# Wave Optics Notes Class 12 | Term 2 | Physics | NCERT

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Wave Optics Notes Class 12 | Term 2 | Physics | NCERT

Nature of light: The phenomenon like interference , diffraction and polarisation establish the wave nature of light . However the phenomenon like black body radiation and photoelectric effect establish the particle nature of light. de Broglie suggested that light has dual nature i.e. it can behave as particle s well as wave .

Wavefront: The locus of all those particles which are vibrating in the same phase at any instant is called wave front. Thus, wave front is a surface having same phase of vibrating particles at any instant at every point on it.

The shape of wave front due to a
(i) Point source is spherical
(ii) Line source is cylindrical
(iii) Source at infinity is a plane.

Phase Speed: Phase speed is the speed with which wave front moves and it is equal to wave speed.

#### Spherical Wavefront

• When the source of light is a point source the wavefront formed will be spherical wavefront.
• Point source means the source of light is so small that it is considered as point. It can be considered as dimensionless.
• For example: – Ripples in water are in the form of concentric circles which are spherical wave fronts.

#### Plane Wavefront

• When the small part of a spherical or cylindrical wavefront originates from a distant source like infinity then the wavefront which is obtained is known as plane wavefront.
• For example: -Rays coming from infinity like Sun.

#### Cylindrical Wavefront

• When the source of disturbance is a slit (i.e. line source) then the wavefront is cylindrical because all the points are equidistant from the source and they lie on the surface of the cylinder.
• For example: – In the figure we can see when rays of lightfall on a lens after coming out of lens, they will converge at a point.
• The waves are bending and converging at a point so the shape of the wavefront is in the form of cylinder.
• Many concentric circles are formed and the wavefront is in the form of cylinder.

Related

#### Hugyens’ Principle of Secondary Wavelets

• Huygens principle states that each point of a wavefront is the source of secondary wavelets (small waves) which spread in all directions with the speed of the wave.
• The new wavefront is formed by drawing a line tangent to all the wavelets.

For Example

• If a stone is thrown in the river, waves will be formed surrounding that point.
• These waves look like concentric circles and they are known as wavefronts.
• When the locus of all the waves is joined which are in the same phase, it will be the same as a sphere, and are known as Primary wavefront.
• Secondary wavefront are formed from each point on the Primary wavefront.
• The common tangential line that envelopes these secondary wavefronts will further give rise to other secondarywavefronts.
• Conclusion: – According to Huygens principle, every point on a wavefront give rise to secondary wavelets which spread out in all the directions with the speed of a wave.

• Using Huygens principle we can determine the new position of the wavefront after time t.
• Mathematically: To calculate the new position of the wavefront after time t.
• Let the initial time ti =0.
• Each wave will travel =vt1where v=speed of the wave.
• Distance travelled by each wave is equal to the radius of the sphere.
• Therefore radius of sphere = v t1.
• Common tangent joining all the spheres will give the position of the new wavefront at time t=t1.
• There are 2 possibilities: – outer tangent and inner tangent.
• The amplitude of the back wave is 0. Therefore the back wave is neglected and the forward wave was considered.
• New wave front will arise from each point on the outer wavefront.
• Therefore the distance covered by these wavefronts = vt2.
• And again spheres will be obtained and by drawing common tangent will tell the position of all the new wavefront after time t2.
• Again back wave is neglected and forward wavefront is considered.
• This shows the wavefront keep on spreading with time.

Effect of frequency, wavelength and speed during refraction

When a light wave travels from one medium to another , its frequency remains unchanged but both its wavelength and speed get changed , depending on the refractive index of the refracting medium.

Laws of reflection on the basis of Huygens’ wave theory

As shown in figure, consider a plane wave front AB incident on the reflecting surface XY, both the wave front and the reflecting surface being perpendicular to the plane of paper.

First the wave front touches the reflecting surface at B and then at the successive points towards C. In accordance with Huygens’ principle, from each point on BC, secondary wavelets start growing with the speed c. During the time the disturbance from A reaches the point C the secondary wavelets from B must have spread over a hemisphere of radius BD = AC = ct, where t is the time taken by the disturbance to travel from A to C. The tangent plane CD drawn from the point C over this hemisphere of radius ct will be the new reflected wave front.

Let angles of incidence and reflection be i and r, respectively . In AABC and ADCB, we have

i.e. the angle of incidence is equal to the angle of reflection. The proves the first law of reflection.
Further, since the incident ray SB, the normal BN and the reflected ray BD are respectively perpendicular to the incident wave front AB, the reflecting surface XY and the reflected wave front CD (all of which are perpendicular to the plane of the paper) therefore, they all lie in the plane of the paper i.e. in the same plane. This proves the second law of reflection.

Law of refraction on this basis of Huygens’ wave Theory

Consider a plane wavefront AB incident on a plane surface XY, separating two media 1 and 2, as shown in Figure.
Let v1 and v2 be the velocities of light in two media, with v1 <v2.

The wave front first strikes at point A and then at the successive points towards C. According to Huygens’ principle, from each point on AC, the secondary wavelets starts growing in the second medium with speed v2. Let the disturbance take time t to travel from B to C, then BC = v1t. During the time the disturbance from B reaches the point C, the secondary wavelets from point A must have spread over a hemisphere of radius AD = v2t in the second medium. The tangent plane CD drawn from point C over this hemisphere of radius v2t will be the new refracted wave front.
Let the angles of incidence and refraction be i and r, respectively.
From right AABC, we have

This proves Snell’s law of refraction. The constant 1μ2 is called the refractive index of the second medium with respect to first medium.
Further, since the incident ray SA, the normal AN and the refracted ray AD are respectively perpendicular to the incident wave front AB, the dividing surface XY and the refracted wave front CD (all perpendicular to the plane of the paper), therefore, they all lie in the plane of the paper, i.e. in the same plane. This proves another law of refraction.

Young’s double slit experiment (YDSE)

The Young’s experiment shows that matter and energy can display both wave and particle characteristics.

Purpose of double slit experiment is as follows:-

1. In order to prove the wave nature of light.
2. To explain the phenomenon of interference.

Two coherent sources of light were taken in order to maintain the 0 or constant phase difference between the sources of light.

• Experimental set-up1
• Young took an ordinary source of light(S) such as light bulb.
• The light was made to pass through a very small slit S (which was comparable with the wavelength of light).
• The light coming from Source S was made to pass through two small slits S1 and S2 which were separated by a very small distance d.
• One screen was kept in front of these 2 sources.

Observation1:

• He observed alternate dark and light bands were formed on the screen.

Setup 2:

• Now he took 2 light bulbs i.e. 2 non coherent sources of light and placed a screen in front of them.

Observation 2:

• He observed there were no alternate bands of light formed on the screen.
• Conclusion:-
• When coherent sources of light were taken then the phenomenon of interference is taking place.
• When non-coherent sources were taken phenomenon of interference was not taking place.
• The source S illuminated the sources S1 and S2 as a result the light from S1 and S2 become coherent.
• S was the source of bothS1 and S2, therefore if there is any change in the phase of the source there will be change in the both sources also.
• Therefore both Sand S2 will be always in phase with each other.

Why alternate sources of light bands were seen

• When the ordinary light source was made to pass through small slit then the wavefront of semicircle shape will be formed.
• Wavefronts are in semicircle shape because of obstacles on both the sides.
• In the figure red bands represents crests and yellow represents troughs.
• When these wavefronts passes through the 2 small slits again the wavefronts will arise.
• There will be points where red and yellow will also overlap with each other.
• Red + Yellow and Yellow + Red à  Destructive interference à Intensity low
• Yellow + Yellow and Red + Red à Constructive interference à Intensity high
• Because of constructive and destructive interference there are alternate dark and light bands seen on the screen.

Conclusion

• When the light wave is originating from the 2 coherent sources they overlap with each other both constructively and destructively.
• As a result alternate light and dark bands of light is shown on the screen.
• This phenomenon of overlapping of light waves giving rise to regions of higher amplitude and regions of lower amplitudeis known as interference.
• Interference is a property related to the wave nature of the light.
• Young’s Double slit experiment proved that the light is of wave nature which was proposed by Huygens principle.

• Important results drawn from the above experiment:-
• Bands were formed as a result of interference and interference was due to the overlapping (either constructive or destructive) of waves.
• Constructive and destructive overlapping depends on the path difference.
• If the path difference is integral multiple of λ = constructive interference.
• If path difference is non integral multiple of λ =destructive interference.

Calculation of path difference

• Let S1 and Sare the sources and consider a point P where we have to calculate the intensity.
• Path difference = S2P-S1P
• Using Pythagoras theorem :- (S1P) 2 = D2 + (x-(d/2)) 2 equation(1)
• Where D=distance of the screen from 2 slits, x= position of the point where intensity has to be found.
• Similarly (S2P)2 = D2+ (x+(d/2)) 2 equation(2)
• Where d=distance between the 2 small slits.
• Subtracting (2) and (1) => (S2P)2 – (S1P) 2 =(x+(d/2)) 2 – (x-(d/2)) 2
• On simplifying (S2P)2 – (S1P) 2 =2xd
• => (S2P+S1P)(S2P- S1P)=2xd
• =>(S2P- S1P) =(2xd)/(S2P+S1P)
• If d<<D ; x<<D(screen is placed quite far away from the slit arrangement)
• => (S2P+S1P) (are almost same as D).
• => S2P+S1P =2D
• Therefore (S2P- S1P) = (2xd)/(2D)

Path difference =(xd)/(D)

Equation(2)

Observation:-

• Light from coherent sources produced alternate dark and bright bands on the screen placed some distance away from it.

Fringe Pattern

• The alternate dark and red bands which are obtained on the screen are known as fringe pattern and the alternate dark and bright bands are known as fringes.

Bright Bands:-

• Bright bands are formed as a result of constructive interference and they are the positions of maximum intensity.
• Condition for maximum intensity:-
• Path difference =n λ.
• =>(xd/D) = nλ using Equation(2)
• =>xn = ((n λ D)/d) where xn=position of nth bright band.
• When n=0 then it will be central bright band.

Dark Bands:-

• Dark bands are formed by the destructive interference and they are the positions of minimum intensity.
• Condition for destructive interference:-
• Path difference =(n+(1/2))λ
• => (xd/D) = (n+ (1/2)) λ.
• =>xn=(n+ (1/2)) (λD/d)
• Where xn=position of nth dark band.

Graphical representation of fringe pattern

Fringe width:

1. Fringe width is the distance between consecutive dark and bright fringes.
2. It is denoted by ‘β’.
3. In case of constructive interference fringe width remains constant throughout.
4. It is also known as linear fringe width.

Angular Fringe width:

1. It is the angle subtended by a dark or bright fringe at the centre of the 2 slits.
2. It is denoted by ‘θ’.
• Mathematical Expression for fringe width(β):-
• xn =((nλD)/d)
• xn+1 =(((n+1)λD)/d)
• β =xn+1 – xn
• =(((n+1) λD)/d) – ((nλD)/d)
• =>β = (λD/d)
• Therefore fringe width depends on:-
• (λ)Wavelength of the light used, (D) distance of the screen from the slits and (d) distance between two slits.
• Mathematical Expression forangular fringe width(θ):-
• θ =(β/D)
• θ = (λD)/(d D)
• θ = (λ/d)

#### Conclusion of Young’s double slit experiment

1. Central fringes gets shifted by – θ if the source gets shifted by θ.
1. If the source S is shifted by some angle θ, there will be no change in fringe pattern. The central fringe will get shifted in the opposite direction.
2. Intensity of the fringes increase if point sources are replaced with slits.
1. If there are slits instead of point source then more light waves will be able to pass through the slits.
2. As a result stronger wavefronts are formed which give rise to even greater intensity fringes.

In interference, maxima is a point where two crests or two troughs of two different waves meet each other and as a result, reinforce each other. On the other hand, minima in interference  is a point where a crest and a trough meet together cancelling out each other.

Coherent & Incoherent Sources

• Coherent sources are those sources of light which emit waves that have same frequency and zero or constant phase difference.
• Suppose if there are 2 sources S1 and S2 are coherent sources if there frequencies are same and also phase difference between them is either 0 or constant.
• Phase difference should not change.

• For example:-Laser light.
• Incoherent sources are those which sources of light which emit waves that have random frequencies and phase differences.
• There is no relationship between the waves in terms of frequencies and phase difference.
• For example:-Electric bulb, night lamp.

Coherent Source

Incoherent Sources

Conditions for sustained interference pattern

1. The two sources of light must be coherent with same frequency or wavelength and with same phase or constant phase difference.
2. The two sources should preferably have the same amplitude.
3. The two sources must be very close to each other.
4. The sources should emit light waves continuously.

Fresnel’s biprism experiment

Fresnel’s biprism experiment is a variation of Young’s double slit experiment. In Fresnel’s biprism experiment, two prisms which are connected through their bases are used. When monochromatic light of a particular wavelength is allowed to fall on this setup, an interference pattern involving bright and dark fringes is observed on a screen kept on the other side of the biprism. It is seen that the interference pattern is concentrated at the centre of the screen and appears to fall from two virtual sources of monochromatic light behind the prism. Thus, this experiment is similar to Young’s double slit experiment, which uses two light sources.

#### Diffraction

• Diffraction is the phenomenon by virtue of which light bends while passing through as lit or an opening.
• The extent of bending depends upon the diameter of the slit.
• Both Interference and Diffraction are closely related to each other.
• Young replaced the two slits by a single slit in his single slit experiment. Therefore this experiment is also referred as Young’s single slit experiment.
• When the light passed through one slit a different type of pattern was observed on the screen.
• The pattern which was observed had a central maximum band and which was very wide as compared to interference pattern.
• There were alternate dark and bright bands and their intensity was decreasing on both the sides.
• The central maxima, was very wide whereas corresponding secondary maxima and minima were reduced in the intensity.
• The change in the pattern was formed due to diffraction instead of interference.

Conclusion:

2. Alternate dark and bright bands on either side.
3. Intensity was decreasing on both sides.

Diffraction Fringe Pattern

• In diffraction there is an incoming wave which passes through a single slit and as a result diffraction pattern was obtained on the screen.

Diffraction pattern Maxima

• The incident wavefront is parallel to the plane of the slit. This shows they are in phase with each other.
• From the given figure:-
• Path difference =NP – LP
• =>NQ = a sinθ where a=width of the slit.
• The slit was divided into smaller parts M1, M2, N and L, and when contribution from each part is added up.
• At the central point θ =0. This implies path difference =0.
• This means all the parts of the slit will contribute completely. Therefore the intensity is the maximum.

Central Maximum occurs at θ =0.

#### Interference and Diffraction

 Interference Diffraction It is the superposition of two waves originating from the narrow slits. It is the superposition of a continuous family of waves originating from each point on a single slit. Pattern has a number of equally spaced bright and dark bands. The central bright maximum is twice as wide as the other maxima. There are 2 point sources. They cannot be split into smaller sources. There is only one source. This source can be split into ‘n’ number of sources.

Interference & Diffraction: Conservation of energy

• Interference and diffraction phenomenon both obey the principle of conservation of energy.
• Light energy is neither lost nor gained. Light energy is redistributed.
• Total energy is conserved.

Resolving Power of Optical instruments

• Resolving power is defined as the inverse of the distance between two objects which can be just resolved when viewed through the optical instrument.
• Optical instruments should be able to display two separate images as distinct two separate images.
• Suppose if we are seeing 2 stars in the sky using telescope, and they are very closely placed to each other, they should not be displaced as one star.
• Instead they should be displaced as two separate stars. The ability to see them as separate 2 stars is known as resolving power.
• For example: –
• Consider 2 rooms one is lit and another is dark.The room which is lit we are able to see everything clearly whereas in second room we are not able to see properly.
• This is because the resolving power with which we are viewing second room is not good.
• If the resolving power is better the image will be clearer.So everything can be seen clearly.
 Resolution Magnification It is possible to view 2 closely spaced objects clearly. It is to magnify the size of objects. It is the distinctness of 2 different objects. It is to enlarge two different objects.

Note:

• Whenever magnification increases the resolution decreases.
• Resolution à Clarity or distinctness of two different objects.
• Magnification à Enlargement of the size of an object.

Resolving Power of Telescope

Telescopes subtend a very small angle for viewing objects like binary stars, individual stars, distant galaxies, quasars, and planets, making it very easy to study and view them. Large apertures are needed to resolve the power of a telescope. We can use Rayleigh’s formula to evaluate the resolving power of the telescope formula.

Based on the Rayleigh’s formula the angular separation between two distant objects should be

Resolving Power = D/d = a/1.22λ

Where,

a = width of the rectangular slit

D = distance of objects of the telescope.

If the diameter d is greater, the resolution will be better. The astronomical optical telescopes usually have a mirror of large diameters, such as 10m, to get the desired resolution. Large wavelengths help in reducing the resolving power of the telescope, and that’s why the radio and microwave telescopes require larger mirrors.

Resolving Power of Microscope:

When it comes to microscopes, the resolving power is inversely proportional to the distance between the two objects. A microscope can be resolved, and this is done by using Abbe’s criterion, which was given by Ernst Abbe in the year 1873.

It states that:

The resolution R depends on the angular aperture (here the resolution is measured in terms of distance, and is not the angular resolution which is considered in the previous part).

The resolving power of microscope formula is given by:

R =

1.22λNAcondenser+NAObjective1.22λNAcondenser+NAObjective

Where NA = n sinθ

Where,

NA = the numerical aperture,

θ = half of the angle α of the lens, which depends on the focal length and the diameter of the lens,

n = refractive index of the medium between the specimen and the lens, and

λ = wavelength of light that is emerging from the object under consideration.

The diffraction of the aperture restricts the resolving power of the light microscope. An optical system cannot be able to form a perfect image of a point due to diffraction. For a resolution to occur, the first-order diffracted beam & direct beam must be collected as an objective of the microscope.

Validity of Wave optics

• Consider a single slit of an aperture of thickness ‘a’.Diffraction pattern will be observed.
• There will be centralmaxima due to diffraction.
• Angular size of central maximum = (λ/a)
• where θ (condition to have central maximum)=(λ/a)
• The distance travelled by the light waves for the spread to occur = ‘z’(the angular spread due to diffraction)
• The spread is given as = (z λ/a)
• Where (z λ/a) = a where a =width of the diffracted beam.
• z ≈ (a2/ λ).This distance is known as Fresnel distance.
• Fresnel distance describes the distance at which spread due to diffraction becomes comparable to the width of the slit or not.
• This is the boundary of ray optics and wave optics.
• For distances << zf :
• Spreading due to diffraction is smaller compared to size of the beam then the ray optics is valid.
• For distances >> zf :
• Spreading due to the diffraction dominates over ray optics then the wave optics is valid.

Problem: Estimate the distance for which ray optics is good approximation for an aperture of 4 mmand wavelength 400 nm.

Answer: Fresnel’s distance (ZF) is the distance for which the ray optics is a good approximation. Itis given by the relation,

ZF = (a2/λ)

Where,

Aperture width, a = 4 mm = 4 ×10−3 m

Wavelength of light, λ = 400 nm = 400 × 10−9 m

ZF = (4×10-3)2/ (400×10-9)

Therefore, the distance for which the ray optics is a good approximation is 40 m.

Polarizing and unpolarized light wave

• Polarized wave: – Polarized wave is defined as the vibration of the wave confined to specific plane.

1. In case of polarized wave the vibrations takes place in one plane not in multiple planes and it is perpendicular to the direction of propagation.

For example:

1. Vibrations of a rubber. It will vibrate randomly in all different directions. But when it is passed through a small slit then it will vibrate only in one plane.

• Unpolarized wave: Unpolarized wave is defined as the plane of vibration which changes randomly in short intervals of time.
• In case of unpolarized wave the vibrations takes place in multiple planes and it is perpendicular to the direction of propagation.

#### What is Polarization?

Polarization is the process of transforming unpolarized light into polarized light.

Methods of polarization:

1. Polarization by Polaroid’s
2. Polarization by Scattering
3. Polarization by Reflection
4. Polarization by Refraction

For example: – 3-D movies, sun glasses, photographic cameras.

Polarization: Transverse vs. Longitudinal waves

• In case of transverse waves there are many different ways in which particles can oscillate.
• Polarization is observed in transverse waves.
• In case of longitudinal waves there is only one direction in which particles oscillate.
• Therefore there is no polarized or unpolarized light in case of longitudinal wave.

Polarization by Polaroid’s

• Polaroids are polarizing materials consisting of long chain of molecules aligned in a particular direction.
• Polaroid’s are in the form of thin sheets.
• Every Polaroid has pass axis. Pass axis is like a gate of aPolaroid which determines how the light will pass through it.
• There is horizontal as well as vertical pass axis.
• Unpolarized light on passing through a Polaroidit gets polarized.

Experiment 1: Polarization by a single polaroid (P1)

• When a beam of light passes through Polaroid was a polarized light.

Observations:

• The intensity was almost halved. As half of light rays were not allowed to enter only few were allowed to pass through it.
• Rotation of thePolaroid P1 has no effect on the intensity of transmitted light.
• There will be many different orientations but all are present in equal amount.
• Whatever is the direction of the Polaroidbut always the intensity of the lightcoming out of the Polaroid will be same.
• Sometimes it will allow vertically polarized light to pass and sometimes horizontal.

Experiment 2:-Polarization by a two polaroid (P1,P2)

• One more Polaroid(P2) was introduced in between the unpolarized light and Polaroid(P1).

Observations:

• Intensity of the light is almost halved after P2.
• When the polarized light enters the second Polaroid (P1) then
• If the pass axis of the polarizer P1 is along the same direction as P2 then the light will come out of the Polaroid P1.
• Intensity will remain same.
• Rotation of the Polaroid P1 has effects on intensity of transmitted polarized light.
• When the two axes are not oriented in the same direction i.e. if the axis of the P1 is perpendicular to the axis of the P2.
• Then P1 won’t allow any light to pass through it as a result net intensity will be 0.
• It concludes that :-
1. Transmitted intensity=0 when pass axis of P1 is perpendicular to P2.
2. Transmitted intensity=maximum when pass axis of P1 is parallel to P2.

Polaroid Experiment: Conclusion

1. Intensity of the polarized light varies with the angle between the pass axes of the two Polaroid.
2. I =0 whenθ =900.
3. I =max when θ=00.
4. As the angle increases from 00 to 900 the intensity keeps on decreasing.

• Malus’ Law: Malus’ law explained when the angle between the pass axes be some arbitrary angle θ.
• Experimentally it was found that intensity varies as cos2θ ;
• I=I0 cos2θ.
• Where θ =angle between the pass axes of two Polaroids, I0=Intensity of polarized light after passing through P1 and I =new intensity after rotation.

Polaroids: Applications

1. Sunglasses:
1. Sunglasses help to reduce the intensity of the sun light.
2. Sunlight is unpolarized light so when it passes through the sun glass it becomes polarized.As a result intensity of sunlight is greatly reduced.
2. Photographic cameras:
1. In this type of cameras Polaroids are attached.
3. Car sun film:
1. Thin film reduces the intensity of the sunlight.
4. 3D glasses:
1. By using 3D glasses objects can be seen in 3 dimensional.
2. There are 2 images slightly displaced with each other are formed by our 2 eyes.
3. These images are then interpreted by the brain.In 3D glasses Polaroids are used.
4. One glass will polarize in such a way that image is formed in particular orientation and other glass will polarize the image in another orientation.
5. As a result these two orientations will be interpreted by the brain and image is formed.
 Unpolarized Polarized Partially polarized Vibrations takeplace in any possible random plane. Vibrations takeplace in one single specific plane. It will have some polarized as well as some unpolarized components. It vibrates in vertical plane but sometimes it vibrates in horizontal plane.

Polarization by Reflection and Refraction

• When an unpolarized beam of light falls on boundary, reflected wave and refracted wave are partially polarized.
• An incident wave is an unpolarized light .Some of the rays will get reflected and some gets refracted.
• Most of the light waves that reflected were polarized waves, which were parallel to the plane.
• Most of the rays along the refracted ray were unpolarized waves with one or two polarized components.
• Refracted Ray à More unpolarized and Reflected rayà more polarized.
• If the angle of incidence varies more and more unpolarized light was able to pass through the surface.
• At one particular value of angle of incidence there was maximumpolarization in the reflected ray.
• Above and below the value of angle of incidence, both reflected and refracted rays were partially polarized.

Brewster’s law:- Brewster’s law states that at any particular angle of incidence, reflected ray is completely polarized; and the angle between reflected and refracted ray is 900.

• At i = iB; where iB = Brewster’s angle of incidence.
• From Snell’s law :- (sin i/sin r)= μ
• => sin iB /(sin (π/2 -iB))
• =>(sin iB /cos iB)=μ
• tan iB = μ

Related

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