Upload
nathanian
View
224
Download
0
Embed Size (px)
Citation preview
7/31/2019 6 2 Formulae Wave Optics
1/12
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)
.
.
.
.
.
.
.
...
.
.
..
.
.
.
7/31/2019 6 2 Formulae Wave Optics
2/12
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
7/31/2019 6 2 Formulae Wave Optics
3/12
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:
7/31/2019 6 2 Formulae Wave Optics
4/12
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)
7/31/2019 6 2 Formulae Wave Optics
5/12
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:
7/31/2019 6 2 Formulae Wave Optics
6/12
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:
7/31/2019 6 2 Formulae Wave Optics
7/12
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
2) At an angle of diffraction = 1:
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:
7/31/2019 6 2 Formulae Wave Optics
8/12
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.
4) At an angle of diffraction = 1:
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.
3) At an angle of diffraction = 2:
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
7/31/2019 6 2 Formulae Wave Optics
9/12
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)
7/31/2019 6 2 Formulae Wave Optics
10/12
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
7/31/2019 6 2 Formulae Wave Optics
11/12
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
7/31/2019 6 2 Formulae Wave Optics
12/12
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.