Optics Course (Phys 311)

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Optics Course (Phys 311)

Wave Optics

Interference (2 of 2)

Lecturer: Dr Zeina Hashim

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Phys 311

1. Continue Interference (1):

A. Circular Fringes.

B. White Light Fringes.

2. Introduction to the Interference of Multiple Beams of Light.

3. Interference of Waves Reflected from a Plane-Parallel Film.

4. Interference of Waves Reflected from a non-Plane-Parallel Film:

A. a Triangular film  Fringes of Equal Thickness.

B. a Lens-Like film  Newton’s Rings.

5. Fabry-Perot Interferometers and Brewster Fringes.

6. Visibility of Fringes.

Objectives covered in this lesson :

Lesson 2 of 2 Slide 1

Wave Optics: Interference

Interference (2)

Interference of Multiple beams of light:

(from Multiple Reflections)

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Phys 311

Circular Fringes :

“Fringes of Equal Inclination”

Circular Fringes obtained by Michelson’s

Interferometer are fringes of equal inclination.

To obtain circular fringes:

1. The light must be monochromatic.

2. The mirrors must be in exact adjustment.

When parallel beams are brought to interfere with each other with a phase difference determined by the angle of inclination  the fringes are called:

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Figure b:

If M₁ and M'₂ are tilted with respect to each other the interference fringes will generally take the shape of conic sections (hyperbolas).

but if M₁ and M'₂ overlap,

the fringes near the axis will be straight, parallel, and equally spaced (fringes of equal thickness).

Figure a:

The optical elements are oriented so that S'₁ and S'₂ are in line with the observer.

the resulting interference pattern consists of circles centered on the normal to M₁ and M'₂ (fringes of equal inclination).

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Phys 311

White Light Fringes :

Q: We studied that the light must be monochromatic. How can fringes be seen with white light?

When using a white light source, the two optical paths must be equal for all wavelengths. To meet this requirement, both the longitudinal and transverse light paths must cross an equal thickness of glass of the same dispersion.

In Figure a: the longitudinal beam crosses the beam splitter three times, while the transverse beam crosses the beam splitter once. To equalize the path lengths, a compensating plate identical to the beam splitter, but without the semireflective coat, is inserted into the path of the transverse beam.

In Figure b: we see that a cube beam splitter is self-compensating.

Fringes can be seen, but they are not well defined patterns. Also:

White light has only a very limited coherence length.

Therefore, white light fringes can only be seen when the path length difference between the two beams is very small (not exceeding a few λ’s).

A compensating plate must be used in the interferometer for non-monochromatic light. Why?

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Phys 311

White Light Fringes :

The Fringes:

A central dark fringe.

Bordered on either side by 8 or 10 colored fringes (with impure colors). Why “impure”?

The central fringes are sharp, but the fringe patterns rapidly become indistinct.

Reason: White light contains wavelengths between (400 – 750 nm).

The fringes of a given color are more widely spaced the greater the wavelength.

After 8 or 10 fringes, the interfering colors are so many  resultant color is white.

To find the Fringes:

It is hard to find the white-light fringes using white light only. Solution: Use a

monochromatic light to locate the fringes and make them straight, then use white light.

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Phys 311

Introduction :

The colors shown by a soap bubble are caused by an interference resulting from the multiple reflections of light between the two surfaces of the thin film of transparent

material.

If the thin film became a thick film  Interference disappears because the coherence of light is spoiled.

Oil Slick (oil spilled in the sea) Soap bubble

Interference (2): Interference of Multiple beams of light: (from Multiple Reflections)

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Phase change:

Phys 311

Introduction :

1. Waves traveling different distances in the same material:

Two wavelets from the same wave will have different phases.

2. Passing two materials (with different n) at the same time:

Wavelets from the same wave will travel with different phases 3. Refraction and transmission through one material:

Never causes a phase shift on the materials’ interface.

4. Reflection: ^𝜃¹

𝜃₂ angle of incidence

angle of reflection

angle of refraction

normal

𝜃₁^′

glass air 𝑛₂

𝑛₁

because the speed changes.

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How does phase shift happen with reflection?

Phys 311

Introduction :

Two ways to explain this:

1. Compare with a wave in a string travelling with a sudden change in speed (caused by a change in string density “this is the place of interface”).

Denser = slower speed  corresponds to a greater n.

Less dense = higher speed  a lower n.

n₁ n₂

n₁> n₂

n₁< n₂

A rope tied to a fixed end (v=0)  a wave encountering a fully reflective surface H.W. Q1:

What is the index of refraction of a shiny metal (fully reflective) such as Aluminum ?

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Phys 311

Introduction :

2. Stokes’ Treatment:

Stoke’s relations are functions of the angles, they should be:

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Phys 311

Introduction :

2. Stokes’ Treatment:

The difference in sign of the amplitudes means

a phase difference = 𝜋 i.e.

If there is no phase change on reflection from above

there must be a phase change of 𝜋 on reflection from below and vice versa.

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The figure: assume angles are very small

Q: Where do we see the interference pattern?

A: On the interface between material 1 and 2.

For rays 1 and 2: we see their interference between (ac):

region (ac) is bright (if they are in phase)

is dark (if they are out-of-phase)

is intermediate between them (between )

Q: What is the phase difference between the rays? A: we have to consider:

1) the reflections. 2) the path difference (≈ 2𝐿) 3) in a different medium (different n)

Phys 311

Interference of Waves Reflected from a Plane-Parallel Film :

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1) The Effect of Reflection:

Ray 1 :

reflected (𝑛₁ < 𝑛₂) phase shift = ½ λ from incident.

Ray 2 :

in (interface a): transmitted  same phase as incident.

in (interface b): reflected (𝑛₂ > 𝑛₁)  same phase.

in (interface c): transmitted  same phase.

𝑛₁

∴ “Reflection” caused a phase difference between ray 1 and 2 of ( ½ wavelength)

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Phys 311

2) and 3 ) The effect of path length difference in a different n: i.e. “optical path difference”

Ray 1 :

Optical path through material 1 is: 𝑛₁( 𝑖 + 𝑟₁ ) Ray 2 :

Optical path through material 1 is: 𝑛₁( 𝑖 + 𝑟₂ ) Optical path through material 2 is: 𝑛₂(2𝐿)

𝑛₁

∆= 𝑛𝑑

same

∴ The “Optical path difference” between rays 1 and 2 is [𝒏_𝟐(𝟐𝑳)]

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Phys 311

2) and 3 ) The effect of path length difference in a different n: i.e. “optical path difference”

If the optical path difference = 𝒎𝝀

 Phase difference = in phase

 𝑛₂ 2𝐿 = mλ  in phase

If the optical path difference = (𝒎 + ^𝟏_𝟐)𝝀

 Phase difference = out-of-phase

 𝑛₂ 2𝐿 = (m + ¹

2)λ  out of phase

𝑛₁

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The total phase difference is then:

𝑛₁ Due to

reflection

Due to

optical path difference 𝑛₂ 2𝐿 = mλ

𝑛₂ 2𝐿 = (m + 1 2)λ

1

2𝜆 Out-of-phase In-phase

Out-of-phase out-of

phase

In phase

1

2𝜆 Out-of-phase

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Phys 311

For non-negligible angles of incidence:

The equations will become: ^𝑛1

2𝐿 sin 𝜃 = 𝑚 _𝑛^𝜆

2 ∶ 𝑚 = 0, 1, 2, … (minima – dark)

2𝐿 sin 𝜃 = (𝑚 + ¹₂) _𝑛^𝜆

2 ∶ 𝑚 = 0, 1, 2, … (maxima – bright)

Q: if part of the ray is transmitted and part is reflected, then the intensity (and amplitude) of ray 2 is less than that of ray 1.

How can they cancel each other and achieve full darkness in their destructive interference??

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To answer the previous question: consider all reflected rays which originated from the incident ray:

Consider the phase change due to reflection only:

Ray (1) and ray (2) are out-of-phase Ray (2) , (3) , (4) , … are all in-phase

So, all rays (except ray (1)) add up constructively.

 Their amplitudes (and intensities) add up.

 The resultant ray (from their interference) will then have an amplitude (and intensity) = that of ray (1).

 Ray (1) and all other rays will have a fully destructive

interference in the cases satisfied by the minima equation.

Phys 311

Q: What about the phase change due to “optical path difference” ?

A: It will not change between every two adjacent rays. So, the equations are the same and the phase difference still depend on 𝑛₂ 2𝐿 (which is the same for all rays)

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If the film thickness is very small: the phase change will only be due to reflection:

Phys 311

If L is much less than 1, for example L < 0.1λ,

then phase difference due to the path difference 2L can be neglected.

Phase difference between r₁ and r₂ will always be ½ wavelength

destructive interference

 film will appear dark when viewed from illuminated side.

r₂ r₁

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Color Shifting by Morpho Butterflies and Paper Currencies :

Phys 311

Furthermore, the “optical path difference” will change when light strikes the surface at different angles, again changing the interference condition for the different wavelengths of light.

For the same path difference, different wavelengths (colors) of light will interfere differently.

For example:

The “optical path difference” could be an integer number of wavelengths for red light but half-integer wavelengths for blue.

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In a plane parallel film:

1. That is so thin.

2. Illuminated with an extended source.

3. Which is Monochromatic.

The fringes are formed from rays reflected with the same angle of reflection

 therefore, the fringes are “fringes of equal inclination”.

Phys 311

Interference of Waves Reflected from a Plane-Parallel Film : Fringes of Equal Inclination :

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Phys 311

Interference of Waves Reflected from a non-Plane-Parallel Film :

Fringes of Equal Thickness

Newton’s Rings

The film’s surfaces make an appreciable angle with each other

One of its surfaces is plane while the other is

convex

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Phys 311

Fabry-Perot Interferometer :

Interference happens from multiple reflections of beams reflected from a plane- parallel film of air.

Fringes are of equal inclination

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Visibility is a measure of the quality of the fringes (i.e. fringe contrast) produced by an interferometric system.

The higher V  the more visible the fringes are.

A change in the irradiance  a change in the visibility.

Q: How can we change the irradiance ? Phys

311

Visibility of Fringes:

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Phys 311

Double-beam interference pattern: Visibility changes with change in the separation between the slits:

d = 0.6 cm V = 0.593

d = 0.8 cm V = 0.361

d = 1 cm V = 0.146

d = 2.3 cm V = 0.035

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Example:

Consider two interfering beams with parallel electric fields that are superposed. Take the electric fields of the individual beams to be:

𝐸₁ = 2 cos(𝑘𝑥₁ − 𝜔𝑡) (kV/m) 𝐸₂ = 5 cos(𝑘𝑥₂ − 𝜔𝑡) (kV/m)

(a) Determine the irradiance contributed by each beam acting alone and that due to their mutual interference (i.e. interference term) at a point where their path difference is such that 𝑘 𝑥₂ − 𝑥₁ = 𝜋/12.

Phys 311

Visibility of Fringes: _{𝑰 = 𝑰}_𝟏 _{+ 𝑰}_𝟐 ₊ ^𝒄𝝐^𝒐

𝟐 𝟐𝑬_𝒐𝟏𝑬_𝒐𝟐𝐜𝐨𝐬 𝜶_𝟏 − 𝜶_𝟐

In the absence of the initial phases (i.e. 𝜀₁ = 𝜀₂ = 0):

 the phase difference = the path difference i.e. 𝛼₂ − 𝛼₁ = 𝑘(𝑥₂ − 𝑥₁)

So, the phase difference here is 𝛿 = 𝛼₂ − 𝛼₁ = 𝜋/12.

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Phys 311

The irradiance contributed by the first beam acting alone is:

𝐼₁ = ^𝑐𝜖^𝑜

2 𝐸_𝑜1² = ¹

2 (3 × 10⁸m/s)(8.85 × 10^;12F/m)(2000 V/m)² = 5310 W/m².

𝑰 = 𝑰_𝟏 + 𝑰_𝟐 + 𝒄𝝐_𝒐

𝟐 𝟐𝑬_𝒐𝟏𝑬_𝒐𝟐𝐜𝐨𝐬 𝜶_𝟏 − 𝜶_𝟐

The irradiance contributed by the second beam acting alone is:

𝐼₂ = ^𝑐𝜖₂^𝑜 𝐸_𝑜2² = ¹₂ (3 × 10⁸m/s)(8.85 × 10^;12F/m)(5000 V/m)² = 33187.5 W/m². The irradiance contributed by their mutual interference is:

𝐼₁₂ = ^𝑐𝜖₂^𝑜 2𝐸_𝑜1𝐸_𝑜2 cos 𝛼₁ − 𝛼₂

= ¹₂(3 × 10⁸m/s)(8.85 × 10^;12F/m)(2)(2000 V/m)(5000 V/m) cos(₁₂^𝜋 ) = 26549.7 W/m².

Or use: 𝐼₁₂ = 2 𝐼₁𝐼₂ cos 𝛿

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Phys 311

Visibility of Fringes: (b) Find the visibility near this point of recombination.

The visibility depends on the maximum and minimum irradiances.

The irradiance is maximum when cos 𝛿 = 1 So, 𝑰_𝒎𝒂𝒙 = 𝑰_𝟏 + 𝑰_𝟐 + 𝟐 𝑰_𝟏𝑰_𝟐

 𝐼_𝑚𝑎𝑥 = 5310 + 33187.5 + 2 5310 × 33187.5 = 65047.5 W/m². The irradiance is minimum when cos 𝛿 = −1

So, 𝑰_𝒎𝒊𝒏 = 𝑰_𝟏 + 𝑰_𝟐 − 𝟐 𝑰_𝟏𝑰_𝟐

 𝐼_𝑚𝑖𝑛 = 5310 + 33187.5 − 2 5310 × 33187.5 = 11947.5 W/m². The visibility is therefore: 𝑉 = ^𝐼_𝐼^𝑚𝑎𝑥^;𝐼^𝑚𝑖𝑛

𝑚𝑎𝑥:𝐼_𝑚𝑖𝑛 = 65047.5;11947.5

65047.5:11947.5 = 0.69

𝑰 = 𝑰_𝟏 + 𝑰_𝟐 + 𝒄𝝐_𝒐

𝟐 𝟐𝑬_𝒐𝟏𝑬_𝒐𝟐 𝐜𝐨𝐬 𝜶_𝟏 − 𝜶_𝟐 𝑰 = 𝑰_𝟏 + 𝑰_𝟐 + 𝟐 𝑰_𝟏𝑰_𝟐 𝒄𝒐𝒔 𝜹

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Phys 311

(c) At which case will the visibility be 1 ? And what does a visibility of 1 mean?

If the amplitudes of the two waves were equal, then:

𝐼_𝑚𝑎𝑥 = 4𝐼 and 𝐼_𝑚𝑖𝑛 = 0 So,

𝑉 = 4𝐼 − 0

4𝐼 + 0 = 1

A visibility of 1 means that the fringes are most visible (i.e. this is the highest possible contrast).

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Phys 311

(d) What is the lowest visibility possible?

The lowest visibility possible is 0.

This means that the fringes are not visible (total destructive interference between the two interfering waves, so that the irradiance of their superposition at that point is 0).

Therefore, the visibility of the fringes vary between 0 and 1.

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Q: (not to submit): Halliday 8^th ed. : Sample Problem (35-5 , 35-6 , 35-7) Q: (not to submit): Halliday 8^th ed. : Checkpoints 5

Q2 :

Phys 311

Homework :

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Q3 :

Q4 :

Phys 311

Homework :

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Q5 :

Phys 311

Homework :

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Q6 :

Phys 311

Homework :

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Q7 :

Phys 311

Homework :

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Q8 :

Q9:

Phys 311

Homework :

To solve this question, you need to study Problem 75 (which is written and solved in the next 2 pages).

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Solution of Problem 75:

Phys 311

Consider the interference pattern formed by waves reflected from the upper and lower surfaces of the air wedge. The wave reflected from the lower surface

undergoes a π rad phase change while the wave reflected from the upper surface does not.

At a place where the thickness of the wedge is d, the condition for a maximum in intensity is:

Where L here is d, and 𝑛₂ = 1 (air)

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Therefore, d = (2m + 1)l/4 The geometry of Fig. 35-47 gives:

where R is the radius of curvature of the lens and r is the radius of a Newton’s ring.

Thus,

First, we rearrange the terms so the equation becomes

Next, we square both sides, rearrange to solve for r², then take the square root. We get

If R is much larger than a wavelength, the first term dominates the second and

Phys 311

d  R R² r² ,

2m + 1 𝜆

4 = 𝑅 − 𝑅² − 𝑟²

𝑅² − 𝑟² = 𝑅 − (2m + 1)𝜆 4

𝑟 = 2𝑚 + 1 𝑅𝜆

2 − (2𝑚 + 1)²𝜆²

16 𝒓 = 𝟐𝒎 + 𝟏 𝑹𝝀

𝟐

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Phys 311

Lesson 2 of 2 Slide 38 (last)

Summary:

Any Questions?

Next lesson will cover:

Diffraction (1 of 2) Wave Optics: Interference

1. Continue Interference (1):

A. Circular Fringes.

B. White Light Fringes.

2. Introduction to the Interference of Multiple Beams of Light.

3. Interference of Waves Reflected from a Plane-Parallel Film.

4. Interference of Waves Reflected from a non-Plane-Parallel Film:

A. a Triangular film  Fringes of Equal Thickness.

B. a Lens-Like film  Newton’s Rings.

5. Fabry-Perot Interferometers and Brewster Fringes.

6. Visibility of Fringes.

Interference (2)

Interference of Multiple beams of light:

(from Multiple Reflections)