Newton's Corpuscular theory1. Every source of light emits large number of tiny particles known as corpuscles.2. These corpuscles are perfectly elastic, rigid and weightless.3. The corpuscles travel in straight line with very high speed, which are different in different media.4. One gets sensation of light when the corpuscles falls on the retina.5. Different colours of lights are due to different size of corpuscles.

Maxwell's electromagnetic theoryLight wave was assumed to be electromagnetic wave.Light waves are associated with changing electric and magnetic field.Light waves are transverse electromagnetic waves.

Max Planck's Quantum theoryLight is propaged in form of packets of light energy called quanta.

Wave theory of light (by Huygen)a. Light travels in a form of longitudinal waveb. Different colour if light is due to different wavelength of wavec. when light enters our eyes, we get sensation of lightd. A material medium is necessary for propagation of longitudinal wave.

Resolving Power:The power of an optical instrument to produce distinctly separate images of two close objects is called resolving power of that instrument.

Resolving power of microscope: The minimum distance by which two point onjects are separated from each other so that their image are produced by the microscope are just seen separate is called limit of resolution.The reciprocal of limit of resolution is called resolving power of microscope.

Rayliegh's Criterion: Two images are said to be just resolved if the position of the central maximum of falls in first minimum of the other and vice versa.Least distance between two objects so that they are just resolved is given by =d=1.22 π

2 sin πΌ$=d=\frac{1.22\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9c\x86}}{2\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9b\u038c}}\phantom{\rule{0.22em}{0ex}}$If the space between the objective and object is filled with oil of refractive index π$\mathrm{\pi \x9d\x9c\x87}$ then,Least distance between two objects so that they are just resolved is given by =d=1.22 π

2 π sin πΌ$=d=\frac{1.22\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9c\x86}}{2\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9c\x87}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9b\u038c}}\phantom{\rule{0.44em}{0ex}}$Resolving power of microscope=1

d=2 π sin πΌ

π$=\frac{1}{d}=\frac{2\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9c\x87}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9b\u038c}}{\mathrm{\pi \x9d\x9c\x86}}\phantom{\rule{0.44em}{0ex}}$2 π sin πΌ$2\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9c\x87}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9b\u038c}$ is called numerical aperture.

Resolving power of telescopeResolving power of telescope is defined as the reciprocal of the least angle subtended at objective by two distinct point objects, which can be distinguished just resolved in the focal plane of the telescope.

Rayleigh's condition for just resolutiona.dπ=π $a.d\mathrm{\pi \x9d\x9c\x83}=\mathrm{\pi \x9d\x9c\x86}\phantom{\rule{0.44em}{0ex}}$βΉdπ=π

a$for\phantom{\rule{0.22em}{0ex}}circular\phantom{\rule{0.22em}{0ex}}aperture,\phantom{\rule{0.22em}{0ex}}d\mathrm{\pi \x9d\x9c\x83}=\frac{1.22\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9c\x86}}{a}\phantom{\rule{0.44em}{0ex}}$Resolving power of telescope=1

dπ=a

1.22π$=\frac{1}{d\mathrm{\pi \x9d\x9c\x83}}=\frac{a}{1.22\mathrm{\pi \x9d\x9c\x86}}$Resolving power of telescope depends upon "telescope aperture diameter" and wavelength.Higher the diameter of aperture (a), better the resolution.

Telescope resolves whereas a microscope magnifies.

Fresnel distance zp=a2

π$\phantom{\rule{0.22em}{0ex}}{z}_{p}=\frac{{a}^{2}}{\mathrm{\pi \x9d\x9c\x86}}$For distances much smaller than zp$z}_{p$ , the spreading due to diffraction is smaller compared to the size of the beam.It becomes comparable when the distance is approximately zp$z}_{p$ . For distances much greater than zp$z}_{p$ , the spreading due to diffraction dominates.

Wave SurfaceThe surface of sphere with source as center and distance traveled by light wave as radius where each wave arrives simultaneously is called wave surface.Wave Front : The locus of all the points of medium to which waves reach simultaneously so that all the points are in same phase is called wavefrontWave Normal: The perpendicular drawn to the surface of a wavefront at any point of a wavefront in the direction of propagation of light is called wave normal.Huygen's Principal1. Every point on a wavefront behaves as if it is a secondary source of light sending secondary waves in all possible direction.2. The new secondary wavelets are more effective in forward direction only.3. The resultant wavefrom in any position is given by tangents to all secondary wavelets at that instrance.Polarization is the property of wave that can oscillate with more than one orientation. A light wave that is vibrating in more than one plane is referred to as unpolarised light.Plane of vibration: The plane in which the vibrations of polarized light take place is called plane of vibration.Plane of polarisation: The plane perpendicular to the plane of vibrations in which there is no vibration of polarised light is called plane of polarisation.Polarized light: In polarized light, oscillations take place in a single direction. In polarized light, at a given time, the oscillations can take place along one direction only.Unpolarized light: In unpolarized light, oscillations can take place in any direction at a given time.Polarisation of Light: The phenomenon of the restrictions of the vibrations of the light waves in particular plane perpendicular to the direction of propagation of wave motion is called polarisation of light waves.Polarisation Angle: It s the angle at which unpolarized light is incident on transperant medium such that the light that is reflected from the surface is perfectly polarized. When unpolarised light is incident on the boundary between two transparent media, the reflected light is polarised with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other. Thus we have seen that when reflected wave is perpendicular to the refracted wave, the reflected wave is perpendicular a totally polarised wave.dots and arrowβ unpolarizeddotsβpolarised perpendicular to plane.

When the incident radiation is polarized in the plane of incidence ( the plane that includes the surface normal and incident ray) , then it is called p-polarization

When light is polarized perpendicular to the plane of incidence then it is called s-polarizedlight. It induces electronic oscillation parallel to the surface.

Unpolarized incident light can be resolved into s and p polarized components. At Brewsterβs angle only the s-polarized component is reflected. So the surface polarizes the light on reflection.

Brewester's Law: When a beam of unpolarised monochromatic light is incident on a plane glass plate, part of light is reflected while the rest is transmitted. The reflected light is partially polarised. At certain angle of incidence, the reflected ray of light is completely plane polarised.When the unpolarised monochromatic light is incident on a transparant medium at polarising angle, then, reflected light is completely polarised and reflected and refracted rays are separed by 90 degrees.The tangent of the polarization angle is equal to the refractive index of the refracting medium at which partial reflection takes place.Refractive index=π=tan ip where ip is polarising angle.$=\mathrm{\pi \x9d\x9c\x87}=tan\phantom{\rule{0.22em}{0ex}}{i}_{p}\phantom{\rule{0.66em}{0ex}}where\phantom{\rule{0.66em}{0ex}}{i}_{p}\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}polarising\phantom{\rule{0.22em}{0ex}}angle.\phantom{\rule{0.22em}{0ex}}$Frequency of light using theory of relativityaFrequency=π'=π (1Β±Vr

c

1-(Vr

c)2) where Vr=radial component of the velocity of the source relative to observer When VrβͺcFrequency=π'=π ( 1Β±Vr

c)βΉπ₯π

π=π₯π

π= Vr

c$\begin{array}{l}Frequency=\mathrm{\pi \x9d\x9c\x88}\text{'}=\mathrm{\pi \x9d\x9c\x88}\phantom{\rule{0.22em}{0ex}}\left(\frac{1{\rm B}\pm \frac{{V}_{r}}{c}}{\sqrt{1-(\frac{{V}_{r}}{c}{)}^{2}}}\phantom{\rule{0.22em}{0ex}}\right)\phantom{\rule{2.2em}{0ex}}\\ \\ \phantom{\rule{1.1em}{0ex}}where\phantom{\rule{0.44em}{0ex}}{V}_{r}=radial\phantom{\rule{0.22em}{0ex}}component\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}the\phantom{\rule{0.22em}{0ex}}velocity\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}the\phantom{\rule{0.22em}{0ex}}source\phantom{\rule{0.22em}{0ex}}\\ \phantom{\rule{5.72em}{0ex}}relative\phantom{\rule{0.22em}{0ex}}to\phantom{\rule{0.22em}{0ex}}observer\\ \\ \phantom{\rule{1.1em}{0ex}}When\phantom{\rule{0.88em}{0ex}}{V}_{r}\beta \x89\u037ac\\ \\ Frequency=\mathrm{\pi \x9d\x9c\x88}\text{'}=\mathrm{\pi \x9d\x9c\x88}\phantom{\rule{0.22em}{0ex}}(\phantom{\rule{0.22em}{0ex}}1{\rm B}\pm \frac{{V}_{r}}{c}\phantom{\rule{0.22em}{0ex}})\phantom{\rule{1.1em}{0ex}}\\ \beta \x9f\u0389\frac{\mathrm{\pi \x9d\x9b\u20af}\mathrm{\pi \x9d\x9c\x88}}{\mathrm{\pi \x9d\x9c\x88}}=\frac{\mathrm{\pi \x9d\x9b\u20af}\mathrm{\pi \x9d\x9c\x86}}{\mathrm{\pi \x9d\x9c\x86}}=\frac{\phantom{\rule{0.22em}{0ex}}{V}_{r\phantom{\rule{0.22em}{0ex}}}}{c}\phantom{\rule{0.22em}{0ex}}\\ \end{array}$Electromagnetic waves exist with an enormous range of frequencies. This continuous range of frequencies is known as the electromagnetic spectrum. PolaroidIt transmits light only in one direction of electric vector.Pass axis of polaroid: It is gate of polaroid that determines how the light will pass thru it.When the molecules of the polaoid are aligned virtically then the pass axis is horizontal.When the molecules of the polaoid are aligned horizontally then the pass axis is virtical.If an unpolarised light wave is incident on a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules this direction is known as the pass-axis of the polaroid.Malus LawWhen completely plane polarized light is incident on the analyzer, the intensity I of the light transmitted by the analyzer is directly proportional to the square of the cosine of angle between the pass axis of the analyzer and the pass axis of polarizer.I=I0cos2π $I={I}_{0}co{s}^{2}\mathrm{\pi \x9d\x9c\x83}\phantom{\rule{0.22em}{0ex}}$DichroismThe property in which doubly refracting crystals absorbs the ordinary rays (O-rays) completely and extraordinary rays whose direction is parallel to the optic axis while passing thru the crystal, is called Dichroism. Example: Tourmaline crystal, HearapathiteNote:1. The ordinary rays experiences a constant refractive index given by snell's law. 2. Extraordinary rays don't have snell 's refractive index instead have variable refractive index.3. When an unpolarized light is incident on a birefringent material it is split into two types of polarized rays one of these rays has polarization in a direction perpendicular to the optic axis (ordinary rays) and the other in the direction of the optic axis of the medium (extraordinary rays).4. Birefringent materials: Crystalline materials may have different indices of refraction associated with different crystallographic directions. A mineral crystals is that there are two distinct indices of refraction, and they are called birefringent materials.5. An optic axis of a crystal is a direction in which a ray of transmitted light suffers no birefringence (double refraction)6. For rays with any other propagation direction, there is one linear polarization that would be perpendicular to the optic axis, and a ray with that polarization is called an ordinary ray and is governed by the same refractive index.Red and Blue shift1. Due to dopler effect, when a source and observer moves away from each other then the wavelength in the middle of the visible spectrum will be shifted towards red.2. Due to dopler effect, when a source and observer approach each other then the wavelength in the middle of the visible spectrum will be shifted towards blue.Principal of SuperpositionWhen 2 or more waves overlap, the resultant displacement at any point and at any instant is equal to the vector sum of instantaneous displacements and would be produced at the point by the individual The resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves.Suppose there are two wave source A and Bay1=a1sinπt y2=a2sin(πt+π)if I1 and I2 are intensity of two interfering wave then resultant intensityIR=I1+I2+2I1 I2 cosπ $\begin{array}{l}{y}_{1}={a}_{1}sin\mathrm{\pi \x9d\x9c\x94}t\phantom{\rule{0.22em}{0ex}}\\ {y}_{2}={a}_{2}sin(\mathrm{\pi \x9d\x9c\x94}t+\mathrm{\pi \x9d\x9c\x99})\phantom{\rule{0.22em}{0ex}}\\ \phantom{\rule{0.22em}{0ex}}\\ if\phantom{\rule{0.22em}{0ex}}{I}_{1}\phantom{\rule{0.22em}{0ex}}and\phantom{\rule{0.22em}{0ex}}{I}_{2}\phantom{\rule{0.22em}{0ex}}are\phantom{\rule{0.22em}{0ex}}intensity\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}two\phantom{\rule{0.22em}{0ex}}interfering\phantom{\rule{0.22em}{0ex}}wave\phantom{\rule{0.22em}{0ex}}then\phantom{\rule{0.22em}{0ex}}resultant\phantom{\rule{0.22em}{0ex}}intensity\\ {I}_{R}={I}_{1}+{I}_{2}+2\sqrt{{I}_{1}\phantom{\rule{0.22em}{0ex}}{I}_{2}\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{0.22em}{0ex}}cos\mathrm{\pi \x9d\x9c\x99}\phantom{\rule{0.88em}{0ex}}\\ \end{array}$Interference of lightThe modification in the intensity of light produced by the superposition of two or more light waves is called interference of light.Interference due to two light sourceCondition for producing steady interference pattern1. Two sources of light must be coherent2. The sources of light must be monochromatic3. The two interfering wave must be in same state of polarization.4. The two sources of light must emit light waves of equal amplitude or intensity.5. The sources of light must be narrow.6. The separation between two light sources must be as small as possible.7. The distance from the screen from the two sources should be large.8. The two interfering wave must travel in same direction.The distance between the center of two adjacent bright or dark bands is called band width or fringe width$}$=πD

d$={\displaystyle \frac{\mathrm{\pi \x9d\x9c\x86}D}{d}}$If the slits are brought closer to each other in Young's double slit experiment then "d" decreases. So fringe width increases.for constructive interference path difference = n π$n\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9c\x86}$for destructive interference path difference = n π

2$\frac{n\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9c\x86}}{2}$Distance of nth$n}^{th$ bright band from center of interference pattern=xn=nπD

d$={x}_{n}=\frac{n\mathrm{\pi \x9d\x9c\x86}D}{d}$Distance of nth$n}^{th$ dark band from center of interference pattern=xn=(2n-1)πD

2d$={x}_{n}=(2n-1)\frac{\mathrm{\pi \x9d\x9c\x86}D}{2d}$if a thin transparent plate of thickness t and refractive index π$\mathrm{\pi \x9d\x9c\x87}$ is placed over upward slit then entire fringe is shifted upward by x0=D

d(π-1) t $x}_{0}=\frac{D}{d}(\mathrm{\pi \x9d\x9c\x87}-1)\phantom{\rule{0.22em}{0ex}}t\phantom{\rule{0.66em}{0ex}$Defraction of Light: The bending of the light near the edges of an obstacle or slit and spreading into the region of geometrical shadow is known as defraction of light.Types of defraction1. Fraunhoper diffraction2. Fresnel diffractionFresnel's diffraction: In this source of light and screen at finite distance from the obstacle. Here no lenses are used for making rays parallel. The wavefront is either spherical or cylindrical.Fresnel distance is the minimum distance which is travelled by a ray of light along linear path before diffraction.Fraunhoper diffraction: Here the source of light and screen placed infinite distance from obstacle. In this case parallel rays and plane wavefronts are produced.Diffraction pattern is obtained as a result of interaction of light coming from different parts of same wave front.Diffraction : Minimum Intensity=π=nπ

a$=\mathrm{\pi \x9d\x9c\x83}=\frac{n\mathrm{\pi \x9d\x9c\x86}}{a}\phantom{\rule{0.22em}{0ex}}$ (where a = slit width)Diffraction : Position of secondary Maximum=π=(2n+1)π

2a$=\mathrm{\pi \x9d\x9c\x83}=(2n+1)\frac{\mathrm{\pi \x9d\x9c\x86}}{2a}\phantom{\rule{0.44em}{0ex}}$Angular width of central central diffraction band =2π

a$=\frac{2\mathrm{\pi \x9d\x9c\x86}}{a}$

Q-When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?A-Wave defracted from the dge of the circular obstacle interfere constructively at the center of the shadow producing bright spot.

QAImportant component of Michelsonβs method to determine speed of light is an octagonal mirror.

Q.10.18Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?A.10.18Distance between towers=40 km = 40000 mThere is a hill between the tower.Distance between hill and tower = 40/2=20 km = 20000 mThe line joining the tower is 50 m above the hill._ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ___ _ _ _ _ | || 50 m || | || | || | || | ||____________________|____________________|Tower-1 Hill Tower-2So wave can maximum deviate by 50 m in 20000 m zp=a2

π$\phantom{\rule{0.22em}{0ex}}{z}_{p}=\frac{{a}^{2}}{\mathrm{\pi \x9d\x9c\x86}}$zp=20000 m $z}_{p}=20000\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}$a = 50 m$a\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}50\phantom{\rule{0.22em}{0ex}}m$ π=a2

zp=502

20000=0.125 m $\phantom{\rule{0.22em}{0ex}}\mathrm{\pi \x9d\x9c\x86}=\frac{{a}^{2}}{{z}_{p}}=\frac{{50}^{2}}{20000}=0.125\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}$

Exam questions

Q.1When an unpolarized light is incident on a plane glass surface, find the expression for theangle of incidence so that the reflected and refracted light rays are perpendicular to each other.What is the state of polarisation, of reflected and refracted light, under this condition?A.1If unpolarized light is incident strikes an interface so that there is a 90Β°$90}^{\mathrm{{\rm B}\xb0}$angle between the reflected and refracted rays, the reflected light will be linearly polarized. The reflected light is completely polarised. The direction of polarization (the way the electric field vectors point) of reflected light is parallel to the plane of the interface.Refracted ray is partly polarized. This is because it has more light with electric field vectors in the plane of the picture than perpendicular to it.

Q2.(a)There are two sets of apparatus of Youngβs double slit experiment. In set A, the phase difference between the two waves emanating from the slits does not change with time, whereas in set B, the phase difference between the two waves from the slits changes rapidly with time. What difference will be observed in the pattern obtained on the screen in the two set ups?(b) Deduce the expression for the resultant intensity in both the above mentioned set ups (A and B), assuming that the waves emanating from the two slits have the same amplitude A and same wavelength Ξ».A.2(a) In Set A, as the phase difference between the two waves emanating from the slits does not change with time so alternate bright and dark bands would be seen on the screen as per Youngβs double slit experiment as two slits are acting as coherent sources.In Set B, the Phase difference changes rapidly with time, the light waves coming out from two independent sources of light will not have any fixed phase relationship and would be incoherent. So when this happens, the intensities on the screen will add up and only bright light will be seen.(b)Let the displacement produced by the source S1$S}_{1$ at the point P is given byy1=a cosπt $y}_{1}=a\phantom{\rule{0.22em}{0ex}}cos\mathrm{\pi \x9d\x9c\x94}t\phantom{\rule{0.22em}{0ex}$Let the displacement produced by the source S2$S}_{2$ at the point P is given byy2=a cosπt $y}_{2}=a\phantom{\rule{0.22em}{0ex}}cos\mathrm{\pi \x9d\x9c\x94}t\phantom{\rule{0.22em}{0ex}$Let at some point G, the phase difference between 2 sources is (π)$(\mathrm{\pi \x9d\x9c\x99})$y1=a cosπt $y}_{1}=a\phantom{\rule{0.22em}{0ex}}cos\mathrm{\pi \x9d\x9c\x94}t\phantom{\rule{0.22em}{0ex}$y2=a cos(πt+π)$y}_{2}=a\phantom{\rule{0.22em}{0ex}}cos(\mathrm{\pi \x9d\x9c\x94}t+\mathrm{\pi \x9d\x9c\x99})\phantom{\rule{0.22em}{0ex}$Resultant displacement=y=y1+y1$Resultant\phantom{\rule{0.22em}{0ex}}displacement=y={y}_{1}+{y}_{1}\phantom{\rule{0.22em}{0ex}}$y=a cosπt+a cos(πt+π)$y=a\phantom{\rule{0.22em}{0ex}}cos\mathrm{\pi \x9d\x9c\x94}t+a\phantom{\rule{0.22em}{0ex}}cos(\mathrm{\pi \x9d\x9c\x94}t+\mathrm{\pi \x9d\x9c\x99})\phantom{\rule{0.22em}{0ex}}$y=2 a cosπ

2 cos(πt+π

2)$y=2\phantom{\rule{0.22em}{0ex}}a\phantom{\rule{0.22em}{0ex}}cos\frac{\mathrm{\pi \x9d\x9c\x99}}{2}\phantom{\rule{0.22em}{0ex}}cos(\mathrm{\pi \x9d\x9c\x94}t+\frac{\mathrm{\pi \x9d\x9c\x99}}{2})\phantom{\rule{0.22em}{0ex}}$So the amplitude of the resultant displacement isA=2 a cosπ

2$A=2\phantom{\rule{0.22em}{0ex}}a\phantom{\rule{0.22em}{0ex}}cos\frac{\mathrm{\pi \x9d\x9c\x99}}{2}\phantom{\rule{0.22em}{0ex}}$We know that intensity is proportional to square of amplitudeSo, I=4 I0 cos2π

2$I=4\phantom{\rule{0.22em}{0ex}}{I}_{0}\phantom{\rule{0.22em}{0ex}}co{s}^{2}\frac{\mathrm{\pi \x9d\x9c\x99}}{2}\phantom{\rule{0.44em}{0ex}}$For maximum intensity : π=0,Β±2π,Β±4π,...$\phantom{\rule{0.44em}{0ex}}\mathrm{\pi \x9d\x9c\x99}=0,\phantom{\rule{0.22em}{0ex}}{\rm B}\pm 2\mathrm{\pi \x9d\x9c\x8b},{\rm B}\pm 4\mathrm{\pi \x9d\x9c\x8b},\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.44em}{0ex}}$For zero intensity : π=Β±π,Β±3π,...$\phantom{\rule{0.44em}{0ex}}\mathrm{\pi \x9d\x9c\x99}={\rm B}\pm \mathrm{\pi \x9d\x9c\x8b},{\rm B}\pm 3\mathrm{\pi \x9d\x9c\x8b},\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.44em}{0ex}}$The two sources are coherent , then the phase difference π$\mathrm{\pi \x9d\x9c\x99}$ at any point will not change with time and we will have a stable interference pattern; i.e., the positions of maxima and minima will not change with time. If phase difference changes with time and is not constant , then the interference pattern will also change with time If the phase difference changes very rapidly with time, the positions of maxima and minima willalso vary rapidly with time and we will see a βtime-averagedβ intensity distribution. When this happens, we will observe an average intensity that will be given by<I>= 4 I0<cos2π

2>$<I>\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}4\phantom{\rule{0.22em}{0ex}}{I}_{0}<co{s}^{2}\frac{\mathrm{\pi \x9d\x9c\x99}}{2}\phantom{\rule{0.22em}{0ex}}>$βΉI=2 I0$\beta \x9f\u0389I=2\phantom{\rule{0.22em}{0ex}}{I}_{0}\phantom{\rule{0.44em}{0ex}}$So when the phase difference between the two vibrating sources changes rapidly with time, we say that the two sources are incoherent and when this happens the intensities just add up. This happens when two separate light sources illuminate a wall.