Interference by division of Amplitude, Physics tutorial

Introduction:

We all have observed the marvelous rainbow colors which come into view in soap bubbles and thin oil films. Whenever a soapy plate drains, a colored reflection frequently arises from it. A similar effect takes place whenever light is reflected from the wet pavement that consists of an oil slick on it. Have you ever questioned what causes the display of colors if light is reflected from this thin oil film or soap bubble?

All such effects are due to the interference of light reflected from the opposite surfaces of the film. Therefore the phenomenon owes its origin to the combination of reflection and interference.

Stroke's analysis of phase change on Reflection:

551_strokes analysis.jpg

An elegant and novel means of looking at reflection and transmission at a boundary was introduced by Stokes. The outcomes will frequently be made use of in interference conversations. According to the principle of optically reversibility, in the absence of any absorption, a light ray which is reflected or refracted will go back over its original path if its direction is reversed.

Now assume a light ray incident on an interface of two media of refractive indices n1 and n2 as shown in figure above. When the amplitude of the incident ray is 'a', the amplitudes of the reflected and refracted beam would be ar1 and at1 correspondingly. Here, r1 and t1 are the reflection and transmission coefficients correspondingly.

In the figure above, the rays are reversed. We assume that a ray of amplitude at1 incident on medium 1 and a ray of amplitude ar1 incident on medium 2. The ray of amplitude at1 will give mount to a reflected ray of amplitude at1r2 and a transmitted ray of amplitude at1t2 where r2 and t2 are the amplitude reflection and transmission coefficients if a ray is incident from medium 2 to medium 1. Likewise, the ray of amplitude ar1 will give mount to rays of ar1r1 and ar1t1.

According to the principle of reversibility, the two rays of amplitudes ar1r1 and at1t2 should join to provide the incident ray.

ar1r2 + at1t2 = 2 or t1t2 = 1 - r12

Moreover, the two rays or amplitudes at1r2 and ar1t1 should cancel one other, that is,

at1r2 + ar1t1 = 0

Or r2 = - r1

From the outcome above, it is observed that the fraction of the intensity reflected is similar for a wave incident from either side of the boundary, as the negative sign disappears on squaring the amplitudes. The difference in sign of the amplitudes on reflection from above and reflection from beneath points out a difference of phase of 'π' between the two cases. (That is, a reversal of sign signifies or represents a displacement in the opposite sense).

It should be pointed out that the amplitude coefficients are functions of the incident angles and thus Stroke's relations might better be represented as:

Here, n1 sin θ1 = n2 sin θ2.

The second equation points out that there is an 180o phase difference between the waves internally and externally reflected that θ1 and θ2 associated by way of Snell's law.

Interference in thin films:

The bubbles of soap and thin layers of oil floating on the surface of water are common illustrations of thin films. Whenever light is reflected from such a thin film, we view various colors.

In thin films of oil and soap, visible color bands are the outcome of interference of light in thin films. The interference in this condition is caused through the interference of light waves reflected from the opposite surfaces of thin films. White light comprises of seven colors each of various wave length, thus, interference might be constructive or destructive.

2408_Interference in thin films.jpg

We will first assume that the interference due to the parallel-sided film as represented in the figure above. Interference can be observed at 'O' by division of amplitude, some light being reflected from each and every side of the film.

Portion of the light beam travels a distance BEP = 2BE whereas the other travels a distance BM.

The path difference is thus = 2nBE - BM

Now, the distance 2BE is in the thin film which consists of a refractive index n and therefore the equivalent path length in a vacuum is 2nBE.

Therefore, path difference = 2nBE - nBN = n (BE + EF - BN) = nNF =     2nt cos r

Here, 'n' is the refractive index of the film of thickness 't'.

Path difference in thin film = 2nt cos r

For soap film in air a phase change will take place at the first face and therefore a maximum will takes place if the path difference is equivalent to an odd number of half wavelengths.

Thus:

Path difference in a thin film = 2 nt cos r = (m + 1/2) λ

Here, m = 0, 1, 2 and so on.

For oil film on water the phase changes will fully based on the refractive index of the oil.

If we now assume that the soap film and normal incidence (that is, cos r = 1) it is simple to see why thick films reflect light however don't explain colored effects and appear transparent and why very thin films appear 'black' that is, don't reflect light.

Interference by a wedge-shaped film:

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Whenever two glass plates are positioned face to face by one end separated through a piece of tissue paper or thin metal foil an air wedge will be formed between them. When monochromatic light is shone on the plates a sequence of straight-line fringes will be observed parallel to the line all along which they touch. This is due to interference through division of amplitude, as by Newton's rings. A few lights are reflected from the bottom surface of the top plate and a few from the top surface of the bottom plate.

To observe the fringes clearly the angle should be small, somewhat similar to 4 minutes of arc. You should as well look for fringes close to the join of the plates where the air gap is smallest, as the fringes are not well stated for the path differences of more than some hundred wavelengths (that is, 0.058 mm for sodium light - compare this by the thickness of the sheet of paper).

Assume that a point a distance x from the join. Path difference = 2e = 2xθ, where θ is the angle between the plates in radians (that is, this angle is small, therefore tan θ = θ in radians).

For the air wedge there is a phase change on reflection at the top surface of the lower plate and therefore:

2e = 2xθ = m λ (for a dark fringe)

2e = 2xθ = (2m + 1) λ/2   (for a bright fringe)

The travelling microscope or the eye should be focused close to the upper surface of the air wedge as this is where the fringes are localized.

Pressing down gently with your finger on the plates will move the interference pattern, as merely a very small movement is required to modify the path difference significantly.

The vertical soap film is a good illustration of the wedge fringes. As the soap drains to the bottom of the film a wedge of extremely small angle is made. If the top portion goes black the film is about to break.   The flatness of a glass surface might be tested by putting it on a test surface that is termed to be flat and illuminating them by monochromatic light; any imperfections will show up as loop-shaped interference fringes around bumps or depressions on the surfaces.

Newton's Ring:

1818_Newtons ring.jpg

If a plano-convex lens having its convex surface of large radius of curvature R is positioned on a plane glass sheet, an air film of gradually rising thickness outward is made between the lens and the sheet. The thickness of film at the point of contact is zero. When monochromatic light is permitted to fall normally on the lens and the film is observed in reflected light, alternate bright and dark concentric rings are observed around the point of contact. These rings were first discovered through Newton that is why they are termed as Newton's Rings.

At any point a distance 'r' from the axis of the lens the path difference will be '2h', where 'h' is the distance between the lens and the plate at that point. The interference fringes are circular as the system is symmetrical about the centre of the lens. The radius of any ring is represented by:

(2R - h) h = r2 therefore r2 = 2Rh - h2

However h2 is small as compared by 2Rh and therefore: 2Rh = r2

The path difference (2h) is thus r2/R

A phase change of π takes place if the light reflects from the top surface of the plate however not at the lower surface of the lens, and thus:

For a bright ring analyzed through reflection: (2m + 1) λ/2 = rm2/R

For a dark ring analyzed through reflection: m λ = rm2/R

Here, m = 0, 1, 2, 3, and so on and rm is the radius of the mth ring.

Whenever a graph is plotted of r2 against m for the dark rings a straight line must be generated by a gradient represented by:

(rm2 - r12)/(m - 1) = λ R

Here, r1 and rm are the radii of the first and mth rings correspondingly.

Whenever doing the experiment, this is much simpler (and more precise) to measure the diameter of the rings and then compute their radius. A dark central spot must be obtained if observed through reflection.

The rings can be observed through transmission by putting the microscope beneath the plate, and when this is done the equations for dark and bright rings must be interchanged as two phase changes will take place, generating an efficient phase difference of 2π. A bright central spot must be obtained.

Whenever white light is employed a few colored rings will be seen due to the various wavelengths of the different colors of light.

Application of the principle of Interference in thin film:

1) Coatings:

We can employ coatings of λ/4 thickness to cancel out reflections (that is, eyeglasses make use of coating to assist in eliminating back glare). Eye glasses can be coated to assist in reducing the UV penetration.

Astronomical instruments and professional lenses make use of coatings to reduce the reflections (that is, a piece of glass can reflection around 4% of the light passing via it, and most of the lenses encompass 4 or more pieces of glass which would reflect this amount).

2) The fringes acquired through a wedge-shaped film have significant practical applications in the testing of optical surfaces for flatness. An air-film is made between a perfectly plane surface and the surface under test. When the latter surface is plane, the fringes will be straight and parallel, and, if-not, these will be unequal in shape.

3) The precision of the grinding of a lens surface can be tested through observing the shape of Newton's rings made between it and a precisely flat glass surface, by employing monochromatic light. Whenever the rings are not completely circular, the grinding is imperfect.

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