Saturday, 5 September 2009

INTEFERENCE

homas Young, an English Physicist in 1801, reported his experiment about the interference of light - when light was made to pass through a pin-hole and then through another two pin-holes side-by-side, a variation of intensity was observed on the screen.

Later when this experiment was performed with sunlight using long parallel slits instead of pin-holes, bands of colour called "interference fringes" were produced. If monochromatic light is used instead of sunlight, dark and white fringes would appear on the screen instead. This is only observed if the two slits in the second set are placed close together, otherwise all that would appear on the screen would be two bright overlapping patches. Young's original experiments were performed with white light and he could deduce the values of wavelengths of extreme red and extreme violet lights from his observations, which were quite close to the accurate ones. The pattern of light and dark fringes is sharply defined only if light of a single wavelength is used.

The above observation can be understood using Grimaldi's concept of diffraction. When sunlight passes through the first slit, it is diffracted, and the emergent light spreads out. This emergent light reaches a screen with two more narrow slits, which act as sources of light, and diffract it once more. As the light waves form each slit spread out, they meet each other at different points in the space. At the points where these light waves are in phase, they add together to give a bright fringe - constructive interference. At those points, where these waves are completely out of phase, they cancel each other out, thus producing a dark area - destructive interference.

A more concrete explanation to the above observation can be given thus:

Let us assume a monochromatic light, which, after suffereing diffraction from two slits - S1 and S2, is incident on a screen at a distance of d units from the slits (refer to diagram).

Let P be a point on the screen. Two light waves, one from each slit, are considered, that reach point P after covering different distances:

S1P = x
S2P = x+ Δx
Hence, the electric fiels at P due to these two light waves are:
E1 = E01 sin (kx - ωt)
E2 = E02 sin (k(x+ Δx) - ωt)
= E02 sin (kx - ωt + δ)
where E01 and E02 are the electric fields at S1 and S2 respectively, and
δ = kΔx = (2πΔx)/λ
The resultant electric field at P can then be written as
E = E0 sin (kx - ωt + &3949;)
where,
E02 = E012 + E022 + 2E01E02 cos δ
and,
tan ε = (E02 sin δ) / (E01+E02 cos δ)
For a fully constructive interference on the screen, the path difference between the two light waves from S1 and S2 should be an integer multiple of wavelengths. Therefore,
Δx = mλ, where m is an integer.
Hence, from the above condition:
δ = 2nπ, where n is an integer. - (1)

For a fuly destructive interference on the screen, the path difference between the two light waves should be an odd-number multiple of half-wavelengths:
Δx = (2m + 1)λ, where m is an integer.
= (m + 1/2)λ, where m is an integer.
The above condition can be expresseed as:
δ = (2n + 1)π, where n is an integer. - (2)

Therefore, at the points where the first condition is satisied, a bright area is observed. Points where the second condition is satisfied, dark fringes appear.

Intensity variation

If the two slits, S1 and S2 are identical,
E01 = E02 = E0'
Hence,
E0'2 = 2E0'2 (1 + cos δ)
But, as the intensity is directly proportional to the square of amplitude,
Ir = 2I' (1 + cos δ) = 2I' (2 cos2(δ/2))
= 4I' cos2(δ/2)
where, Ir is the resultant intensity at the screen, and I' is the intensity at each slit.

Fringe-width

Fringe-width is defined as the sparation between two consecutive dark or bright fringes on the screen.

In the same diagram as before, a perpendicular from S1 is dropped on S2P. Let the foot of this perpendicular be A. Let O be the point exactly between the two slits. Let d be the distance between the slits and the screen, and s the distance between the two slits, such that s<1P is nearly parallel to S2P, and S1A is perpendicular to S1P, S2P and OP. Since both OP and S1S2 are perpendicular to OB, where B is the point where a perpendicular from O meets the screen,
angle S2S1A = angle POB = θ
Approximating the path difference Δx, which is (PS2 - PS1) as S2A,
Δx = S2A = s sinθ
Since sinθ can be approximated as tanθ,
Δx = s tanθ = s (y/d)
where y is the distance BP. But, as shown before, at the centre of a bright fringe,
Δx = mλ, where m = ...-3,-2,-1,0,1,2,3,...
Equating the two relations
y = mdλ/s, where m = ...-3,-2,-1,0,1,2,3,...
Hence, bright fringes will occur at:
y = +dλ/s, -dλ/s, +2dλ/s, -2dλ/s, +3dλ/s, -3dλ/s ...
Similarly, for dark fringes,
y = (m + 1/2)dλ/s, where m = ...-3,-2,-1,0,1,2,3,...
Hence, dark fringes will occur at:
y = +dλ/2s, -dλ/2s, +3dλ/2s, -3dλ/2s, +5dλ/2s, -5dλ/2s ...
Therefore, the fringe-width = dλ/s

This also explains why the fringes are only observed when the slits are near and not when they are too far from each other. If s becomes much larger than λ, the fringe-width will be very small. Hence the light and dark fringes will be so closely spaced that the screen will appear to have just one bright patch. The above equation can also be used to measure the wavelength of light in an experiment, which can be calculated by measuring d, s and the fringe-width.

Interference from thin films

This is a very common phenomenon which can easily be observed in daily life. The colours that are seen on a soap bubble or in an oil slick are caused due to interference of light from these thin films, whose thickness is of the order of wavelength of the light involved.

Let us assume a thin transparent film of thickness = d with plane parallel faces (refer to diagram). A parallel beam of light is incident on one of these surfaces at an angle of incidence = i. One of the light rays of this incident beam is considered. A part of this light ray is reflected back into the same medium, while the major part suffers refraction at the surface of the film. As the wave that is inside the film now touches the other surface, it is again split into two parts - one that is reflected and the other, which suffers refraction. The amplitude of each of these emergent and reflected waves is different. Considering only the emergent rays first, we focus all of them at one point with the help of a convex lens. According to their phase-differences they will be producing constructive or destructive interference at the focal point. If the optical path difference between two such rays = 2μd, the phase difference:
δ = (2πΔx)/λ = (4πμd)/λ
Now, all these waves are in phase if
δ = 2nπ, where n is an integer.
If the above condition is satisfied, constructive interference occurs.

However, if
δ = (2n + 1)π, where n is an integer.
The adjacent waves are out of phase, and destructive interference occurs. One can assume that the same happens for the waves that are reflected. However, one case that needs to be explined here is when the conditions are such that destructive/constructive interference occer for both reflected and emergent rays, since, zero or maximum intensity in both reflection and transmission is not possible. This can be explained by using the transmission of a transverse wave through a composite string, half of which is thinner than the other half (refer to diagram). If a transverse wave is initiated at the thinner end, a part of it is reflected back at the junction, whereas the other is transmitted into the thicker string, the reflected one being inverted as compared to the transmitted part. However, when the wave is initiated from the thicker end, what occurs is a sudden phase change at the junction in the reflected part. Same applies for light waves.

When light is incident from an optically rarer medium, like air, to an optically denser medium, like a soap film, the reflected ray suffers a sudden phase change. But the next wave with which it interferes does not suffer such a change. Therefore, if 2μd = nλ (condition for minimum intensity in reflected rays), the second wave has to be out of phase with the first one which has suffered the phase change. Similar is the case when 2μd = (n + 1/2)λ (condition for maximum intensity in reflected rays). Using the above concepts, it can be also explained why the top of a vertically placed soap film, which is thinner, appears dark (as light rays reflected there suffer destructive interference), whereas coloured interference fringes decorate the rest of the film.


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