appreciably reduced.

• Initially the coating were made by depositing several monomolecular layers of an organic substance on glass plates.

• Now it is done by either evaporating calcium or magnesium fluoride on the surface in vacuum or by chemical treatment of the surface with acids which leave a thin layer of silica on the surface.

• No light is destroyed by non reflecting film, but there is redistribution means decrease in reflection results increase in transmission.

ar

t

12

air(a)

film(f)

glass(g)

Thickness of nonreflecting thin film

g f> a

Two interfering beams will interfere constuctively if

2ftcosr = nRays will interfere destructively if

2ftcosr = (2n+1)/2For normal incidence <r=900

2ft = (2n+1)/2

So 2ft = /2 for min thickness, n = 0

f4t

If a film having thickness of /4f and having refractive index less than that of the glass is coated on glass, then waves reflected from the upper surface of the film destructively interfere with the waves reflected from thelower surface of the film. Such a film known as a non reflecting film.

ar

ar’tt’

1 2

na

nf

ng

a

atar’t

Amplitude of ray 1 = arAmplitude of ray 2 = ar’tt’

r, t

r’, t’

fa

a

fa

fa

nn

n2t,

nn

nnr

fa

f

gf

gf

nn

n2't,

nn

nn'r

ann

n2at,a

nn

nnar

fa

a

fa

fa

ann

nn

)nn(

nn2

ann

n2

nn

n2

nn

nn'tt'ar

gf

gf2

fa

fa

fa

f

fa

a

gf

gf

For complete destructive interference ray 1 and 2 mustHave the same amplitude, i.e.

ann

nn

nn

nn4a

nn

nn

gf

gf2

fa

af

fa

fa

unity toqeualnearly very isnn

nn42

fa

af

gf2fgafa

gf2fgafa

gf

gf

fa

fa

nnnnnnn

nnnnnnn

ann

nna

nn

nn

gaf

ga2f

nnn

nnn

This equation gives the estimate of refractive index ofThis film which should be coated on a surface to reduceIts reflectivity. If na= 1 (for air) and ng = refractive index of glass then

gf

g

f

nn

n

film thin ofindex refractiven