Interference (Amplitude)

In order to cause light to show Interference, a coherent (same Frequency, Wavelength, Speed and  constant Phase difference) light source must be used.

Division of Amplitude

A light wave that has been "split" in terms of amplitude is simply the separation of the wave into two (or more) waves whose energy, the sum of which is the Energy of the original beam. 

SQA Exam Answer :  Division of the energy of a wave being partially transmitted and partially reflected when incident on a surface between two media of different refractive index.

Thin Film Interference

In the above diagram, a single beam of monochromatic light is incident on the glass at an angle to the normal. 

Note -  As a single light beam is used, all the reflected and transmitted rays will be coherent. Also, this diagram is not representative of size, the actual thickness of material is equivalent to oil on water.

At the air-glass boundary :-

Some light is reflected from the surface and follows a path towards A and some light is refracted into the glass and follows a path D1-D2

At the second glass-air boundary ( D2 ) :-

Some light is reflected back into the glass and some of the light is refracted and follows a path towards B 

At the original air-glass boundary :-

The reflected ray is  partially refracted and follows a path towards A and the rest of the ray is reflected  back into the glass.

At the second glass-air boundary :-

The doubly reflected ray is refracted and follows a path towards B

This means that an observer at both A and B has two rays approaching it, with a path difference between them, and may observe an interference pattern. 

Maxima and Minima for Thin Film Interference - Reflected Rays

The above diagram shows only light that would arrive at A the original full diagram. 

Each ray (1 and 2) undergo different paths and different types of reflection such that :-

Ray 1 - This ray is reflected at the transition to a higher refractive index, and therefore undergoes a phase change of π

Ray 2 - This ray is reflected at the transition to a lower refractive index, and therefore undergoes no phase change. This ray does travel a longer optical path length of 2nt, where n is the refractive index of the material.

In order for constructive Interference to be seen, the two separate rays must arrive in phase. In order for this to happen, the increased Optical Path Length must cancel out the π phase change caused by reflection :-

2nt = (m + 1/2) λ                     Where m is an integer 

In order for destructive Interference to be seen, the two separate rays must arrive out of phase. In order for this to happen, the increased Optical Path Length must not cancel out the π phase change caused by reflection :-

2nt = m λ                                 Where m is an integer 

Note - This is the OPPOSITE  of what we would normally expect for constructive and destructive bands, make sure you understand why!!

Maxima and Minima for Thin Film Interference - Transmitted Rays

The above diagram shows only light that would arrive at B the original full diagram. 

Each ray ( 3 and 4 ) undergo different paths and different types of reflection / refraction such that :-

Ray 3 - This ray passes through the block, refracting twice, with a phase change of 0

Ray 4 - This ray reflects twice within the block, but as the transitions are to a lower refractive index, and therefore undergoes no phase change. The ray does travel an additional Optical Path length of 2nt, however.

In order for constructive Interference to be seen, the two separate rays must arrive in phase. In order for this to happen, the increased Optical Path Length must give zero phase change :- 

2nt = m λ                                 Where m is an integer 

In order for destructive Interference to be seen, the two separate rays must arrive out of phase. In order for this to happen, the increased Optical Path Length must cause a phase change of π  :-

2nt = (m + 1/2) λ                     Where m is an integer

Note - This is the EXACTLY what we would normally expect for constructive and destructive bands, make sure you understand why!!

By comparing the above conditions for maxima and minima, it can be shown that for a set thickness (t), the transmitted ray shows always the opposite of the reflected ray.

This is because energy must be conserved, and as the Amplitude of the wave is being "split", when on ray is at maximum, the other must be at minimum. 

Example 1 - 

A sheet of Mica is 4.8x10-6 m thick. Light of Wavelength 512 nm is shone onto the sheet of Mica. When viewed from above, which of the three choices below will be seen :-

1.Total constructive interference

2. Total destructive interference 

3. Partial constructive interference

The refractive index for Mica is 

1.58 for Red light

1.60 for Green light

1.62 for Violet light 

In order to complete this question we must first decide what type of rays are being observed (reflected or transmitted) then apply the correct formula :-

In this question as the light source and observer are both above the Mica, it is the reflected rays that are being observed.

For reflected rays destructive interference is given by -

2nt = m λ

2 x 1.6 x 4.8x10-6  =  m x 512x10-9 

m = 30

as m is an integer destructive interference is observed.       

The embedded website below allows a simulation of thin film interfence on trasmission through a soap film :-

Non-Reflective Coatings

When a high quality camera lens is viewed under bright lights, it often has a slightly purple hue to it. The lens has been treated with a non-reflective coating in order to reduce the visible reflected light by utilising thin film reflection. The process of coating a lens in this way is called Blooming, giving a "Bloomed Lens". 

A very thin layer of  (generally) magnesium fluoride (n ~ 1.38) has been added onto the lens (n ~ 1.50). 

Examinable Derivation - Non- Reflective Coating

In the above diagram the ray of light is incident first upon the magnesium fluoride, then the transmitted ray is incident upon the glass lens itself.

As both the surfaces have a higher refractive index, both reflected rays  have a phase change of π. 

Following the above formula for destructive interference in reflected rays, it can be shown that :-

path difference = λ/2

Optical Path in Fluoride = 2nd

Where n = refractive index of coating

therefore :

2nd = λ/2

If this is rearranged to solve for the thickness of the coating (d), then the following formula can be found :-

Note - as can be seen from the formula, for a given thickness, only one specific Wavelength is completely removed through destructive interference. Partial cancellation occurs for the other visible Wavelengths. As green is the colour chosen to be completely removed (centre of the visible spectrum), the remaining red and blue light being reflected combine to give a slightly purple hue to the lens.

Example 1 -

What thickness of magnesium fluoride (n = 1.38) must be applied to a glass lens  (n = 1.50) in order to make the lens non-reflective at λ = 520 nm ? 

Wedge Fringes

The diagram below shows the set up for an experiment to view Wedge Fringes, a series of linear maxima and minima generated by interference of light passing through two flat pieces of glass at an angle to one another :-

Note - The angle between the two plates has been greatly increased in the above diagrams for clarity, in most experiment work, approximately a sheet of paper thickness is the difference between the plates. In order to derive the formulae below, it is assumed that the angle between the plates is approximately zero. 

Non Examinable Derivation - Wedge Fringes

In the above diagrams light is incident from above onto two thin glass plates, separated by a very small thickness (t). When viewed from above, this gives the Optical path difference between the rays as equal to 2t. There is a Phase change only at the reflection at P (a boundary of higher refractive index), therefore this ray has a Phase change of π. 

As there is a phase change, the formula to express  Destructive interference is given by :-

 2t = m λ

where it is assumed that the material between the plates is air (n = ) 

By looking at the right hand diagram above, it can be shown that the separation between two sets of dark fringes can be expressed by λ/2. 

This gives the formula for fringe separation as :-

If we substitute :-

Where :-

D - the separation of the the plates

L - the length of the plates

The fringe spacing can then be shown as :-

Example 2 - 

Two glass slides of length 60mm are placed on top of each other and separated by a sheet of paper at one end. Monochromatic light of 512nm is shone onto the plates and an interference pattern is observed. A Travelling microscope is used to measure the separation of 10 fringes, giving a value of 2.56x10-3 m. What is the thickness of the sheet of paper? 

Δx = (2.56x10-3) / 10 = 2.56x10-4 m

λ = 512x10-9 m

L = 60mm = 60x10-3 m

D = ?

Δx = λL / 2D

(2.56x10-4) = ((512x10-9 )x(60x10-3)) / (2xD)

D = 0.06x10-3 m

Application of Wedge Fringes

By the use of the above formula, it is possible to use Wedge Fringes to measure the size of very small objects accurately :-

1. By using Microscope slides and a human hair to separate them, it is possible to find the diameter of a human hair.

2. If a crystal is placed between the ends of the plates and heated, it is possible to measure the thermal expansion by observing the increase in fringe separation.