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## Superposition of Waves II, Physics tutorial

The phenomenon of diffraction that results because of superposition of many waves of same amplitude and frequency, but differing slightly in phase, is generally referred to as bending of waves round corners. Due to this phenomenon, we are sometimes inclined to think as if waves do not travel in straight lines. There are two classes of diffraction patterns, known as Fresnel and Fraunhofer classes of diffraction.

Both interference and diffraction are very significant phenomena. They have contributed hugely in justifying wave nature of light. Difference between two is fairly subtle. Interference arises due to superposition of waves originating from two (or more) narrow sources, derived from same source. Diffraction arises from superposition of wavelets from different several parts of same wavefront.

Let us consider the superposition of the following two waves:

y

_{1}= asin(ωt - kx) and y_{2}= asin(ωt - kx + Φ)These two waves have same angular frequency ω and same wave vector k, and are travelling along same direction. They have phase difference Φ which stays constant with time. Energy is found to be maximum at certain points and minimum (or probably zero) at others. This kind of energy distribution in space is called as interference pattern.

Coherent Sources:We find from experience that to have the stable and a well defined interference pattern, two sources should emit waves either with zero, or with constant difference of phase, say, Φ. If sources emit waves with zero or constant difference of phase, they are known as coherent sources. The simplest method to get coherent sources is to get them from same original source.

One method to get such sources in optics is to put the opaque screen containing two slits in path of waves emitted by the single source.

Interference of Waves from Two Slits:In the present section you will study the formation of an interference pattern due to such a system. Let S be a point source of the waves. S

_{1}and S_{2}are two narrow slits that are equidistant from source. The screen MN is kept parallel to plane of slits. Screen is at a distance D from mid-point of slits. As slits are equidistant from source, wavefront reaches the slits S_{1}and S_{2}at same time, i.e., with zero phase difference.Waves from slits S

_{1}and S_{2}, separated by the distance d, are in phase with each other. Whatever phase difference gets created subsequently between waves from two sources is because of their travelling different distances. Slits S_{1}and S_{2}act like coherent sources and waves of angular frequency ω and amplitude A. Let us consider the point P at the distance of xi from S and x_{2}from x2. Let distances be adequately large compared to d. Let displacement at P because of waves from S, bey

_{1}= Asin(ωt - kx_{1})Then the displacement at the same point due to source S2 will be

y

_{2}= Asin(ωt - kx_{2})The path difference (i.e., difference between paths covered) between two waves at P is given by (x

_{1}- x_{2}). This will lead to a phase difference ofδ = (2π/λ)(x

_{2}- x_{1})This is due to phase difference is always related with path difference according to relation phase difference/2π = path difference/λ

Due to the superposition of waves at P we get

y = y

_{1}+ y_{2}Terms in parentheses are constant in time. Let us write

A(cosk

_{1}+ cosk_{2}) = A1cosΦ andA(sink

_{1}+ sink_{2}) = A_{1}sinΦSo that Equation can be stated as

y = A

_{1}sinωtcosΦ(-cosωtsinΦ) = A_{1}sin(ωt - Φ) or Φ = tan^{-1}k((x_{1 }+ x_{2})/2) expression for intensity of resultant wave at point P as I ∝ A_{1}^{2}= 2A^{2}(1 + cosδ)or I ∝ 4A

^{2}cos^{2}δ/2Clearly, when cosδ/2 = ±1 the intensity I ∝ 4A

^{2}. This is maximum intensity and may be denoted by I_{max}._{ }Calculate position of maxima. Let d be distance between centres of two slits, θ be angle at which we observe beams and (x_{2}- x_{1}) path difference between sinθ = x_{2}- x_{1}/d or dsinθ = (x_{2}- x_{1})Maxima in intensity are attained whenever this path difference is the integral multiple of λ, wavelength of waves used. Therefore for maxima,

dsinθ = nλ with n = 0, 1, 2,.....

Interference minima take place whenever this path difference becomes odd integral multiple of λ/2, i.e.,

dsinθ = (2n + 1)λ/2 with n = 0, 1, 2,....

Intensity Distribution in Interference Pattern:Two waves of amplitudes a

_{1}and a_{2}and having the phase difference, δ, are superposed, resulting intensity I , is provided byI = A

^{2}= a_{1}^{2}+ a_{2}^{2}+ 2a_{1}a_{2}cosδMaximum intensity

I

_{max }= A^{2}= a_{1}^{2 }+ a_{2}^{2}+ 2a_{1}a_{2}On other hand, whenever phase differences is π, 3π, 5π, etc., cosδ = -1. We then have minimum of intensity, i.e.

I

_{max}= A^{2}= a_{1}^{2}+ a_{2}^{2}- 2a_{1}a_{2}(a

_{1}- a_{2})^{2 }waves of equal amplitude, intensity I = 2a^{2}(1 + cosδ) = 4a^{2}cos^{2}δ/2The path difference between waves reaching P from S

_{1}and S_{2}= (x_{2}- x_{1}) = dsinθ. If θ is very small, and is estimated in radians we can use approximationsinθ ≈ tanθ ≈ θ

We can write for maxima

dsinθ = dy

_{0}/D = nλGiving

y

_{n}= nDλ/d, where n = 0, 1, 2,...Where y

_{n}is distance of n^{th}maxima from point where perpendicular bisector or line joining two slits meets screen.Writing positions of two adjacent maxima as

y

_{n }= nDλ/d and y_{n+1}= (n + 1)Dλ/dThe separation Δy between any two consecutive maxima (or fringe widths B) is

B = y

_{n+1}- y_{n}= Dλ/dIf θ is very small, separation between two consecutive maxima of intensity is independent of n, i.e., maxima are equally spaced. Likewise, it can be shown that separation between two adjacent minima is also equal to Dλ/d, and that they too are evenly spaced.

Interference in Thin Films:Consider the ray of light AB incident on the thin film of uniform thickness t and refractive index μ as shown in figure. A part of this is reflected along BC while remaining part is refracted along BD in film. At D it is again partly reflected along DE. Ray DE partially emerges in air along EF, which is parallel to BC. Incident ray therefore divides at B in two beams of different amplitudes, out of which refracted beam suffers multiple reflections at D, E, etc. EH is perpendicular from E on BC. Path difference between rays BC and EF in reflected system is

μ(BD + DE) - BH

Interference in thin films BD is extended to I so that BD = DI. Path difference between reflected ray BC and EF is μ(BD x DE) = μGI = 2μcosr

Path difference = 2μtcosr

Where r is angle of refraction in film, a phase change of π occurs on reflection at denser medium. This is equal to the path difference of λ/2. Ray BC is because of reflection at denser medium.

Therefore net path difference between reflected rays BC and EF is given by:

Path difference = 2μtcosr - λ/2

Film appears bright when 2μtcosr - λ/2 = nλ and dark when 2μtcosr - λ/2 = (2n + 1)λ/2.

Monochromatic light alternate bright and dark fringes are attained. With white light, which is a mixture of numerous colors, colored fringes are attained.

:DiffractionDiffraction is slight bending of light as it passes around edge of the object. Amount of bending depends on relative size of wavelength of light to size of the opening. If opening is much larger than light's wavelength, bending will be approximately invisible. Though, if two are closer in size or equal, amount of bending is significant, and easily seen with naked eye.

Different kinds of Diffraction: Fraunhofer and Fresnel:Diffraction of light is generally categorized into two kinds: Fraunhofer and Fresnel diffractions. In Fraunhofer diffraction (or far-field diffraction), diffracting system (i.e. obstacle, or aperture) is so far away from source that waves producing pattern may be considered as plane. This can be attained in laboratory by making rays of light parallel by placing source at focus of the convex lens. In Fresnel (or near-field diffraction), on other hand, source of waves is so close to diffracting system that waves producing pattern still retain their curved features. This means that in Fresnel diffraction convex lens is not utilized, and wavefront remains spherical or cylindrical depending on nature of source.

Fraunhofer Diffraction by a Single Slit:Let us examine diffraction pattern generated by plane waves passing through single slit. Given slit or aperture, howsoever small or narrow it may be, has a finite size. According to Huygens's principle, every point in it serves as a source of secondary wavelets. This fact gives rise to interference between waves from different regions of same slit. Figure given below represents enlarged diagram of the narrow slit of width, d. Let us suppose that as plane wavefront reaches slit, all points in it emit secondary wavelets in same phase. Therefore if disturbance is observed at point P on far side of slit at angle θ to normal, then there is net path difference of dsinθ between waves from two edges of the slit AB. This corresponds to phase difference of 2πdsinθ/λ.

Now imagine slit AB is divided in the large number of strips of equal width, Δd. Each of the strips sends secondary wavelets and path difference between waves arriving at the point P from two adjacent strips is equal to Δd sinθ. Equivalent phase difference δ is given by

δ = 2πΔd sinθ/λ

If we divide slit AS in a total number of N strips, then clearly d = NΔd, and total phase difference.

2Δdsinθ/λ = 2π/λNΔdsinθ = Nδ

Assume that amplitude of secondary wave from each strip is denoted by A

_{s}. Then resultant disturbance at P is attained by superposition of waves from all these strips. In other words,Y = A

_{s}sin(ωt - Φ) + A_{s}sin(ωt - Φ - δ) + A_{s}sin(ωt - Φ - 2δ)+.... (up to N terms) where Φ = 2πr/π is phase difference corresponding to distance r from first slit to point P.The resultant amplitude is given by

A = A

_{s}(sin(Nδ/2)/sin(δ/2))This subdivision of the slit in a finite number of sub-slits (or strips) is artificial. We take limit as N→∞, d→0. In this situation we have a continuous variation of phase. Resultant amplitude is then

A = A

_{0}(sinα)/α with A_{0}= RNδ, and α = πdsinθ/λIntensity I

_{0}of light at any angle θ with respect to incident direction is given by:I = I

_{0}sin2α/α2For the single slit Fraunhofer diffraction pattern, minima of intensity are observed at angles θ

_{n}from incident direction, wherenλ = dsinθ

_{n}Here n = 1, 2, 3, etc. is the number of the diffraction dark band, starting from the central maximum.

Angular spread of intensity curve is given by sinθ = λ/d

This shows that as wavelength λ increases, or width of slit decreases, angular spread increases. That is, the narrower the slit, the wider the diffraction pattern. Likewise, greater the wavelength, the more widely spread is the pattern. In terms of distance D between the slit and screen the width of the central maximum, Δy, on the screen is given by

Δy = Dλ/d

Central peak in intensity curve is known as primary maximum, whereas other peaks are known as secondary maxima. Height of primary maximum is much more than any of secondary maxima.

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