6 2 Formulae Wave Optics

7/31/2019 6 2 Formulae Wave Optics

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Wavefront:A wavelet is the point of disturbance due to propagation of light.A wavefront is the locus of points (wavelets) having the same phase of oscillations.A line perpendicular to a wavefrontis called a ray.

SphericalWavefrontfrom a pointsource

Cylindrical

Wavefrontfrom a linearsource

PlaneWavefront

Pink Dots WaveletsBlue EnvelopeWavefrontRed Line Ray

Each point on a wavefront acts as a fresh source of disturbance of light.

The new wavefront at any time later is obtained by taking the forwardenvelope of all the secondary wavelets at that time.

Note:Backward wavefront is rejected. Why?

Amplitude of secondary wavelet is proportional to (1+cos). Obviously, forthe backward wavelet = 180and (1+cos ) is 0.

(Wavelets - Red dots on the wavefront)

6.2 WAVE OPTICS

Huygens Construction or Huygens Principle of Secondary Wavelets:

S

New Wavefront(Spherical)

NewWave-front(Plane)

.

.

.

.

.

.

.

...

.

.

..

.

.

.


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i r

A

B

C

D

N N

AB Incident wavefrontCD Reflected wavefrontXY Reflecting surface

E

F

GEF

t =c

+FG

c

AF sin it =

c+

FC sin r

c

For rays of light from different parts on the incident wavefront, the values of AF aredifferent. But light from different points of the incident wavefront should take thesame time to reach the corresponding points on the reflected wavefront.So, t should not depend upon AF. This is possible only if sin i sin r = 0.

i.e. sin i = sin r or i = r

X Y

AC sin r + AF (sin i sin r)t =

c

ri

Laws of Refraction at a Plane Surface (On Huygens Principle):

ii

r

r

A

B

C

D

N N

AB Incident wavefrontCD Refracted wavefrontXY Refracting surface

E

F

G

X Y

c, 1

v, 2Denser

Rarer

If c be the speed of light, t bethe time taken by light to gofrom B to C or A to D or E toG through F, then

EFt =

c+

FG

v

AF sin it =

c+

FC sin r

v

AC sin rt =

cv

sin i+ AF (

v

sin r- )

Laws of Reflection at a Plane Surface (On Huygens Principle):If c be the speed of light, tbe the time taken by light togo from B to C or A to D or Eto G through F, then

For rays of light from different parts on the incident wavefront, the values of AFare different. But light from different points of the incident wavefront should takethe same time to reach the corresponding points on the refracted wavefront.So, t should not depend upon AF. This is possible onlyif

c v = 0sin i sin r

-c v

sin i sin r=

c

v

sin i

sin r= = or or


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Constructive Interference E= E1+E2

Destructive Interference E = E1-E2

S1

S2

Bright Band

Dark Band

Dark Band

Bright Band

Bright Band

CrestTroughBright BandDark Band

1st Wave (E1)

2nd

Wave (E2)Resultant WaveReference Line

E1

E2

E2

E1

The phenomenon of one wave interfering withanother and the resulting redistribution of energy inthe space around the two sources of disturbance iscalledinterference of waves.

E1 + E2

E1 - E2

Theory of Interference of Waves:

E1 = a sin tE2 = b sin (t + )

The waves are with same speed, wavelength, frequency, timeperiod, nearly equal amplitudes, travelling in the same directionwith constant phase difference of .

is the angular frequency of the waves, a,b are the amplitudes andE1, E2 are the instantaneous values of Electric displacement.

Applying superposition principle, the magnitude of the resultant displacement of thewaves is E = E1 + E2

E = a sin t + b sin (t + )

E = (a + b cos ) sin t + b sin cos t

Putting a + b cos = A cos b sin = A sin

a

bA

b cos

b sin

A cos

A sin

We get E = A sin (t + )

(where E is the resultantdisplacement, A is theresultant amplitude and is the resultant phasedifference)

A = (a + b + 2ab cos ) tan =b sin

a + b cos

Interference of Waves:


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A = (a2 + b2 + 2ab cos )

Intensity I is proportional to the square of the amplitude of the wave.

So, I A2 i.e. I (a2 + b2 + 2ab cos )

Condition for Constructive Interference of Waves:

For constructive interference, I should be maximum which is possible

only if cos = +1.i.e. = 2n

Corresponding path difference is = ( / 2 ) x 2n

where n = 0, 1, 2, 3, .

= n

Condition for Destructive Interference of Waves:

For destructive interference, I should be minimum which is possibleonly ifcos = - 1.

i.e. = (2n + 1)

Corresponding path difference is = ( / 2 ) x (2n + 1)

where n = 0, 1, 2, 3, .

= (2n + 1) / 2

Imax (a + b)2

Imin (a - b)2

Relation between Intensity (I), Amplitude (a) of the wave and Width (w) ofthe slit:

I a

a w

I1

I2=

(a1)2

(a2)2

=w1

w2

Comparison of intensities of maxima and minima:

Imax (a + b)2

Imin (a - b)2

Imax

Imin=

(a + b)2

(a - b)2

(a/b + 1)2

(a/b - 1)2=

Imax

Imin=

(r + 1)2

(r - 1)2where r = a / b (ratio of the amplitudes)


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S

S O

P

D

S1

S2

d

y

d/2

d/2

Single Slit Double Slit

Screen

The waves from S1 and S2 reach the point P with some phase difference and hencepath difference = S2P S1P

S2P2

S1P2

= [D2

+ {y + (d/2)}2

] - [D2

+ {y - (d/2)}2

](S2P S1P) (S2P + S1P) = 2 yd

(2D) = 2 yd = yd / D

Positions of Bright Fringes:

For a bright fringe at P, = yd / D = n

where n = 0, 1, 2, 3,

For n = 0, y0 = 0For n = 1, y1 = D / dFor n = 2, y2 = 2 D / d For n = n, yn = n D / d

y = n D / d

Positions of Dark Fringes:

For a dark fringe at P, = yd / D = (2n+1)/2

where n = 0, 1, 2, 3,

For n = 0, y0 = D / 2dFor n = 1, y1 = 3D / 2dFor n = 2, y2 = 5D / 2d ..For n = n, yn = (2n+1)D / 2d

y = (2n+1) D / 2d

Expression for Dark Fringe Width:

D = yn yn-1= n D / d (n 1) D / d= D / d

Expression for Bright Fringe Width:

B = yn yn-1= (2n+1) D / 2d {2(n-1)+1} D / 2d= D / d

The expressions for fringe width show that the fringes are equally spaced onthe screen.

Youngs Double Slit Experiment:


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Intensity

0 yy

Suppose the two interfering waveshave same amplitude say a, thenImax (a+a)

2 i.e. Imax 4a2

All the bright fringes have this sameintensity.Imin = 0All the dark fringes have zero intensity.

Conditions for sustained interference:

The two sources producing interference must be coherent.

The two interfering wave trains must have the same plane of polarisation.

The two sources must be very close to each other and the pattern must beobserved at a larger distance to have sufficient width of the fringe. (D / d)

The sources must be monochromatic. Otherwise, the fringes of differentcolours will overlap.

The two waves must be having same amplitude for better contrast between

bright and dark fringes.

Colours in Thin Films:

P Q

R S

r

i

t

It can be proved that the path differencebetween the light partially reflected fromPQ and that from partially transmitted andthen reflected from RS is

= 2t cos rO

Since there is a reflection at O, the rayOA suffers an additional phasedifference of and hence thecorresponding path difference of /2.

A

B

C

For the rays OA and BC to interfereconstructively (Bright fringe), thepath difference must be (n + ) So, 2t cos r = (n + )

When white light from the sun falls on thin layer of oil spread over water in therainy season, beautiful rainbow colours are formed due to interference of light.

For the rays OA and BC tointerferedestructively (Dark fringe), the pathdifference must benSo, 2t cos r = n

Distribution of Intensity:


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1) At an angle of diffraction = 0:

O

0123456789

101112

A

B

= 0

Slit

Screen

PlaneWavefront

The wavelets from the single wavefront reach the centre O on the screen in same phaseand hence interfere constructively to give Central or Primary Maximum (Bright fringe).

Bright

D

d

B

A

Slit

Screen


O

0123456789101112

P1

N

1

/2

PlaneWavefront

Bright

Dark

The slit is imagined to be divided into 2 equal halves.

The wavelets from the single wavefront diffract at an angle 1 such that BN is and reach thepoint P1. The pairs (0,6), (1,7), (2,8), (3,9), (4,10), (5,11) and (6,12) interfere destructively withpath difference /2 and give First Secondary Minimum (Dark fringe).

Diffraction of light at a single slit:


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The wavelets from the single wavefront diffract at an angle 2 such that BN is 2 and reachthe point P2. The pairs (0,3), (1,4), (2,5), (3,6),(4,7), (5,8), (6,9), (7,10), (8,11) and (9,12) interferedestructively with path difference /2 and give Second Secondary Minimum (Dark fringe).

The slit is imagined to be divided into 4 equal parts.


Bright

Dark

Bright

The slit is imagined to be divided into 3 equal parts.

The wavelets from the single wavefront diffract at an angle 1 such that BN is 3/2 and reachthe point P1.The pairs (0,8), (1,9), (2,10), (3,11) and(4,12) interfere constructively with pathdifference and (0,4), (1,5), (2,6), and (8,12) interfere destructively with path difference/2. However due to a few wavelets interfering constructively First Secondary Maximum(Bright fringe) is formed.


P1

O

0123456789101112

P1

N

A

B 2

/2

P2

3/2

Slit

Screen

PlaneWavefront

Bright

Dark

Dark

O

0123456789101112

P1

N

A

B

1

/2

P1

3/2

1

1

P2

Slit

Screen

PlaneWavefront


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Since n is very small,d n = nn = n / d (n = 1, 2, 3, )

Since n is very small,d n = (2n + 1) / 2n = (2n + 1) / 2d (n = 1, 2, 3, )

To establish the condition for secondary minima, the slit is divided into 2, 4, 6, equalparts such that corresponding wavelets from successive regions interfere with pathdifference of /2.

Or for nth secondary minimum, the slit can be divided into 2n equal parts.

For 1, d sin 1 = For 2, d sin 2 = 2For n, d sin n = n

To establish the condition for secondary maxima, the slit is divided into 3, 5, 7, equalparts such that corresponding wavelets from alternate regions interfere with pathdifference of .Or for nth secondary minimum, the slit can be divided into (2n + 1) equal parts.

0 = 2D / d

y1 = D / dSince the Central Maximum isspread on either side of O, thewidth is

Theory:

The path difference between the 0th wavelet and 12th wavelet is BN.If is the angle of diffraction and d is the slit width, then BN = d sin

For 1, d sin 1 = 3/2For 2, d sin 2 = 5/2For n, d sin n = (2n + 1)/2

Width of Central Maximum:

O

012345

6789

101112

P1

N

A

B

/2

Slit

Screen

PlaneWavefront

Bright

Dark

D

dy1

tan 1 = y1 / Dor 1 = y1 / D (since 1 is very small)

d sin 1 = or1 = / d (since 1 is very small)


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Fresnels Distance:Fresnels distance is that distance from the slit at which the spreading of lightdue to diffraction becomes equal to the size of the slit.

y1 = D / d

At Fresnels distance, y1 = d and D = DFSo, DF / d = d or DF = d

2 /

If the distance D between the slit and the screen is less than Fresnels distance DF, then the

diffraction effects may be regarded as absent.So, ray optics may be regarded as a limiting case of wave optics.

Polarisation of Transverse Mechanical Waves:

Transverse

disturbance(up and down) Narrow Slit

Transverse

disturbance(up and down) Narrow Slit

Narrow Slit

90


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90

No light

Polariser

Unpolarisedlight

PlanePolarisedlightAnalyser

Plane of Vibration Plane of Polarisation

90

Polarisation of Light Waves:

Light waves are electromagnetic waves with electric and magnetic fieldsoscillating at right angles to each other and also to the direction ofpropagation of wave. Therefore, the light waves can be polarised.

PolariserTourmalineCrystal

AnalyserTourmalineCrystal

Unpolarisedlight

PlanePolarisedlight

PlanePolarisedlight

Optic Axis

Unpolarisedlight

PlanePolarisedlight


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When a beam of plane polarised light is incident on an analyser, the intensity I oflight transmitted from the analyser varies directly as the square of the cosine of theangle between the planes of transmission of analyser and polariser.

I cos2

P A

a cos a sin If a be the amplitude of the electricvector transmitted by the polariser,

then only the component a cos willbe transmitted by the analyser.

Intensity of transmitted light fromthe analyser is

I = k (a cos )2or I = k a2 cos2

I = I0 cos2

Case I : When = 0or 180, I = I0Case II : When = 90, I = 0Case III: When unpolarised light is incidenton the analyser the intensity of thetransmitted light is one-half of the intensity of

incident light. (Since average value of cos2

is )

a

(where I0 = k a is the intensity

of light transmitted from thepolariser)

Polarisation by Reflection and Brewsters Law:

r

P

90

P + r = 90 or r = 90- P

sin Pab =

sin r

a

b

sin Pab =sin 90- P

ab = tan P

Malus Law:

The incident light wave is made ofparallel vibrations ( components)on the plane of incidence andperpendicular vibrations ( components : perpendicular to plane

of incidence).

At a particular angle P, the parallelcomponents completely refractedwhereas the perpendicularcomponents partially get refracted andpartially get reflected.

i.e. the reflected components are all inperpendicular plane of vibration andhence plane polarised.

The intensity of transmitted lightthrough the medium is greater thanthat of plane polarised (reflected) light.

6 2 Formulae Wave Optics

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