Lecture Physics A2: Interference

Interference

Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Science Ho Chi Minh University of Technology

A Single Oscillating Wave

The formula

y(x, t) = Acos(kx !t)

describes a harmonic plane wave of amplitude A moving in the +x direction

For a wave on a string, each point on the wave oscillates in the x direction with simple harmonic motion of angular frequency ω

The speed/velocity

The wavelength

v = f = =

! k 2⇡ k

A2 I

The intensity is proportional to the square of the amplitude

Multiple Waves: Superposition

The principle of superposition states that when two or more waves of the same type cross at a point, the resultant displacement at that point is equal to the sum of the displacements due to each individual wave.

For inequal intensities, the maximum and minimum intensities are:

Imax = |A1 + A2|2 Imin = |A1 - A2|2

Multiple Waves: Superposition

Constructive “Superposition”

Destructive “Superposition”

Superposing sine waves

Superposing Sinusoidal Waves

If you added the two sinusoidal waves shown, what would the result look like?

If we added the two sinusoidal waves shown, what would the result look like?

1 . 0 0

0 . 5 0

0 . 0 0

0 0 1

0 0 2

0 0 3

0 0 4

0 0 5

0 0 6

0 0 7

0 0 8

0 0 9

0 0 0 1

- 0 . 5 0

- 1 . 0 0

The sum of two sines having the same frequency is another sine with the same frequency. Its amplitude depends on their relative phases.

2 .0 0

1 .5 0

1 .0 0

0 .5 0

0 .0 0

- 0 .5 0

- 1 .0 0

- 1 .5 0

- 2 .0 0

Let’s see how this works.

Superposing sine waves

If you added the two sinusoidal waves shown,

Superposing sine waves

what would the result look like?

1 . 0 0

0 . 5 0

0 . 0 0

0 0 1

0 0 2

0 0 3

0 0 4

0 0 5

0 0 6

0 0 7

0 0 8

0 0 9

0 0 0 1

Superposing Sinusoidal Waves

- 0 . 5 0

If you added the two sinusoidal waves shown, what would the result look like?

- 1 . 0 0

If we added the two sinusoidal waves shown, what would the result look like?

1 . 0 0

0 . 5 0

0 . 0 0

0 0 7

0 0 8

0 0 9

0 0 1

0 0 2

0 0 3

0 0 4

0 0 5

0 0 6

0 0 0 1

- 0 . 5 0

- 1 . 0 0

The sum of two sines having the same frequency is another sine with the same frequency. Its amplitude depends on their relative phases.

The sum of two sines having the same frequency is another sine with the same frequency → Its amplitude depends on their relative phases

2 .0 0

1 .5 0

1 .0 0

0 .5 0

0 .0 0

The sum of two sines having the same frequency is another sine with the same frequency. Its amplitude depends on their relative phases.

- 0 .5 0

0 0 1

0 0 2

0 0 3

0 0 4

0 0 5

0 0 6

0 0 7

0 0 8

0 0 9

0 0 0 1

- 1 .0 0

- 1 .5 0 2 .0 0 - 2 .0 0 1 .5 0

1 .0 0

0 .5 0

0 .0 0

- 0 .5 0

- 1 .0 0

Let’s see how this works.

- 1 .5 0

- 2 .0 0

Let’s see how this works.

Adding Sine Waves with Diﬀerent Phases Suppose we have two sinusoidal waves with the same A1, ω and k:

!t) !t + )

and

y1 = A1cos(kx y2 = A1cos(kx

One starts at phase 𝜙 after the other

Spatial dependence of 2 waves at t = 0

Resulting wave:

y = y1 + y2

↵

A1(cos↵ + cos) = 2A1cos

2 + ↵ 2

✓ ◆ ✓ y1 + y2

◆ !t + /2) (/2) y = 2A1cos(/2)cos(kx

!t + /2) y = 2A1cos(/2)cos(kx

Amplitude

Oscillation

Interference of Waves

What happens when two waves are present at the same place? Interference of Waves

For equal A and ω:

Always add amplitudes (pressures or electric fields). What happens when two waves are present at the same place? However, we observe intensity (power). Always add amplitude (e.g. pressures or electric ﬁelds) However, we observe intensity (i.e. power)

/ 2)

2 4I cos (

/ 2)

⇒

2A cos( 1

A For equal A and ω:

A = 2A1cos(𝜙/2) (cid:15482) I = 4I1cos2(𝜙/2)

Example: Stereo speakers: Stereo speakers: Listener:

Listener:

Terminology: Constructive interference: Terminology: waves are “in phase” Constructive interference: (φ = 0, 2π, 4π, ..) waves are “in phase” Destructive interference: (𝜙 = 0, 2π, 4π,…)

waves are “out of phase” (φ = π, 3π, 5π, …) Destructive interference: waves are “out of phase” (𝜙 = π, 3π, 5π,…)

Of course, φ can take on an infinite number of values. We won’t use terms like “mostly constructive” or “slightly destructive”.

Lecture 2, p.6

Quiz

Each speaker alone produces an intensity of I1 = 1 W/m2 at the listener:

Example: Changing phase of the Source Example: Changing phase of the Source

Each speaker alone produces an intensity of I1 = 1 W/m2 at the listener:

2 = 1 W/m2

I = I1 = A1

I = I1 = A12 = 1 W/m2

Example: Changing phase of the Source

2 = 1 W/m2

I = I1 = A1

Each speaker alone produces an intensity of I1 = 1 W/m2 at the listener:

Drive the speakers in phase. What is the intensity I at the listener?

2 = 1 W/m2

Drive the speakers in phase. What is the intensity I at the listener?

I = I1 = A1

I =

Drive the speakers in phase. What is the intensity I at the listener?

I =

I =?

Now shift phase of one speaker by 90o.What is the intensity I at the listener?

I =

Now shift phase of one speaker by 90o.What is the intensity I at the listener?

I =

Now shift phase of one speaker by 90o.What is the intensity I at the listener?

I =

Now shift phase of one speaker by 90°. What is the intensity I at the listener? φφφφ

φφφφ

Lecture 2, p.7

I =

I =?

φφφφ

Lecture 2, p.7

Quiz

Each speaker alone produces an intensity of I1 = 1 W/m2 at the listener:

Example: Changing phase of the Source Example: Changing phase of the Source

Each speaker alone produces an intensity of I1 = 1 W/m2 at the listener:

I = I1 = A1

Example: Changing phase of the Source

I = I1 = A12 = 1 W/m2 2 = 1 W/m2 2 = 1 W/m2

I = I1 = A1

Each speaker alone produces an intensity of I1 = 1 W/m2 at the listener:

Drive the speakers in phase. What is the intensity I at the listener?

2 = 1 W/m2

Drive the speakers in phase. What is the intensity I at the listener?

I = I1 = A1

I =

Drive the speakers in phase. What is the intensity I at the listener?

I =

I = (2A1)2 = 4I1 = 4 W/m2

Now shift phase of one speaker by 90o.What is the intensity I at the listener?

I =

Now shift phase of one speaker by 90o.What is the intensity I at the listener?

I =

Now shift phase of one speaker by 90o.What is the intensity I at the listener?

I =

Now shift phase of one speaker by 90°. What is the intensity I at the listener? φφφφ

φφφφ

Lecture 2, p.7

I =

I =?

φφφφ

Lecture 2, p.7

Quiz

Each speaker alone produces an intensity of I1 = 1 W/m2 at the listener:

Example: Changing phase of the Source Example: Changing phase of the Source

Each speaker alone produces an intensity of I1 = 1 W/m2 at the listener:

I = I1 = A1

Example: Changing phase of the Source

I = I1 = A12 = 1 W/m2 2 = 1 W/m2 2 = 1 W/m2

I = I1 = A1

Each speaker alone produces an intensity of I1 = 1 W/m2 at the listener:

Drive the speakers in phase. What is the intensity I at the listener?

2 = 1 W/m2

Drive the speakers in phase. What is the intensity I at the listener?

I = I1 = A1

I =

Drive the speakers in phase. What is the intensity I at the listener?

I =

I = (2A1)2 = 4I1 = 4 W/m2

Now shift phase of one speaker by 90o.What is the intensity I at the listener?

I =

Now shift phase of one speaker by 90o.What is the intensity I at the listener?

I =

Now shift phase of one speaker by 90o.What is the intensity I at the listener?

I =

Now shift phase of one speaker by 90°. What is the intensity I at the listener? φφφφ

φφφφ

Lecture 2, p.7

I =

I = 4I1cos2(45°) = 2 I1 = 2 W/m2

φφφφ

Lecture 2, p.7

ACT 1: Noise-cancelling Headphones Noise Cancelling Headphones

Noise-canceling headphones work using Noise-canceling headphones working using interference. A microphone on the interference. A microphone on the earpiece earpiece monitors the instantaneous monitors the instantaneous amplitude of the amplitude of the external sound wave, external sound wave, and a speaker on the inside and a speaker on the inside of the of the earpiece produces a sound wave cancel it. earpiece produces a sound wave to cancel it.

1. What must be the phase of the signal from the speaker relative to the

external noise?

1. What must be the phase of the signal from the speaker relative to the external noise?

d. -180˚

b. 90˚

a. 0

c. π a. 0 b. 90 c. π d. 180 e. 2π

e. 2π

2. What must be the intensity Is of the signal from the speaker relative to the

2. What must the intensity Is of the signal from the speaker relative to the external noise? a. Is = In b. Is < In c. Is > In

external noise In? a. Is = In

b. Is < In

c. Is > In

Lecture 2, p.10

ACT 1: Noise-cancelling Headphones Noise Cancelling Headphones

Noise-canceling headphones work using Noise-canceling headphones working using interference. A microphone on the interference. A microphone on the earpiece earpiece monitors the instantaneous monitors the instantaneous amplitude of the amplitude of the external sound wave, external sound wave, and a speaker on the inside and a speaker on the inside of the of the earpiece produces a sound wave cancel it. earpiece produces a sound wave to cancel it.

1. What must be the phase of the signal from the speaker relative to the

external noise?

1. What must be the phase of the signal from the speaker relative to the external noise?

d. -180˚

b. 90˚

a. 0

c. π a. 0 b. 90° c. π d. 180° e. 2π

e. 2π

Destructive interference occurs when the waves are ±180° out of phase (=π radians)

2. What must be the intensity Is of the signal from the speaker relative to the

2. What must the intensity Is of the signal from the speaker relative to the external noise? a. Is = In b. Is < In c. Is > In

We want A = As - An = 0. Note that I is never negative.

external noise In? a. Is = In

b. Is < In

c. Is > In

Lecture 2, p.10

Interference of Light

Review: Lights of Diﬀerent Colors

Long wavelength (Low frequency)

Short wavelength (High frequency)

Violet light has the highest energy in visible region

Interference of Light Interference is a phenomenon in which two coherent light waves superpose to form a resultant wave of greater or lower amplitude

(a) Constructive interference: If a crest of one wave meets a crest of another wave, the resultant intensity increases.

(a) Destructive interference: If a crest of one wave meets a trough of another wave, the resultant intensity decreases.

Light Polarization

Conditions for Interference

✓ Two interfering waves should be coherent, i.e. the phase diﬀerence

between them must remain constant with time.

✓ Two waves should have same frequency.

✓ If interfering waves are polarized, they must be in same state of

polarization.

✓ The separation between the light sources should be as small as

possible.

✓ The distance of the screen from the sources should be quite large.

✓ The amplitude of the interfering waves should be equal or at least

very nearly equal.

✓ The two sources should be narrow.

✓ The two sources should give monochromatic or very nearly

monochromatic.

Young’s Double Split Experiment

• In 1801, Young admitted the sunlight through a single pinhole and

then directed the emerging light onto two pinholes.

• The spherical waves emerging from the pinholes interfered with

each other and a few colored fringes were observed on the screen.

Calculation of Optical Path Diﬀerence between Two Waves

As slits (S1 and S2) are equidistant from source (S), the phase of the wave at S1 will be same as the phase of the wave at S2 and therefore S1, S2 act as coherent sources. The waves leaving from S1 and S interfere and produce alternative bright and dark bands on the screen.

2d is the distance between S1 and S2 θ is the angle between MO and MP

Let P is an arbitrary point on screen, which is at a distance D from source

x is the distance between O and P S1N is the normal on to the line S2P S1M = S2M = GO = OH = d

Calculation of Optical Path Diﬀerence between Two Waves

The optical paths are identical with the geometrical paths, if the experiment is carried out in air The path diﬀerence between two waves is S2P - S1P = S2N Let S1G and S2H are perpendicular on the screen and S2HP forms triangle

(S2P)2 = (S2H)2 + (HP)2 = D2 + (x + d)2

1 +

(S2P)2 = D2

(x + d)2 D2



Calculation of Optical Path Diﬀerence between Two Waves Divergence Operator

1 +

(S2P)2 = D2

(x + d)2 D2



1/2

1 +

S2P = D

(x + d)2 D2



Since D >> (x + d), (x + d)2/D2 is very small. After expansion,

1 +

S2P = D



D +

S2P =

(x + d)2 D2 (x + d)2 D



d)2

Similarly,

D +

S1P =

1 2 1 2 1 2



Path diﬀerence,

(x + d)2 (x d)2 = S2P S1P =

1 2D 2xd D

⇥ ⇤

Conditions for Observing Fringes

The nature of the inference of the two waves at P depends simply on what waves are contained in the length path diﬀerence (S2N).

• If the path diﬀerence (S2N) contains an integral number of

wavelengths, then the two waves interfere constructively producing a maximum in the intensity of light on the screen at P.

• If the path diﬀerence (S2N) contains an odd number of half-

wavelengths, then the two waves interfere destructively and produce minimum intensity of light on the screen at P.

Conditions for Observing Fringes

For bright fringes (maxima),

= n where n = 0, 1, 2,… S2N =

2xd D

x = n

For dark fringes (minima),

where n = 0, 1, 2,…

D 2d

= (2n + 1) S2N =

2xd D 2

x = (2n + 1)

Let xn and xn+1 denote the distances of nth and (n+1)th bright fringes Then,

2 D 2d

(cid:15482)

(n + 1) n xn = n xn = D 2d D 2d x(n+1) D 2d

Continued….

Spacing between nth and (n+1)th bright fringe is

(n + 1) n xn =

D 2d x(n+1)

Then the distance between (n+1)th and nth bright fringes is

Let xn and xn+1 denote the distances of the nth and (n+1)th bright fringes. D 2d D 2d

* This is independent of n

Hence, the spacing between any consecutive bright fringe is the same. * The distance between any two consecutive bright fringe is the same i.e. Similarly, the spacing between two dark fringes is (Dλ/2d). Dλ/2d. `

The spacing between the fringes is independent of n.

* Similarly, the distance between any two consecutive dark fringes is the same i.e. Dλ/2d.

The spacing between any two consecutive bright or dark fringe is called the “fringe width” and is denoted by X

* The distance Dλ/2d is called the “ Fringe-width” and is denoted by

X =

D 2d

The wavelength of unknown light (λ) can be calculated by measuring the values of D, 2d, and .

Resultant Intensity due to superposition of two interfering waves

• Let S be a narrow slit illuminated by a monochromatic (single wavelength)

source.

• S1 and S2 are two similar parallel slits (S1 and S2 are equidistant from S

and very close to each other)

• Suppose the waves from S reach S1 and S2 in the same phase (coherent).

• Beyond S1 and S2, the wave proceed as if they started from S1 and S2.

Resultant Intensity due to superposition of two interfering waves

• Let us calculate the resultant intensity of light of wavelength (λ) at

point P on a screen placed parallel to S1 and S2.

• Let A1 and A2 be the amplitude at P due to the waves from S1 and S2,

respectively.

• The waves arrive at P, having transversed diﬀerent paths S1P and S2P.

• Hence, they are superposed at P with a phase diﬀerence δ

= path di↵erence

= (S2P - S1P)

The displacement at P due to the simple harmonic waves from S1 and S2 can be represented by

2⇡ ⇥ 2⇡ ⇥

y1 = A1sin!t y2 = A2sin(!t + )

Resultant Intensity due to superposition of two interfering waves

By the principle of superposition, when two or more waves simultaneously reach at point, the resultant displacement is equal to the sum of the displacements of all the waves.

Hence the resultant displacement is: y = y1 + y2

= A1sin!t + A2sin(!t + )

= A1sin!t + A2sin!tcos + A2cos!tsin

Let us deﬁne:

= sin!t(A1 + A2cos) + cos!t(A2sin)

A1 + A2cos = Rcos✓ A2sin = Rsin✓

Resultant Intensity due to superposition of two interfering waves

The resultant displacement is

Hence, the resultant displacement at P is simple harmonic and of amplitude R

y = sin!t(Rcos✓) + cos!t(Rsin✓) y = Rsin(!t + ✓)

A1 + A2cos = Rcos✓ A2sin = Rsin✓

Adding the above two equations,

(A1 + A2cos)2 = R2cos2✓ (A2cos)2 = R2sin2✓

R2cos2✓ + R2sin2✓ = (A1 + A2cos)2 + (A2sin)2

1 + A2

2 + 2A1A2cos

R2 = A2

Resultant Intensity due to superposition of two interfering waves

1 + A2

2 + 2A1A2cos

The resultant intensity I at point P, which is proportional to the square of the resultant amplitude, is given by

R2 = A2

I = R2

By assuming proportionality constant = 1,

1 + A2

2 + 2A1A2cos

1 = A2

where, = path di↵erence = (S2P - S1P)

2⇡ ⇥ 2⇡ ⇥

In the case of interference, the resultant intensity at P is not just the sum of the intensities of the individual waves.

Conditions for Maxima and Minima Intensity

The resultant intensity I at point P,

2 + 2A1A2cos

1 + A2 The intensity I is a maximum, when cosδ = +1 or δ = 2nπ; n = 0, 1, 2,..

1 = A2 I =

Phase diﬀerence δ = 2nπ; n = 0, 1, 2,..

1 + A2

Imax = A2

Path diﬀerence (S2P - S1P)= 2nπ x λ/2π = nλ 2 + 2A1A2 = (A1 + A2)2

The maximum intensity is greater than the sum of two separate intensities

The intensity I is a minimum, when cosδ = -1 or δ = (2n + 1)π; n = 0, 1, 2,..

Phase diﬀerence δ = (2n + 1) π; n = 0, 1, 2,..

Path diﬀerence (S2P - S1P)= (2n + 1)π x λ/2π = (2n + 1)λ

1 + A2

Imin = A2 A2)2

2A1A2 = (A1

The minimum intensity is less than the sum of two separate intensities

Techniques for Producing Interference

The phase relation between the waves emitted by two independent light sources rapidly changes with time and therefore they can never be coherent, even though sources are identical in all respects.

• If two sources are derived from a single source by some devices, then any phase change occurring in single source is simultaneously accompanied by the same phase change in the other source.

Therefore, the phase diﬀerence between waves emerging from the two sources remains constant and the sources are coherent.

• The techniques used for creating coherent sources of light are

(a) Wavefront splitting

(b) Amplitude splitting

(a) Wavefront Splitting

One of the method consists of dividing a light wavefront, emerging from a narrow slit, by passing it through two slits closely spaced side by side.

The two parts of the same wavefront travel through diﬀerent paths and reunite on a screen to produce fringe pattern. This is known as interference due to the division of wavefront.

This method is useful only with narrow sources.

Example: Young’s double split; Fresnel’s double mirror; Fresnel biprism; Lloyd’s mirror; etc.

Fresnel’s Biprism

BP

A

S1

B

2d

S

C

S2

E

D

* S is a narrow vertical slit illuminated by a monochromatic light.

* The light from S is allowed to fall symmetrically on the biprism BP.

* The light beams emerging from the upper and lower halves of the

prism appears to start from two virtual images of S, namely S1 and S2.

* S1 and S2 act as coherent sources.

* The cones of light BS1E and AS2C, diverging from S1 and S2 are

superposed and the interference fringes are formed in the overlapping

region BC of the screen.

(b) Amplitude Splitting

The amplitude (intensity) of a light wave is divided into two parts, namely reﬂected and transmitted components, by partial reﬂection at a surface.

The two parts travel through diﬀerent paths and reunite to produce interference fringes. This is known as interference due to division of amplitude.

Optical elements such as beam splitters, mirrors are used for achieving amplitude division.

Examples: Newton’s ring experiment; Michelson’s interferometer; etc.

Interference of Thin Film (inverse phase)

Incoming wave

Transmitted wave

Reﬂected wave

(same phase)

Incoming wave

Transmitted wave

Reﬂected wave

Interference of Thin Film

Assuming,

I n

y(x, t) = Acos(kx !t)

cid

e

n

t r

a

Phase change,

y R e ﬂ e cte d ra y Φ = π )

(∆

R e ﬂ e cte d ra y Φ = 0)

(∆

= k(2h) ⇡

For zero reﬂection, = (2n

1)⇡

For maximum reﬂection,

= 2n⇡

Interference of Thin Film

I n

⇡ = (2n 1)⇡

For zero reﬂection, = 2hk

cid

e

n

t r

a

4⇡h

y R e ﬂ e cte d ra y Φ = π )

(∆

⇡ = (2n 1)⇡

R e ﬂ e cte d ra y Φ = 0)

(∆

k =

where

2⇡

✓

◆

Cancellation of rays

OR

h =

n 2

2h n

Interference of Thin Film

For maximum reﬂection,

I n

cid

= 2hk ⇡ = 2n⇡

e

n

t r

a

4⇡h

y R e ﬂ e cte d ra y Φ = π )

(∆

⇡ = 2n⇡

R e ﬂ e cte d ra y Φ = 0)

(∆

= (2n + 1)⇡

Maxima of rays

n = 0, 1, 2,…

h =

(2n + 1) 4

n = 1, 2,…

(2n 1)

h =

The color of soap bubble is due to maximum reﬂected rays

Interference of Newton’s Ring

OY2 + XY2 = 4r2

ON2 + NY2 = OY2

(x-t)2 + x2 = OY2

NX2 + NY2 = XY2

t2 + x2 = XY2

(2r-t)2 + x2 + t2 + x2 = 4r2

4r2 - 4rt + t2 + x2 + t2 + x2 = 4r2

∟

t2 + x2 = 2rt

Interference of Newton’s Ring

t2 + x2 = 2rt

Thin lens approximation: “a thin lens is a lens with a thickness (distance along the optical axis between the two surfaces of the lens) that is negligible compared to the radii of curvature of the lens surfaces.”

x2 ≈ 2rt (cid:15482)

r >> t (cid:15482) neglect t2 x2 2r

⇡

Condition for Destructive Interference:

∟

(cid:15482)

∟

x = pnr t =

n 2

Always dark fringes at the center

Condition for Constructive Interference:

(cid:15482)

x =

n +

1 2

t =

◆

s✓

(2n + 1) 4

Interference of Newton’s Ring

t2 + x2 = 2rt

Thin lens approximation: “a thin lens is a lens with a thickness (distance along the optical axis between the two surfaces of the lens) that is negligible compared to the radii of curvature of the lens surfaces.”

x2 ≈ 2rt (cid:15482)

r >> t (cid:15482) neglect t2 x2 2r

⇡

Interference of Newton’s ring

Condition for Destructive Interference:

(cid:15482)

x = pnr t =

n 2

Always dark fringes at the center

Condition for Constructive Interference:

(cid:15482)

x =

n +

1 2

t =

◆

s✓

(2n + 1) 4

Lecture Physics A2: Interference - PhD. Pham Tan Thi