Coherent and Incoherent Addition of Waves

You must have seen light coming out from the laser. Let us carry out a

small activity. Take two needles and touch the needles on the surface

of the water. Here if both the needles move with the same speed then

they are said to be coherent. Let us learn more about coherent waves.

Coherent and Incoherent Addition of waves

Suppose there is a surface of the water and you take a needle and

touch the surface of the water. What will happen? Yes, ripples are

formed. Now if you take two needles and you touch the surface of the

water with the needles. What do you think will happen?

You will see a pattern. That pattern is the interference pattern. When

you touch both the needles at the surface of the water at the same time,

both the needles are in the same phase. Needle 1 will produce a wave.

Also, needle 2 will produce its own ripples and they will intersect with

waves of the first needle.

Now, if both the needles are moving with the same velocity, the wave

formed here are coherent. If the velocity of a 1st needle and 2nd

needle are not steady they won’t intersect. This is because one is at a

steady speed and other is at variable speed.

Coherent Waves

If the potential difference between two waves is zero or is constant

w.r.t time, then the two ways are said to be coherent.

Non-coherent Waves

The waves are non-coherent if the potential difference between the

two ways keeps on changing. Lightbulb, study lamp are the examples

of the coherent waves. They emit waves at random potential

difference.

Video on Wave Optics

Explanation

Now let us consider there are two needles say S1 and S2 moving up

and down on the surface of the water and are pointing at point P. So

the path difference here is given as S1P – S2P. Now the displacement

by two needles and S1 S2 are:

y1 = A cos wt ……………… (1)

y2 = A cos wt …………….. (2)

So the resultant displacement at point P is, y = y1 + y2. When we

substitute the value of y1 and y2 we write,

y = A cos wt + A cos wt

y = 2A cos wt……………….. (3)

Now, we know the intensity is proportional to the square of the amplitude waves.

I0

∝

A²

Where I0 is the initial intensity and A² is the amplitude of the wave.

From equation 3, we say that A = 2A. So,

I0

∝

(2A)² or I0

∝

4 A²

I = 4 I0

Now, if two needles that are S1 are S2 are in the same phase, the

potential difference is,

S1P – S2P = nλ

Where n = 0, 1, 2,3 ……… and λ = the wavelength of the wave. If the

two needles S1 and S2 are vibrating at its destructive interference

then, the potential difference is

S1P – S2P = (n + 1/2) λ

Now if the potential difference of the waves is Φ then,

y1 = α cos wt

y2 = α cos wt

The individual intensity of each wave is I0 , we get,

y = y1 + y2

= α cos wt + α cos (wt +Φ)

y = 2 α cos(Φ/2) cos (wt + Φ/2)

Since, the intensity is I0

∝

A²

I0

∝

4α² cos² (Φ/2)

I = 4 I0 cos² (Φ/2)

Well, the time-averaged value of cos²(Φt/2) is 1/2. So, the resultant

intensity will be I = 2 I0 at all the points.

Solved Questions For You

Q. Two coherent sources of light can be obtained by

A. Two different lamps

B. Different lamps having the same power

C. Two different lamps of the same power and having the same

color

D. None of the above

Answer: D. The coherent source cannot be obtained from two

different light

Diffraction

Suppose there is a dark room, a completely dark room and through the

window, there is a small hole. When light enters through that tiny

hole, what happens? We see that through the small hole light enters

but instead of just bright light, we see a region of light and dark bands.

This is nothing but the diffraction of light. Let us study diffraction in

detail.

Diffraction

Every one of us knows what diffraction is. It is the bending of light

around the corner of an obstacle. Reflected light produces fridges of

light, dark or colored bands. At times diffraction of sunlight in clouds

produces a multitude of colors. Example of diffraction in nature is

diamond rays in the solar eclipse.

Types of Diffraction

There are two types of diffractions

● Fresnel Diffraction

● Fraunhofer Diffraction

From the above figure, we observe that the source is located at a finite

distance from the slit, and the screen is also at a finite distance from

the slit. The source and the screen are not very far from each other. So

this is a Fresnel diffraction. Here, if suppose the ray of light comes

exactly at the edge of the obstacles, the path of the light is changed. So

the light bends a little and meets the screen.

A beam of width α travels a distance of α2/λ , called the Fresnel

distance before it starts to spread out due to diffraction. But when the

source and the screen are far away from each other, and when the

source is located at the infinite position, then the ray of light coming

from that infinite source are parallel rays of light. So this is Fraunhofer

diffraction.

Here we have to make use of the lens. But why do we use the lens?

Because in Fraunhofer diffraction, the source is at infinity so the rays

of light which pass through the slit are parallel rays of light.

So in order to make these rays parallel to focus on the screen, we,

make use of the converging lens. The zone which we get in front of

the slit is the central maxima. On either side of central maxima, there

is bright zone i.e 1st maxima.

Fresnel Diffraction Fraunhofer Diffraction

Here the wavefront used is spherical. Here the wavefront used is plane.

An image is formed at a finite distance.

An image is formed at an infinite distance.

A lens is not required. A lens is required.

Questions For You

Q1. In a single slit diffraction λ= 500mmwith and a lens of diameter

0.01mm, width of central maxima, obtain on screen at a distance of 1

m will be

A. 5mm

B. 1mm

C. 10mm

D. 2.5mm

Answer: C. The angle subtended by two minima in the slit = α = 2λ/w,

where w is the slit width. Here the lens diameter would act as slit

width. The width of central maxima is the distance between the two

minima = dα, where d is the distance between slit and screen = 1m.

Thus the width of central maxima = 10mm

Q2. Yellow light is used in single slit experiment with slit width 0.6

mm. If the yellow light is replaced by X-rays, then the pattern will

reveal

A. no pattern

B. that the central maxima narrower

C. less number of fringes

D. more number of fringes

Answer: D. A wavelength of X rays ranges from 0.1 to 10mm and

yellow light is the range 570 to 590mm. Fringes are formed at nλD/d.

Hence the fringes of x-rays are formed close to each other.

Huygen’s Principle

When you open your window in a room, the light enters through the

window and spreads throughout the room. Do you why does this

happen? This is because light has got some wave nature, that spreads

in the room in all the directions. To understand this in a better way let

us study the Huygen’s Principle.

Huygen’s Principle

It states that each point of the wavefront is the source of the secondary

wavelets which spread out in all direction with the speed of a wave.

So if we consider a point source, it will emit its wavefront and nature

of the wavefront will be spherical one.

As per the Huygen’s principle, all the points on the wavefront are

going to become a secondary source. So the wavefronts will in the

forward direction. All the secondary sources emit wavelets. Tangent

drawn to all the wavelets is the new position of the waveform.

This means that, suppose you are standing on the mountain and you

throw a stone in the water from a height. What do you observe? You

see that the stone strikes the surface of the water and waves are seen

surrounding that point. Every point on the surface of water starts

oscillating.

The waves spread in all the direction. Earlier the water was at rest. But

the moment we throw the stone in the water, within a few fractions of

seconds the disturbance spreads in all directions. There are ripples

formed in the water. The ripples form the concentric circle around the

disturbance and spread out.

These ripples are nothing but the wavefront. The wavefronts gradually

spread in all the directions. So at every point, we have a wave coming

out. The primary wavefront is formed and again from the primary

wavefront, a secondary waveform is formed and so on. The

disturbance does not last for a long time. It fades gradually because

more and more waveforms are formed.

Solved Questions For You

Q1. Ray optics is valid when characteristic dimensions are:

A. of the same order as the wavelength of light.

B. much smaller than the wavelength of light.

C. much larger than the wavelength of light.

D. of the order 1mm

Answer: C. Ray optics is valid when characteristic dimensions are

much larger than the wavelength of light so that the rectilinear

property of light can be used.

Q2. According to Huygen’s Principle, the ether medium pervading

entire universe is:

A. Less elastic and denser

B. Highly elastic and less dense

C. Not elastic

D. Much heavier

Ans: B. Huygen considered light needs a medium to propagate called

either as highly elastic and less dense.

Q3. Huygens’s concept of secondary wave

A. allow us to find the focal length of the thick lens

B. it is a geometrical method to find the wavelength

C. used to determine a velocity of light

D. it is used to explain polarisation

Ans: B. Huygens’s concept of secondary wave is a geometrical

method to find the wavelength.

Huygen’s principle states that every point on the wavefront may be

considered a source of secondary spherical wavelets which spread out

in the forward direction at the speed of light. The new wavefront is the

tangential surface to all these secondary wavelets. Thus, it is a

geometrical method to find the wavelength.

Interference of Light Waves and Young’s Experiment

Suppose you are busy studying in your room and one of your friends

calls you out for cricket. So what happens here is that he has interfered

you with what you were doing. So that was kind of disturbance for

you. Let us now study about something called as an interference of

light waves.

Constructive and Destructive Interference of Light Waves

We know that there are two kinds of interference of light waves,

which are:

● Constructive Interference: Suppose if the crest of one wave

falls on the crest of another wave, then the amplitude is

maximum. This is constructive interference. Here both the

waves have the same displacement and the waves are in phase.

● Destructive interference: Suppose if the crest of one wave falls

on the trough of another wave, then the amplitude here is

minimum. This is destructive interference. Here the waves do

not have the same displacement and the waves are out of phase.

Condition of a Steady Interference Pattern

i. A1 = A2 . The amplitude of two waves must be equal.

ii. λ1 = λ2. The two waves interfering must have same color i.e

they must be of the same wavelength.

iii.Sources must be narrow.

iv.The distance between source should be less.

v. Source and screen should be at large distance.

vi.We should get coherent sources.

Path difference for constructive and destructive interference

Suppose there are two coherent sources S1 and S2. There is also a

point source P. The point source P is located at the same distance from

the sources S1 and S2. When both the sources are in the same phase,

the constructive path difference will be 0, λ, 2λ……. The destructive

path difference will be λ/2, 3λ/2, 5λ/2……

Young’s experiment

Young’s performed an experiment to prove the wave nature of light by

explaining the phenomenon of interference. He used two coherent

sources to perform in this experiment. He used a light bulb and two

small slits, S1 and S2 and source S.

Here in the above figure, we can see that the slits are placed very close

to each other and are separated by the distance ‘ d ‘. There is a screen

placed in front of this setup. He observed that alternate dark and light

bands were formed on the screen. Why did this kind of pattern come

on the screen?

The source S illuminate the source S1 and S2 .Therefore the light from

S1 and S2 becomes coherent. Why did they become coherent? This is

because both S1 and S2 receive their light from the same source S. So

if there is any change in the phase, the change will reflect in both S1

and S2. When both the slits are open fringes are formed.

Suppose you have two taps in your house and water that comes in both

the taps is coming from both the source. So if you put some mud at the

source of the water, the moment you open the taps, you will see that

water from both the taps will be muddy. Whatever changes you make

in the source, the same changes are seen in the taps too.

∴ we can say that S1 and S2 will always remain in phase.

In the double-slit experiment consecutive bright as well as dark

fringes are seen on the screen as a consequence of the type of

interference of light waves.

● The interference fringe maxima occur for path difference = n λ

● The interference fringe minima occur for path difference =

(2n+1) λ

Solved Question For You

Q. A YDSE uses a monochromatic source. The shape of the fringe

formed on the screen is

A. Hyperbola

B. Circle

C. Straight line

D. Parabola

Answer: C. If the light consisted of ordinary particles, and these

particles were fired in the straight line through the slit and allowed to

strike a screen, we would expect to see a pattern of the size and shape

of the slit. Here the shape of the fringe forms on the screen is the

straight line.

Polarisation

Suppose on a hot sunny day you step out of your house, you

definitely use sunglasses. Also, everyone must have watched 3D

movies. Here, polarisation plays a very important role. Let us see what

polarisation is and study them in detail.

Polarisation

Polarisation is nothing but transforming unpolarised light into

polarised light. Unpolarised light is the light in which particles vibrate

in all different planes.

( Source: Wikipedia )

Ways of Polarising the Light

Polarisation by Polaroids

( Source: s-cool )

In the above figure, we see that there is plane of vibration parrel to the

plane and there is a plane of vibration perpendicular to the plane. The

first image is of unpolarised one. The second image is polarised which

is either perpendicular or parallel. So let us start understanding

polarisation by polaroids. Polaroids are the polarising materials

consisting molecules aligned in particular direction.

Every Polaroid has a pass axis. It will allow light to pass only through

the pass axis. A polaroid can have horizontal pass axis as well as

vertical pass axis. These determine how the light will pass through it.

So when an unpolarized light passes through a polaroid, it gets

polarised.

Polarisation by Scattering

Suppose we are having a molecule or an atom when the light gets

incident on this particle light energy will be absorbed by the particles

and will be re-emitted by it in different directions. This is scattering by

polarisation. The emitted light travels in different directions.

Suppose if we have an unpolarized light and that light is incident on a

particle or element or molecule, we will get the scattered light. The

scattered light will be present everywhere in all the directions. So

when the unpolarized light passes through the molecule, it is observed

that the light is polarised in the direction perpendicular to the incident

ray, along these directions we get polarised light.

This is how we get polarisation by scattering of light. The scattered

light in a direction perpendicular to the direction of the incident ray is

completely polarised whereas light passing through molecules is

partially polarised.

Polarisation by Reflection and Refraction

( Source: Wikipedia )

In the above figure, we can see the incident ray, reflected ray and the

refracted ray. On the incident, we see the unpolarized light. The

unpolarized light is denoted as shown in the above figure, where dot

represents perpendicular directions and lines indicate the parallel

direction.

It is observed that most of the light in the reflected ray is polarised

parallel to the plane with a very few unpolarised ones. Whereas in the

refracted ray, most of the light is unpolarised with one or two

polarised components. Thus we see that both the reflected and the

refracted ray are partially polarised.

Brewster’s law

The law states that at a particular angle of incidence, the reflected ray

is completely polarised and the angle between the reflected ray and

refracted ray is 90°. At i = iB, when the angle of incidence is equal to

Brewster’s angle then, total angle = 90°

By Snell’s law, we say that

Sini

Sinr

= μ

∠iB + r = 90°

r = 90°- iB

⇒ \( \frac{Sin iB}{cos iB} \) = μ

⇒ tan iB = μ

Solved Questions For You

Q1. A ray of light is incident on the surface of the plate of glass of

refractive index 1.5 at the polarising angle. The angle of refraction of

the ray will be.

A. 33.7°

B. 43.7°

C. 23.7°

D. 53.7°

Answer: A. If iP be the polarising angle, then, tan iP = μ = 1.5 or iP =

56.3°

Angle of refraction, r = 90° – iP = 90° – 56.3° = 33. 7°

Q2. In case of linearly polarised light, the magnitude of the electric

field vector.

A. Varies periodically with time

B. Does not change with time

C. Increases or decreases with linearly with time

D. Is parallel to the direction of propagation

Answer: A. In any type of light whether polarised or unpolarized, the

magnitude of electric field vector always varies periodically with time.

Refraction and Reflection of Waves Using Huygen’s Principle

As we know that when light falls on an object, it bends and move

through the material, this is what refraction is. Also when the light

bounces off the medium it is called a reflection. Let us know study

reflection and refraction of waves by Huygen’s principle.

Reflection using Huygens Principle

We can see a ray of light is incident on this surface and another ray

which is parallel to this ray is also incident on this surface. Plane AB

is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays

are incident from the surface, so we call it incident ray. If we draw a

perpendicular from point ‘A’ to this ray of light, Point A, and point B

will have a line joining them and this is called as wavefront and this

wavefront is incident on the surface.

These incident wavefront is carrying two points, point A and point B,

so we can say that from point B to point C light is travelling a

distance. If ‘ v ‘ represents the speed of the wave in the medium and if

‘ r ‘ represents the time taken by the wavefront from the point B to C

then the distance

BC = vr

In order the construct the reflected wavefront we draw a sphere of

radius vr from the point A. Let CE represent the tangent plane drawn

from the point C to this sphere. So,

AE = BC = vr

If we now consider the triangles EAC and BAC we will find that they

are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal.

This is the law of reflection

Refraction using Huygen’s principle

We know that when a light travels from one transparent medium to

another transparent medium its path changes. So the laws of refraction

state that the angle of incidence is the angle between the incident ray

and the normal and the angle of refraction is the angle between the

refracted ray and the normal.

The incident ray, reflected ray and the normal, to the interface of any

two given mediums all lie in the same plane. We also know that the

ratio of the sine of the angle of incidence and sine of the angle of

refraction is constant.

We can see a ray of light is incident on this surface and another ray

which is parallel to this ray is also incident on this surface. As these

rays are incident from the surface, so we call it incident ray.

Let PP’ represent the medium 1 and medium 2. The speed of the light

in this medium is represented by v1 and v2. If we draw a perpendicular

from point ‘A’ to this ray of light, Point A, and point B will have a

line joining them and this is called as wavefront and this wavefront is

incident on the surface.

If ‘ r ‘ represents the time taken by the wavefront from the point B to

C then the distance,

BC = v1 r

So to determine the shape of the refracted wavefront, we draw a

sphere of radius v2r from the point A in the second medium. Let CE

represent a tangent plane drawn from the point C on to the sphere.

Then, AE = v2r, and CE would represent the refracted wavefront. If

we now consider the triangles ABC and AEC, we readily obtain

sin i =

BC

AC

=

v

1

r

AC

sin r =

AE

AC

=

v

2

r

AC

where’ i ‘ and ‘ r ‘ are the angles of incidence and refraction,

respectively. Substituting the values of v1 and v2 in terms of we get

the Snell’s Law,

n1 sin i = n2 sin r

Solved Question for you on Reflection and Refraction of light using Huygens Principle

Q. The phase change in reflected wave, when lightwave suffers

reflection at the interface from air to glass is

A. 0

B. π/2

C. π

D. 2π

Answer: C. When a light is reflected from denser to rarer medium,

there is no phase change in the light but when a light is reflected from

rarer to denser medium, there is a phase change of π. So, here the air is

rarer than glass and there is a phase change of π.