PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE Form …day.sfc-plk.edu.hk/Physics/notes_cyp/Microsoft Word - Waveph.pdfAL Physics/Wave Phenomena/P.3 - The new position of the wavefront

AL Physics/Wave Phenomena/P.1

PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGEForm Seven AL Physics: Wave Phenomena

Wave phenomena Familiarity with ripple tank experiments isassumed from lower form work.

Huygens’ principle Explanation of laws of reflection and refraction.Reflection Examples to include brief discussion of radar,

sonar and long distance propagation of radiowaves by reflection from the ionosphere. Phasechange on reflection, illustrated for example,using a slinky spring.

Refraction Refraction as a result of change in wave speeds.Refractive index in terms of speeds.

Polarization Polarization by selective absorption, reflectionand scattering. Practical applications to includepolaroid spectacles, VHF and UHF antennas(briefly).

E6. Polarising light by(a) reflection from a shiny surface;(b) absorption using a sheet of

polaroid; and(c) scattering using cloudy water

(NAP 8.15/J 21d).Superposition Mathematical treatment not required. E7. Superposition of transverse waves on

a slinky spring.Beats Qualitative treatment. Use in tuning. E8. Observation of beats on a CRO.Diffraction Diffraction of light at apertures (simple

qualitative treatment only).E9. Looking at a lamp through a slit or a

pin-hole to study how the diffractionpatterns depend on(a) the shape of the aperture;(b) the size of the aperture; and(c) the wavelength of light.(NAP 8.1a/J 1)

Interference Two-source interference with quantitativetreatment for maxima and minima. Conditionsfor observable interference. Practicalapplications of interference to include theblooming of lenses and the testing of theflatness of a surface (very briefly). Quantitativetreatment of interference effects at normalincidence in parallel-sided and wedge-shapedthin films. Every day examples to include thecolours of oil films and soap bubbles.Newton’s rings (qualitatively).Plane transmission grating as an interferencesystem. Use of the formula d sin θ = nλ.Proportionality between intensity and square ofthe amplitude (by analogy with harmonicoscillator and energy delivered by analternating current). Energy distribution ininterference patterns.

E10. Estimation of the wavelength of lightusing(a) double slit, and(b) plane diffraction grating.(Nat Phil Workbook 5)

E11. Observation of Newton’s rings andinterference fringes in soap film.

E12. Investigation of the amplitude andenergy distribution in an interferencepattern of sound waves (NAP 8.7/J4).

A. Huygen’s Principle- enables the new position of a wavefront to be found, knowing its position at some previous

instant.- can be used to explain the reflection, refraction, etc.- Huygen’s Principle: every point on a wavefront may be regarded as a source of secondary

spherical wavelets which spread out with the wave velocity. The new wavefront is theenvelope of these secondary wavelets, i.e. the surface which touches all the wavelets.


1. Propagation of straight/plane waves- Given the wavefront of a plane wave in free space, where will the wavefront be at time t

later?

2. Propagation of circular/spherical waves

3. Law of Reflection- the wavefront AB is incident obliquely on the reflecting surface and A has just reached it.- to find the new position of the wavefront when B is about to be reflected at B’,- a secondary wavelet with centre A and radius BB’ is drawn.- The tangent B’A’ from B’ to this wavelet is the required reflected wavefront.

In ∆AA’B’ and ∆ABB’∠ AA’B = ∠ ABB’ = 90ºAB’ is commonAA’ = BB’ (by construction)therefore, ∆AA’B’ ≡ ∆ABB’ (RHS)⇒ ∠ BAB’ = ∠ A’B’Ai.e. i = r (Law of reflection)

4. Law of Refraction- the end A of a plane wavefront AB about to cross the boundary between media ! and " in

which its speeds are v1 and v2 respectively.

rays showingthe directionof propagation

ct

ct

ct

ctSecondarysource

BD

A CConstructed

wavefront

First positionof wavefront

Secondarywavefront

A

B

C

D

fromSecondary wavelet

incident wavefront Reflected wavefrontD

B'A

rri i

CB A'

A


- The new position of the wavefront at time t later when B has travelled a distance BB’ = v1tand reached B’ is found by drawing a secondary wavelet with centre A and radius AA’ = v2t.

- The tangent B’A’ from B’ to this wavelet is the required wavefront and the ray AA’, whichis normal to the wavefront, is a refracted ray.

∆ BAB’ sin'

iv tAB1

1=

∆ A’AB’ sin'

iv tAB2

2=

Hence, sinsin

//

ii

vv

c vc v

nn

1

2

1

2

2

1

2

1

= = = (Snell’s Law)

where n1 and n2 are absolute refractive index of themedia.

** Refraction occurs when a wave passes from onemedium into another in which it has a different speed.

B. Reflection1. Application of Reflectiona. Radar (Radio detection and ranging)

- radio waves (EM wave, λ: 30 km - 3 cm) are emitted in short pulses by a transmitter andpicked up after reflection from the object.

- By measuring the time between the transmitted pulse and the reflected pulse, distance fromthe transmitting station to the ‘object’ is then found.

b. Sonar (Sound navigation and ranging)- similar to radar but employs ultrasonic waves (Sound waves with frequency above 20 kHz).- measure the depth of the sea, detecting shoals of fish, etc.

c. Reflection from the ionosphere- stretching from about 80 km above the earth to 500 km- ionosphere, positively charged through the removal of electrons by the sun’s ultraviolet

radiation.- E.M. wave with frequency below 30 MHz cannot penetrate the ionosphere but reflected.- Short waves (SW): 6 to 20 MHz ‘bounced’ from the ionosphere and are widely used for

long distance broadcasts.

Refracted wavefront

Incidentwavefront

C

A

B

B'

A'

N

N'

v1 > v2

v1

v2

i1

i2

!

"

i1

i2

Secodnary wavelet from A


2. Phase Inversiona. - Consider a long narrow spring with the right-hand end of the spring being fixed. A

transverse ‘upward’ pulse travels towards the fixedend and is reflected.Observation: The reflected pulse is inverted relativeto the incident pulse!!i.e. a phase change of 180º or π rad has occurred.

- now the right-hand end of the spring becomes free,the transverse ‘upward’ pulse travelling towards it isreflected.Observation: The whole of the incident pulse isreflected the right way up.i.e. no phase change occurs!!

- Why??

b. Principle of SuperpositionWhen two waves disturbances meet, the resultant displacement at any point is the vector sumof the separate displacements due to the two waves.

** If the two disturbances travel in different directions then when they separate, each moves onas if nothing has occurred.

c. - A pulse P is travelling along the rope AOwith O being fixed.We can imagine that there is another ropeOA’ which is the mirror image of OA and thereflected pulse is coming from OA’.At the time the pulse P reaches the reflectingsurface, the image P’ reaches the interfacesimultaneously. The displacement of thepoint O is the sum of the individual

free end

fixed end

(i)

(ii)

(iii)

(iv)

(v)

(b) Crest and trough(a) Crest and crest

(a)

(b)

(c)

A

P

P' A'

A A'

A'AP'

P

P

P'

O

O

O


displacement of P and P’ at O. However, as O is fixed, displacement at O must be equal tozero. Hence, the displacement of P’ must be negative of that of P or else the sum of thetwo pulses could not be zero. Therefore, there must be a phase difference π between theincident pulse and the reflected pulse.

- If O is not fixed, i.e. a free end, the constraint on the displacement of the particle O will beremoved and no phase inversion will occur.

d. - When a transverse wave on a spring is reflected at a ‘denser’ medium (e.g. a fixed end or aheavier spring) there is a phase change of 180º (or π rad or λ/2).

- Phase changes also occur when longitudinal waves are reflected. At a fixed end acompression is reflected as a compression, at a free end a compression is reflected as ararefaction.

- When lights hits a medium of higher refractive index, it is reflected with a phase change of π; when it hits a medium of lower refractive index, it is reflected with no phase change.

C. Refractionsee law of refraction in Huygen’s Principle

D. Polarization1. Polarization of Waves

- A transverse wave due to vibrations in one plane is said to be plane-polarized.The figure shows a plane-polarized wave due to vibrations in the vertical plane yOx (y-polarized) and another plane-polarized wave due to vibrations in the perpendicular planezOx (z-polarized).

Heavy spring Light spring Heavy spring Light spring

"

!

x

x

R C

C R

ξ

ξ


- Longitudinal waves such as sound cannot be polarized.

2. - Consider a horizontal rope AD attached to a fixed point D at one end. Unpolarizedtransverse waves due to vibrations in many different planes can be set up along A byholding the end A in the hand and moving it up and down in all directions perpendicular toAD, as illustrated by the arrows in the plane X.

- Suppose there are two parallel slits B and C between A and D as shown.- A wave then emerges along BC, but unlike the waves along the part AB of the rope (not

shown), which are due to vibrations in many different planes, the wave along BC is dueonly to vibrations parallel to the slit B. This plane-polarized wave passes through theparallel slit C.

- But when C is turned so that it is perpendicular to B, as shown in figure (ii), no wave isnow obtained beyond C.

3. Polarization of microwaves- A grille of parallel metal rods is rotated between (a) a source T of 3 cm plane-polarized

E.M. wave (i.e. microwave) as shown; (b) detector, a probe, with a meter connected to it.- When the rods are horizontal the meter reading is high, figure (i). So a wave travels past

the grille.- When the grille is turned round so that the rods are vertical, there is no deflection in the

meter, figure (ii). Thus the wave does not travel past the grille.- If the electric field is along the metal rods, the electrons in the rods will be driven into

motion. Energy is transferred from the microwave to the electrons. The microwave isabsorbed and cannot pass on. If the electric field is perpendicular to the rods, the electronscannot go into a corresponding motion. The microwave is not absorbed; it propagatesthrough the ‘comb’.

x

y

z

O

A

X

B

polarizedwave

polarizedwave

D

A

X

B

polarizedwave

D

C

C

(i)

(ii)


4. Polarization of light- Ordinary light from a lamp or the sun is unpolarized.- Producing polarized light:-

a. Polarization by Selective absorption (by Polaroid)Polaroid contains long chain-like molecules, arranged to lie parallel to each other.Electrons can move freely along these chains, but not perpendicular to the chains. Lightpolarized perpendicular to the chains will pass through while that polarized parallel tothe chains will be absorbed. (The effect is like the metal grid in polarization ofmicrowaves.)Example:-(i) Amplitude of the Electric field E after passing through the polarizer = ??

** In fact what our eyes detect is the light intensity, which is the energy per unit areaper unit time, but not the amplitude of the electric field. However, the intensity ofa propagation wave is directly proportional to the square of the amplitude of thewave. i.e. I ∝ A0

2

wave E wave E highreading

Pmetalrods

wave E

no wave

90º rotation

zeroP

(i)

(ii)

T

!E0

β

!E E= 0 cos ββ

polarizer

plane-polarized

light

E0 sin β

E0 cos β

!E0

β


(ii) Intensity of the light after passing through the polariser = ??

Amplitude of the incident plane-polarised light E0 = √I0

(Let the proportional constant be 1)after passing through the polariser, the amplitude of the light wave = E0 cos β

= √I0 cos βhence intensity of the light observed = square of the amplitude = I0 cos2β

(iii) I = ???

- after passing through the polariser A, the unpolarised incident light becomesplane polarised with light intensity I0/2 (and the amplitude of the wave = √(I0/2),let the proportional constant = 1)

- passing through the polariser B, amplitude of the E-field becomes √(I0/2) cos β,hence the intensity of the light observed = (I0/2) cos2 β

b. Polarization by reflection(i) - Looking at the reflection of a lamp

(unpolarised) from a glass platethrough a Polaroid.

- the observed intensity changes whenthe Polaroid is rotated.

- imply that the reflected light is partially polarized, i.e. there is more linearpolarization of one kind and less of the other.

- why?? (ii) - mechanism of producing reflected waves:-

when the light ray passes into the glass,electrons in the glass are set into vibrationsalong the electric field, i.e. along the direction

I 0

β

polarizer

plane-polarizedlight with intensity

I = ??

β

A

B

I

I 0

*glass plate

polaroid

light source

e-


of polarization. The oscillating electrons radiate E.M. waves, which form thereflected waves.** Notice that there is no waves generated in the direction of oscillation and theamplitude of waves generated is small for those propagating close to thedirection of oscillation.

(iii) - for polarization which is perpendicular to the plane of paper, electrons vibrate inand out of the paper, and radiate efficientlyin the reflected direction so reflectedintensity is strong.

- for polarization which is in the plane ofpaper, the polarization of the transmittedwave is parallel to the direction of vibrationof the electrons. The reflected direction isclose to the vibration direction of electron,so the reflected intensity is relatively low.

- Hence the reflected light is preferentiallypolarized perpendicular to the plane ofreflection.

- For the case that the reflected ray isperpendicular to the transmitted ray, thereflected ray becomes completely polarized (perpendicular to the plane ofpaper). The incident angle at which this occurs is called Brewster’s angle.

- Example: find the Brewster’s angle for glass,given that n = 1.5By Snell’s law na sin θb = ng sin r

= ng sin (90º - θb)= ng cos θb

Hence tan θb = ng/na = 1.5 θb = 56º

c. Polarization by scattering(i) The light ray from the scattering of sunlight by air is

partially polarized. Why??(ii) - a transparent tank with cloudy water (make by

adding a drop of milk).- illuminate the tank by a narrow beam of

unpolarised light.- direction OA: the direction of observation OA is

parallel to the vibration along the x-axis, hence only the vibration along the y-axis can be observed. The light is y-polarized.

strongi r

t

(a)

weaki r

t

(b)

θ

θ’

θ

r

b

ng

na = 1

incident reflected

transmitted

scatteredlight

polaroid


direction OB: the direction of observation OB is parallel to the vibration alongthe y-axis, hence only the vibration along the x-axis can be observed. The lightis x-polarized.direction OC: unpolarised.direction OD: partiallypolarized, preferentiallypolarized in the horizontaldirection.

5. Applicationa. Reducing glare

The Sun’s rays reflected from glass, water or scattered by the sky are preferentiallypolarized in the horizontal direction. By making use of suitably oriented Polaroid discs,the glare can be reduced.

b. VHF and UHF antennas- FM radio broadcast in the VHF band makes use of horizontal polarization. FM

antennas should be horizontal.- Television broadcast in the UHF band may either be on horizontal or vertical

polarization. The antennas should be oriented according to direction of polarization ofthe received wave.

E. SuperpositionSee principle of superposition in phase inversion of reflection.

F. Beats- When two sound waves of slightly different frequencies but similar amplitude are sounded

together, the loudness increases and decreases periodically and beats are said to be heard.- By principle of superposition, the displacement of the resultant wave is the sum of

displacement of individual wave.- At an instant A or C, the waves from the sources arrive in phase and reinforce to produce a

loud sound.- At an instant B, the waves from the sources are 180º out of phase. The two waves cancel

each other and little sound is heard.- Beat frequency:

Let T be the beat period (i.e. the time between two successive maxima).Within T, one wave-train of frequency f1 makes one cycle more than the other frequency f2.i.e. f1T - f2T = 1 f1 - f2 = 1/T

hence, beat frequency = f1 - f2 = ∆f

B

A

C

D

O


- use: Tuning of musical instrument such as violin.

G. Diffraction

- The spreading of waves when they pass through an opening or round an obstacle is calleddiffraction.

- Diffraction is noticeable when the width, a, of the opening is comparable to the wavelengthof the waves (i.e. λ ≥ a) or very small when the width is large compared to the wavelength.

- Sound: long wavelengths, hence can diffract after passing through doorways.visible light : short wavelength (6000 × 10-10 m), hence diffract appreciably only throughvery small opening.

- Parallel plane wavefronts are diffractedat a rectangle slit AB of width a.The light passing through the slit isreceived on a screen S.Alternate bright and dark fringes areobserved.Most of light is concentrated in thecentral bright band.At some points, the intensity diminishes to a minimum (e.g. Q1, Q2, ...)Position of the minimum is given by

a sin θ = m λ (where m = 1, 2, 3, ...)

Time

Time

Displacement

ResultantDisplacement

(a)

(b)

0

0

f 1 f 2

A B CVariation of

amplitude

Beat period = T

Hence, Beat frequency = 1/ T = f 1 f 2-

central bright band

brightnessO

darkθ1

λ__a=

θ2λ__

a= 2

a

A

B

Q1

Q2

θ

S


The angular width of the central bright fringe (which reflects how serious the spreading ofthe waves is) is 2θ, where θ is the angle between the direction of maximum intensity andthe direction of first minimum intensity.The angle is given by a sin θ = m λ with m = 1. i.e. sin θ = λ/aWhen the slit is wide, a becomes large compared with λ, then sin θ is very small and hence θ is very small, no spreading occurs.When the slit is small, a becomes small compared with λ, sin θ is very large and hence θ isvery large and spreading (diffraction) is appreciable.

H. Interference1. Interference:- in a region where wave-trains from coherent sources cross, superposition

occurs giving reinforcement of the waves at some points and cancellation at others.

** Coherent Sources:- sources have a constant phase difference, which means they must havethe same frequency.

The coherent sources required in Interference can be obtained by:-(1) division of wavefront, e.g. Young’s double slit, Fresnel’s biprism, Lloyd’s mirror, etc.(2) division of amplitude, e.g. wedge fringes, Newton’s rings, etc.

2. Division of wavefronta. Young’s double slit experiment

Principle:-- monochromatic light (i.e. of one

color) from a narrow vertical slitS falls on two other narrow slitsS1 and S2.

- S1 and S2 act as two coherentsources (both being derived fromS, by division of wavefront).

- diffraction causes the emerging beams to spread into the region beyond the slits.- superposition occurs in the shaded area, where the diffracted beams overlap and interfere.

1

2

1 + 2

t

t

t

Reinforcement

1 + 2

2

1t

t

t

Cancellation

interferenceeffects in region wherebeams overlapMonochromatic

light source

(narrow)Single slit Double

slit

S

S1S1

S2

Diffracted beamfrom S1

Diffracted beamfrom S2


- alternate bright and dark equally spaced interference fringes can be observed on the screen.

Theory:-- The path difference between waves reaching O from S1 and S2 is zero, i.e. S1O = S2O,

hence, they arrive in phase, it is a bright fringe at O.- At any point P on the screen, the path difference between waves reaching from S1 and S2 is

S2A which is approximately given byS2A = d sin θ

for small angle θ, sin θ ~ tan θ = x/L,hence the path difference S2A = dx/L

maxima: There will be a bright fringe if thepath difference is a whole number ofwavelength

i.e. S2A = nλ = d sin θd sin θ = n λ (where n = 0, 1, 2, ...)

minima: There will be a dark fringe if the path difference is (n + ½) λi.e. d sin θ = (n + ½) λ (where n = 0, 1, 2, ...)

- Separation of the fringes:-position of maxima is given by d sin θ = n λor d x/L = n λhence separation ∆x between the nth and (n + 1)th bright fringes is given d (∆x/L) = λi.e. ∆x = λ L /d

Appearance of Young’s interference fringes:-(i) ∆x ∝ 1/d if λ and L constant (therefore the slit separation should be small in order to

increase the separation);∆x ∝ L if λ and d constant (therefore the fringes should be viewed from a distance);∆x ∝ λ if L and d constant.

(ii) If the source slit S is moved nearer the double slits the separation of the fringes isunaffected but their brightness increases.

(iii) If the source slit S is widened the fringes gradually disappear. The slit S is thenequivalent to a large number of narrow slits, each producing its own fringe system atdifferent places. The bright and dark fringes of different systems therefore overlap,giving rise to uniform illumination. It can be shown that, to produce interferencefringes which are recognizable, the slit width of S must be less than λD’/d, where D’ isthe distance of S from the two slits S1, S2.

(iv) If one of the slits, S1 or S2, is covered up, the fringes disappeared.

L

x

d θ

θ

P

OA

S1

S2


(v) If white light is used the central fringe is white, and the fringes either side are colored.Blue is the color nearer to the central fringe and red is farther away. The pathdifference to a point O on the perpendicular bisector of the two slits S1, S2 is zero for allcolors, and consequently each color produces a bright fringe here. As they overlap, awhite fringe is formed. Farther away from O, in a direction parallel to the slits, theshortest visible wavelength, blue, produce a bright fringe first.

(vi) The number of fringes obtained depends on the amount of diffraction occurring at theslits and this in turn depends on their width. The narrower the slits, the greater will bethe number of fringes due to the increased diffraction but the fainter they will be, sinceless light gets through.

Necessary conditions for observable interference:-(i) The two waves must have the same polarization.(ii) The two waves must have a constant phase difference.

If the phase difference changes continuously, then there is sometimes constructiveinterference and sometime destructive interference. The average would be nointerference at all.

If we use two different sodium lamps to illuminate the two slit, interference will not beseen. The reason is that a sodium lamp emits light not continuously but in ‘bursts’,which we call wave packets or wave trains. There is no definite phase differencebetween the wave trains from the first lamp and those from the second lamp. Henceinterference is not observed.

(iii) The two waves must have exactly the same frequency.(iv) The path difference must not be too large.

Consider two waves from the same source, but traveling over slightly differentdistances to meet at a certain point. If the path difference is larger than the length(coherent length) of wave train, the received waves will come from different wavetrains, which are unrelated and cannot be coherent.** The coherent length can be several meters for a laser, but less than 1 mm for aconventional light source.

Examples:- S1 and S2 are two coherent light sources in a Young’s two-slit experimentseparated by a distance 0.50 mm and O is a point equidistant from S1 and S2. O is on a screenA which is 0.80 m from the slits.When a thin parallel-sided piece of glass G of thickness is 3.6 × 10-6 m is placed near S1 asshown, the center of the fringe system moves from O to a point P. Calculate OP if the

wave trains

coherent length


wavelength of the monochromatic light from the two slits is 6.0 × 10-7 m and the refractiveindex of the glass is 1.5.

Solution:- If P is the center of the fringe system, the number of waves is S1P = the number of waves in

S2P.- Since light travels slower in the glass G than in air, the number of wavelengths in G is more

than in air of the same thickness as G.- The Extra path difference due to the inserting of G = nt - t = (n-1)t **- This extra path difference should be compensated by the extra length of S2P, i.e. dsinθ

(= d × OP/L)

- hence, dOPL

n t× = −( )1

OPn t

dL=

−( )1 = 2.88 mm

** No. of waves in air of thickness t = t/λNo. of waves in G of thickness t = t/λ’ = nt/λ(i.e. = no. of waves in air of thickness nt)hence, a glass of thickness t can be viewed as air of thickness nt.

Variation of intensity of the fringe pattern:-

- Let the intensity due to each individual source be I0.- hence amplitude of the wave from each individual source = √I0

- for bright fringes: constructive interference,the amplitude of the resultant wave = 2√I0

hence, intensity of the bright fringes = (2√I0)2 = 4 I0

max max4

1 2 y/s

I/I0

min min0

1

2

1 + 2

t

t

t

constructive interference

1 + 2

2

1t

t

t

destructive interference

A A

A A

A = 02A

d= 0.50 mm θ

dsinθS2

S1G

L = 0.80 mO

A

P


- for dark fringes: destructive interference,the amplitude of the resultant wave = 0hence, intensity of the dark fringes = 0.

- If the two sources are independent, no interference pattern will be observed. On the screenthe light intensity is uniform and is equal to 2I0.

b. Lloyd’s mirror- A plane mirror is illuminated at almost

grazing incidence by light from a slit S1,parallel to the mirror.

- A virtual image S2 of S1 is formed close to S1

by reflection and these two act as coherentsources.

- In the shaded area direct waves from S1 crossreflected waves which appear to come from S2

and interference fringes can be seen.- The expression giving the fringe spacing is

the same as for the Young’s double slitexperiment.

- However, in Lloyd’s mirror, if the point P, for example, is such that the path difference S2P- S1P is a whole number of wavelengths, the fringe at P is dark, not bright. This is due tothe 180º phase change which occurs when light is reflected from the mirror.

c. Fresnel’s biprism- Monochromatic light from a narrow slit S

falls on a double glass prisms arranged asshown.

- Two virtual images S1 and S2 are formedof S, one by refraction at each half of theprism, and these act as coherent sources.

- An interference pattern is obtained by theshaded region where the two refractedbeams overlap.

- The theory and the expression for the fringe spacing are the same as for Young’s method.

3. Division of Amplitudea. Wedge Fringes

- An air wedge can be formed from two microscope slides clamped at one end and separatedby a thin piece of paper at the other end.

- monochromatic light from an extended source is partially reflected vertically downwards bythe glass plate G. When the microscope is focused on the wedge, bright and dark equallyspaced fringes are seen, parallel to the edge of contact of the wedge.

Light from S1 should fall at grazing incidence onthe glass plate; the incidence is exaggerated here

P

C

S1

S2

*

Monochromaticsource Single

slit

Cross-wire ofeyepiece or screen

A

Glass plate

Cross-wireof eyepieceor screen

d

a

BiprismSingleslit

Interference effects

*

Monochromaticsource

S1

S2

S


- Theory:-Some of the light falling on the wedge is reflected(i) upwards from the bottom surface of the top slide

and(ii) some of the rest, (which is transmitted through the

air wedge) is reflected upwards from the top surfaceof the bottom slide.

If l is the thickness of the air wedge at P then the pathdifference between the rays at P = 2l.However, owing to the 180º phase change of the wavereflected at the top surface of the bottom slide, there isan extra path difference of half a wavelength.Hence, the total path difference between the twowavetrains at P = 2l + ½λfor constructive interference: o.p.d. = nλ

i.e. 2l + ½λ = nλ2l = (n - ½)λ (bright fringe)

for destructive interference: o.p.d. = (n + ½)λ i.e. 2l + ½λ = (n + ½)λ

2l = nλ (dark fringe)separation of the fringes: 2ln = nλ

2ln+1 = (n + 1)λhence, 2(ln+1 - ln) = λ

2 ∆l = λfor small angle θ, θ = ∆l/∆x

hence, ∆l = θ ∆xhence, 2 θ ∆x = λ

∆x = λ/(2θ)** Separation ∆x increases, as λ increases. Separation ∆x increases, as θ decreases.

- Question:What would happen if(i) water is introduced into the region between the slides?(ii) the contact side of the upper slide is uplifted slowly?

b. Newton’s rings- Monochromatic light (e.g. from a sodium lamp) is reflected by the glass plate G so it falls

normally on the air film formed between the convex lens of long focal length and the flatglass plate.

- Interference occurs between light reflected from(i) the lower surface ABC of the lens and

Sodium lamp

Microscope

G

Air wedge

Monochromaticlight

Glassplates

PQ

O

θ

l

∆x

∆l

Travelling


(ii) the upper surface DBE of the plate.- Optical path difference between the two ray at P =

2PQ + ½λ (the extra o.p.d. ½λ is due to the phaseinversion when light is reflected from the uppersurface DBE of the plate.)

- Constructive interference:if the o.p.d. = mλ (bright ring)

Destructive interference:if the o.p.d. = (m + ½)λ (dark ring)

- A series of bright and dark rings is observed.- The rings are fringes of ‘equal thickness’.- As the radii of the rings increase, the separation decreases,

why?- At the center B of the fringe system where the geometrical

path difference between the two waves is zero, there is adark spot, why?

- In what way would the Newton’s rings pattern change if theconvex lens is moved upwards slowly?

4. Using interferencea. Testing of optical surfaces

- Fringes of equal thickness are useful for testing optical components.- Example:

(i) In the making of optical ‘flats’, the plate under test is made to form air wedge with astandard plane glass surface. Any uneven parts of the surface will show up asirregularities in what should be a parallel, equally spaced straight set of fringes.

(ii) The grinding of a lens surface may be checked if it is placed on an optical flat andNewton’s rings observed in monochromatic light. The rings should be exactly circularif the lens is spherical.

b. Non-reflecting glass (Blooming of lenses)- In optical instruments containing lenses or prisms light is

lost by reflection at each refracting surface and results inreduced brightness of the final image.

- The amount of light reflected at a surface can be reducedby coating it with a film of transparent material (e.g.magnesium fluoride) to a thickness of ¼λ’ of light in thefilm.

- Light reflected from the top (ray 1) and bottom (ray 2)surfaces of the film interfere.

- as the o.p.d. between the two rays= 2l = 2(¼λ’) = ½λ’

Sodiumlamp

Travellingmicroscope

Flat glass plate

Plano-convexlens Air film

D EB

A C

G

L

R

A C

BS P

lQ

Orn rn

Air

Film

Glass

! "

l


( l = ¼λ’ = 14

λn

,where λ’ is the wavelength of light in the film,

λ is the wavelength of light in air,n is the refractive index of the material of the film.)

** There is no need to consider the phase inversion. The refractive index of the film isless than that of glass and so each reflection (both the top and bottom reflections)suffers a 180º phase change. The net effect of the phase changes on the pathdifference is thus zero.

- The two rays interfere destructively, i.e. there is no reflected light.- The interference is complete for one wavelength only, usually taken to be that at the center

of the visible range. For red and blue light the reflection is weakened but not eliminatedand a coated or ‘bloomed’ lens appears purple in white light.

- e.g. λ = 550 nm and n = 1.38, then l = 100 nm.

5. Everyday examples of interferencea. Colors of oil films on water

- Interference occurs between two wave-trains:(i) one reflected from the surface of the oil and(ii) the other from the oil-water interface.

- When the path difference gives constructiveinterference for light of one wavelength, thecorresponding colour is seen in the film.

- As the path difference varies with the thickness of the film and angle of viewing, differencecolour is seen at different parts of the film.

b. Vertical Soap film(i) a vertical soap film illuminated by monochromatic light:- At first the film appears uniformly coloured.- As the soap drains to the bottom, a wedge-shaped film of

liquid forms, the top of the film being thinner than thebottom. Horizontal bright and dark fringes are observedacross the film.

- When the upper part of the film becomes extremely thin(small than λ’/4), a black fringe is observed at the top (theblack fringe is due to the 180º phase change by reflection).And the film breaks shortly afterwards.

(ii) a vertical soap film illuminated by white light:- a succession of broad coloured fringes, violet to red, is first observed.- The fringes widen as the film drains, and just before it breaks a black fringe is obtained at

the top.

Air

Oil

Water

*

Soap film


c. Pulsing of the picture on a television receiver- When an aircraft passes low overhead, the signal

travelling directly from the transmitting to thereceiving aerial interferes with that reflected from theaircraft.

6. Diffraction gratinga. A diffraction grating is a large number of close parallel

equidistant slits, ruled on glass or metal (e.g. 300 lines permm).

b. Theory:- a plane wave of monochromatic light of wavelength λ

falls on a transmission grating in which the slitseparation (called the grating spacing) is d.

- consider the wave trains coming form the slits andtravelling at an angle θ to the direction of the incidentbeam.

- The path difference is dsinθ for all pair of wavesfrom any two successive slits.

- If d sinθ = nλ, where n is an integer,then reinforcement of the wave trains occurs indirection θ and a maximum will be obtained.

- Notice that the condition for maximum indiffraction grating is the same as that in Young’sdouble slit experiment.

- Example:a diffraction grating of 300 lines per mm, lightof wavelength λ = 6 × 10-7 m. What is themaximum order of bright fringe that can beviewed?grating spacing d = 1 × 10-3/300maximum possible optical path difference isd.d/λ = 5.5, hence the maximum order that canbe viewed is 5.

c. Variation of intensity- For a diffraction grating, the bright fringe is

narrow and the intensity is very high(compared with that in Young’s double slitsexperiment).

λ

θ

d

d sin θ

n = 0

n = 1

n = 2n = 3

n = 1

n = 2n = 3

θ3

θ2θ1

Grating

Incidentbeam

1/2-1/2 0

N = 2

y/s1-1

0 1/3 2/3 1-1/3-2/3-1

N = 3

0-1 1

N = 5

y/s

y/s

I / I0

I / I0

I / I0

∆x

∆x


- In fact, the width of the bright fringe, ∆x, is inversely proportional to the number of slits N.i.e. ∆x ∝ 1/N.the intensity of the bright fringe I is directly proportional to the square of the number ofslits N. i.e. I ∝ N².

d. Application(i) Determination of wavelength using a diffraction grating- because the interference maxima are much sharper and brighter, a diffraction grating

permits a much better determination of wavelength.- Put a sodium lamp (with a vertical slit) on the

bench. Place a meter rule A pointing at thelight source about 3 m away, and anothermetre rule B perpendicularly at one end of A.Place a fine diffraction grating with thegrating slits vertical. View the illuminated slitthrough the grating and mark the angularpositions of the images by moving a pencil Palong B.

- Since the light source is sufficiently far away from the grating, which is perpendicular tothe direction of incident light, the situation is one of normal incidence.

- The corresponding distance of the pencil fromA is ym. Then the angular position of the mthorder maximum is given by

tan θ = ym

and the wavelength λ of the light can becalculated by

d sin θ = m λprovided that the grating spacing d is known.

(ii) Spectra Analysis- If white light is incident normally on a diffraction

grating several colored spectra are observed on eitherside of the normal.

- All the colors coincide in the center, forming a whiteline.

- As violet has a shorter wavelength than red, θ is lessfor violet than for red. Consequently the spectrumcolors on either side of the incident white light areviolet to red.

- The second order band is wider than the first orderdiffraction band; the third order diffraction band iswider than the second diffraction band, and so on (i.e.

metre rulesodium lamp B

diffractiongrating

θ

P

A

(a)

1 m

θθ

ym

P

(b)

W V1 R1 V2 V3 R2 R3

01

2

3


the dispersion increases with the order). Therefore ahigh resolution of the spectrum can be obtained if ahigh order pattern is observed.(Remark: The dispersion is also inverselyproportional to the grating spacing a. Hence in orderto have high resolution, a high order pattern should beobserved through a fine grating with small gratingspacing a.)

- The second and third order bands overlaps; the third and fourth order bands overlap, etc.The condition for two colors in two different bands to overlap is given by

a sin θ = m1λ1 = m2λ2

where m1, m2 are order numbers and λ1, λ2 are the wavelengths of the two colorsconcerned.

m1 = 2, m2 = 3 and the wavelength for red, λ1 = 700 nm.By the condition above, the minimum wavelengths of the light that will overlap with thered light is 467 nm. As wavelength for violet is about 400 nm which is shorter than 467nm, the second and the third must overlap.

- Conditions for a good diffraction grating: (i) small grating spacing a, (high dispersion).(ii) large total number of slit, N (sharper and brighter line).

red

violet

sin θ

sin θ

12

34

00.7/a 1.4/a 2.1/a

0.4/a 0.8/a 1.2/a 1.6/a0


** Supplementary Notes for diffraction

1. Suppose parallel plane wavefronts from adistant object are diffracted at a rectangular slitAB of width a.

2. Consider a plane wavefront which reaches theopening AB. All points on it between A and Bare inphase, that is, they are coherent. Theircombined effect at any distant point can befound by summing the numerous waves arrivingthere, from the Principle of Superposition.

3. Central bright image- Consider a point O on the screen which lies on

the normal to the slit passing through itsmidpoint C. O is the center of the diffractionimage. It corresponding to a direction θ = 0.

- The waves from the secondary sources such asA, X, Y, B have no path difference.

- All the waves arrive in phase at O. The centerof the diffraction image is therefore brightest.

- In a direction very slightly inclined to θ = 0,the waves from all the sources between A andB arrive slightly out of phase at thecorresponding point P of the image near thecenter. So the brightness decreases.

4. First minimum- In a particular direction θ1, we reach the dark edge Q of the central image. This direction

corresponds to a path difference of λ between the two sources at the edges A and B of theslit.

- Divide the wavefront AB into two halves.- The top point A of the upper half CA, and the top point C of the lower half BC, send out

waves to Q which have a path difference λ/2, from above. Thus the resultant amplitude atQ is zero.

- All other pairs of corresponding points in the two halves of AB, for example, X and Y whereCX = BY and the two bottom points C and B, also have a path difference λ/2. So thebrightness at Q is zero.

- Condition for the first minimum is d sin θ = λ.


5. Secondary fringes- Secondary bright are also obtained on the screen beyond Q.- Point T which lies in a direction θ3 where the path difference of waves starting from A and

B is 3λ/2.- The wavefront AB divided into three equal parts. The waves from the extreme ends of the

upper two parts have a path difference λ. Thus, these two parts of the wavefront producedarkness at T. The third part produces a fringe of light at T much less bright than thecentral fringe.

6. Minimum brightness of the secondary fringes- The directions of the minimum brightness of the secondary bright fringes can be found by

dividing the diffracted wavefront at the slit into four (or six, and so on) equal parts in thesame way as the central bright fringe.

- The directions for the successive minima are given byd sin θ = m λ (where m = 1, 2, 3, ...)

Documents

PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE Form …day.sfc-plk.edu.hk/Physics/notes_cyp/Microsoft Word - Waveph.pdfAL Physics/Wave Phenomena/P.3 - The new position of the wavefront