Wednesday, April 8th, 2020

Interference by Reflected and Transmitted light using Thin Film

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Last updated on Tuesday, May 21st, 2024

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Interference of light by reflected and transmitted light rays from a thin transparent film. We will observe in this topic. How it happens and what are the requirements?. What are the condition for the thin film? Light that we are using is from the white light source or monochromatic source,etc?

Interference of Light by Reflected Light Rays

Table of Contents

Do you know some prior concept which are important to derive the results for the interference by reflected light rays? Here, I have mentioned some supporting physical concepts (SPCs), these are required to understand the topic.

Supporting Physical Concepts (SPC’s)

Relationship between path difference and phase difference $\text{phase difference}=\frac{2 \pi}{\lambda} \text{path difference}$
When a light ray reflect from a denser medium its phase changes by π or you can say its path difference will change by λ/2.
Condition for maxima and minima; for constructive interference

$\text{path difference}= 2n \times \frac{\lambda}{2}$ [integral multiple of λ/2]

For destructive interference

$\text{path difference}= (2n-1) \times \frac{\lambda}{2}$ [integral multiple of λ/2]

triangle properties that develops in the thin film for the total path covered by the light ray

5. and finally the path difference beteen first and second light rays

These were the points which you should keep in mind in advance to know how the derivation goes.

Assumption for the derivation

Consider a thin parallel transparent film, thickness ” t ” and refractive index ” μ “, here μ is greater than one. It means denser than air.
Source is monochromatic and its wavelength is ” λ ” [source: either sodium lamp or laser light]
Incident angle ” i ” made by the incident light from the source to the thin film
Incident light ray will partially reflected and refracted (because medium is transparent)
The refracted ray in the thin film will reflect and transmit again
The first reflected ray and second transmitted ray (ray travelled the path through thin film) intensity will be different
Both light rays origin are from the single source, so following coherent condition to be superimpose.

interference of light in thin film-apniphysics — Figure 1.

In the above figure you can see, SA is the incident light ray, AB is reflected from the surface of the thin film that will satisfy the SPC’s second condition (mentioned above in SPCs).

The incident light Ray SA will partly reflected in air and partly refracted in the thin film medium. This refracted light ray in the thin film again will again reflect and transmit in the same medium.

You can see SA incident at point A, then partly reflected ray is AB, further SA will reach at the point C where it will again reflec CA and transmit CD in the same medium. By this way it will lose intensity.

Now CE is the transmitted ray which is originated from the same source S. So, these two rays AB and CD are coherent.

Interference of Light and Path Difference

Path Difference Between AB and CD rays will be determined by some logical concept that is;

draw a perpendicular on CE light ray at the point D. It will meet at the point of L on the AB light ray. Logically this is starting points for both the rays in air.
We have to find the optical paths for both the rays before these points and then difference between them.
The second light ray travel from AC to CD in the medium (i.e., thin film) while first ray optical path is AB.
Remember when we determine the optical path we multiply the distance by the refractive index of the medium. Here we will determine the optical path difference.

So the path difference before the starting points (L and D) is;

pd = optical path ACD – Optical path AL

pd = μ (AC+CD) – AL ————————–(1)

This is our first equation.

Now using the geometry in figure 1. determine the distance (or find the AC, CD and AL).

For it, first we will assume a triangle Δ ACN;

in this triangle

cos r = NC/AC

so AC = NC / cos r ; here NC is the thickness of the thin film, so it will becomes;

AC = t / cos r ———-(2)

Similarly we can determine the CD by using the triangle ΔCND;

in this triangle,

cos r = CN/CD

or CD = CN/cos r ; where CN is the thickness of the thin film, so it will be;

CD= t/ cos r ————-(3)

Now the third triangle Δ ADL, from it we have to find the AL. So,

Sin i = AL/AD

or AL = AD Sin i

AL = (AN + ND) Sin i ; AD can be written as AN + ND ………..(4)

in the above equation we have to find the AN and ND by using the triangles Δ ANC and Δ CND

But in Δ ANC

tan r = AN/NC or AN= NC tan r = t tan r (because NC= t)

in the triangle Δ CND

tan r =ND/CN => ND = CN tanr => t tanr

Therefore from the equation (4)

AL = (AN + ND) Sin i ; by putting the value of AN and ND;

AL= (t tanr + t tan r) sin i =>2 tan r sin i

tan r = sin r/ cos r

so, AL =2 tan r sin i => 2 sin r sin i /cos r

from the snell’s law μ = sin i/sin r => so, sin i = μ sin r, putting this value in above equation.

$AL= 2t \frac{\sin r}{\cos r}.\mu \sin r\\ = 2 \mu t \frac{\sin^{2}r}{\cos r}$

Now substitute the value of AC, CD and AL in the equation number 1. it becomes

$p = \mu\left [ \frac{t}{\cos r}+\frac{t}{\cos r} \right ]-2 \mu t \frac{\sin^{2}r}{cos r}$

by adding the first two terms it will be 2t/cos r

so,

$= \frac{2 \mu t}{\cos r} - 2 \mu t \frac{\sin^{2} r}{\cos r}\\ = \frac {2 \mu t}{\cos r}[1-\sin^{2}r]\\ =\frac{2 \mu t}{\cos r}.\cos^{2}r\\ \\ =2 \mu t \cos r \\$

This is the optical path difference and now total path difference between AB and DE, as per our assumptions, it will be

path difference (pd)= 2 μt cos r – λ/2

Now you can apply the maxima and minma conditions on this path difference.

Condition of Maxima or Constructive interference

For the constructive interference the path difference should be an even multiple of λ/2, so;

$2 \mu t \cos r +\frac{\lambda}{2}= 2n.\frac{\lambda}{2}$ or this can be written as

$2 \mu t \cos r= 2n.\frac{\lambda}{2}-\frac{\lambda}{2}$

or $2 \mu t \cos r=(2n-1)\frac{\lambda}{2}$

Click Here: Read more about Theory of Interfrence

Condition of Minima or Destructive interference

For the destructive interference the path difference should be an odd multiple of λ/2, so;

$2 \mu t \cos r -\frac{\lambda}{2}= (2n-1).\frac{\lambda}{2}$

or this can be written as

$2 \mu t \cos r= 2n.\frac{\lambda}{2}=n \lambda$

n=1,2,3….for both the cases.

Interference of Light by Transmitted Light Rays

In this topic we will find the interference by transmitted rays. How, using the transmitted raysone can find the path difference and satisfy the minima and maxima conditions.

transmitted rays in thin film-apniphysics — Figure 2

We have to determine the value of path difference between two emitted rays BP and DQ. Our assumptions are same as we have made earlier except to some points, they are mentioned here…

The incident ray refracted at the point A and strike at the point B. Where it again partially reflected BC and transmitted BP.
Again at the point C it will partially reflects and partially transmitted. Hence, the CD reflected light ray transmitted DQ.
Here, second transmitted light ray travel more as compared to the first one. So how much is that difference. This will be decided by the optical path difference as we did earlier.

Let draw a normal CM to the BD and DN normal on the line BP. Have you noticed why we did draw the normals? If you see here BP light ray. The time taken by this ray to travel the distance BN is same as the second ray travel the distance BC +CD. So, starting points are DN beyond that they will travel together.

Now, question is how to find the optical path diffference between these two transmitted rays. For it we have to apply same approach as we did earlier. So;

path difference (pd) = Optical path BCD – optical path BN

pd= μ (BC+CD) – BN ——-(1)

Now, as you have determined earlier the values for BC, CD and BN using the triangle property. Find in the same way. This will be same. So, you can write the result.

pd= 2 μ t cos r ————(2)

No additional path difference is introduced here, why? As earlier we have added λ/2 in the result.

The reason of it is, because both light rays take place in rare medium after the refractions. In earlier case first was reflected and second was transmitted. Here both are transmitted rays.

Condition of Maxima or Constructive interference transmitted rays

In this case,

$2 \mu t \cos r=n \lambda$

n=0,1,2,3….for both the cases

Condition of minima or Destructive interference transmitted rays

$2 \mu t \cos r= (2n-1) \frac{\lambda}{2}$

n=0,1,2,3….for both the cases

Conclusion:

You can observe the results of brightness and darkness for both the cases, they are opposite to each other.

	Interference by Reflected Light Rays	Interference by Transmitted Light Rays
Condition for maxima	$2 \mu t \cos r= (2n-1) \frac{\lambda}{2}$	$2 \mu t \cos r=n \lambda$
Condition for minima	$2 \mu t \cos r=n \lambda$	$2 \mu t \cos r= (2n-1) \frac{\lambda}{2}$

So we can conclude, if the film appears bright for the reflected rays, it will appear dark for the transmitted light rays.

Interference of Light Case 1:

When monochromatic light is used; we will get dark and bright fringes.

Interference of Light Case 2:

If we will use white light source we will get coloured fringes.

Question is why it happens?

Path difference and condition of maxima minima includes some parameters, these are;

μ, t , r and λ [ refractive index of the medium, thickness of the thin film, refractive or reflection angle and the operating wavelength].

For a specific position of the eye if t and r are constants then the path difference depends only on the refractive index μ.

What is refractive index? we define the refractive index by the two ways;

$\mu = \frac{\sin i}{\sin r}$ Snell’s law and
$\mu= \frac{c}{v}$ ratio of speed of light in vaccum to the speed of light in any medium.

If we see further $c = \nu . \lambda$ where ν is the frequency of the wave and λ is the wavelength. If you says that speed of light changes in any medium does it means the frequency changes or wavelength changes?

Yes, its wavelength changes in that medium, so speed of light changes.

From this result we can conclude that refractive index is the function of the wavelength. It will show different behaviour for different wavelength.

SO, if we are using white light for the interference, it means for each wavelength of seven colors the only refractive index changes. Hence, path difference will changes.

$2 \mu t \cos r$

So for each wavelength there is a possibility of maxima and minima condition, for any wavelength if that satisfy the maxima condition you will see color.

By this way we observe the color pattern or fringes when we use the white light.

Interference by Reflected and Transmitted light using Thin Film