Interference Slide

PHY-1109 Physics

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Interference of Light

The phenomenon of interference of light has proved the validity of the wave theory of light. Thomas Young successfully demonstrated his experiment on interference of light in 1802. When two wave trains act simultaneously on any particle in a medium, the displacement of the particle at any instant is due to the superposition of all the wave trains.

Image/Diagram Note: Diagram showing wave propagation from two sources (A and B) resulting in crests and troughs intersecting on a screen.

Young’s Experiment

In the year 1802, Young demonstrated the experiment on the interference of light. He allowed sunlight to fall on a pinhole S and then at some distance away on two pinholes A and B.

Image/Diagram Note: Diagram showing Young’s setup with light incident on pinhole S, wave fronts spreading, and then passing through pinholes A and B to create an interference pattern on a screen.

A and B are equidistant from S and are close to each other. Spherical waves spread out from S. Spherical waves also spread out from A and B. Interference bands are produced which are alternatively dark and bright.

Image/Diagram Note: Diagram showing the single slit (S), double slits ($S_1$ and $S_2$), spherical wave fronts spreading, and the resulting alternating maximum (MAX) and minimum (MIN) interference pattern shown as green bars on a screen.

It is not possible to show interference due to two independent sources of light, because a large number of difficulties are involved. The two sources may emit light waves of largely different amplitude and wavelength and the phase difference between the two may change with time.

Coherent Sources of Light

Two sources are said to be coherent if they emit light waves of the same frequency, nearly the same amplitude and are always in phase with each other. It means that the two sources must emit radiation of the same color (wavelength). In actual practice it is not possible to have two independent sources which are coherent.

But for experimental purposes, two virtual sources formed from a single source can act as coherent sources.

Methods have been devised where:

(i) interference of light takes place between the waves from the real source and a virtual source. (ii) interference of light takes place between waves from two sources formed due to a single source.

In all such cases, the two sources will act, as if they are perfectly similar in all respects.

Phase Difference and Path Difference

If the path difference between the two waves is $\lambda$, the phase difference $=2\pi$.

Suppose for a path difference $x$, the phase difference is $\delta$.

For a path difference $x$, the phase difference $=2\pi x/\lambda$.

$\text{Phase difference }\delta =\frac{2\pi x}{\lambda }=\frac{2\pi }{\lambda }\times (\text{Path difference})$

Analytical Treatment of Interference

Consider a monochromatic source of light S emitting wave of wavelength $\lambda$, and two narrow pinholes A and B. A and B are equi-distant from S and act as two virtual coherent sources.

Image/Diagram Note: Diagram showing source S, pinholes A and B (separated by $d$), and waves reaching point P on a screen D.

Let $a$ be the amplitude of the waves. The phase difference between the two waves reaching the point P, at any instant is $\delta$.

If $y_1$ and $y_2$ are the displacements

$y_1=a\sin \omega t$ $y_2=a\sin (\omega t+\delta )$

$\therefore y=y_1+y_2=a\sin \omega t+a\sin (\omega t+\delta )$

$\text{or }y=a\sin \omega t+a\sin \omega t\cos \delta +a\cos \omega t\sin \delta$

$=a\sin \omega t(1+\cos \delta )+a\cos \omega t\sin \delta$

Taking $a(1+\cos \delta )=R\cos \Theta \text{ (1)}$ and $a\sin \delta =R\sin \Theta \text{ (2)}$

$y=R\sin \omega t\cos \Theta +R\cos \omega t\sin \Theta$

$\therefore y=R\sin (\omega t+\Theta )\text{ (3)}$

which represents the equation of simple harmonic vibration of amplitude R.

Squaring eqns. (1) and (2) and adding,

$R^2{\sin }^2\Theta +R^2{\cos }^2\Theta =a^2{\sin }^2\delta +a^2(1+\cos \delta {)}^2$

$\text{Or }{R}^2=a^2{\sin }^2\delta +a^2(1+{\cos }^2\delta +2\cos \delta )$

$=a^2{\sin }^2\delta +a^2+a^2{\cos }^2\delta +2a^2\cos \delta$

$=2a^2+2a^2\cos \delta$

$=2a^2(1+\cos \delta )$

$=4a^2{\cos }^2\delta /2$

The intensity at a point is given by the square of the amplitude.

$\therefore I=R^2$

$\text{Or }I=4a^2{\cos }^2\delta /2\text{ (4)}$

Special cases

When the phase difference $\delta =0,2\pi ,2(2\pi ),\dots n(2\pi )$ Or the path difference $x=0,\lambda ,2\lambda ,\dots n\lambda$

$I=4a^2$

Intensity is maximum when the phase difference is a whole number multiple of $2\pi$ or the path difference is a whole number multiple of wavelength.

When the phase difference $\delta =\pi ,3\pi ,5\pi ,\dots (2n+1)\pi$ Or the path difference $x=\frac{\lambda }{2},\frac{3\lambda }{2},\frac{5\lambda }{2},\dots (2n+1)\frac{\lambda }{2}$

$I=0$

Intensity is minimum when the path difference is an odd number multiple of half wavelength.

Energy distribution

it is found that the intensity at bright points is $4a^2$ and at dark points it is zero. According to the law of conservation of energy, the energy cannot be destroyed. Here also the energy is not destroyed but only transferred from the points of minimum intensity to the points of maximum intensity.

Image/Diagram Note: Graph showing Intensity (I) versus Phase Difference ($\delta$), oscillating between 0 and $4a^2$, with minimum points at $\pm \pi ,\pm 3\pi ,\pm 5\pi$, and maximum points at $0,\pm 2\pi ,\pm 4\pi$.

For, at bright points, the intensity due to the two waves should be $2a^2$ but actually it is $4a^2$. But, the intensity varies from 0 to $4a^2$, and the average is still $2a^2$.

Theory of Interference Fringes

Consider a narrow monochromatic source S and two pinholes A and B equidistant from S. A and B act as two coherent sources separated by a distance $d$. Let a screen be placed at a distance D from the coherent sources. The point C on the screen is equidistant from A and B. Therefore, the path difference between the two waves is zero. Thus the point C has maximum intensity.

Image/Diagram Note: Diagram showing the geometry for interference fringes (Source S, slits A and B separated by $d$, screen D away, central point C, and point P at distance $x$ from C). The diagram also illustrates the geometric segments Q and R, and a visual representation of the alternating DARK and BRIGHT fringes, noting FRINGE WIDTH $\beta$.

Consider a point P at a distance $x$ from C. The waves reach at the point P from A and B.

Here,

$PQ=x-\frac{d}{2}$ $PR=x+\frac{d}{2}$

$(BP{)}^2-(AP{)}^2=[D^2+(x+\frac{d}{2}{)}^2]-[D^2+(x-\frac{d}{2}{)}^2]$

$(BP{)}^2-(AP{)}^2=2xd$

$BP-AP=\frac{2xd}{BP+AP}$

But $BP\approx AP\approx D$ (approximately).

$\therefore \text{Path difference }=BP-AP=\frac{2xd}{2D}=\frac{xd}{D}\text{ (5)}$

$\text{Path difference }=\frac{2\pi }{\lambda }(\frac{xd}{D})\text{ (6)}$

Bright fringes: If the path difference is a whole number multiple of wavelength $\lambda$, the point P is bright.

$\therefore \frac{xd}{D}=n\lambda$

$\text{where }n=0,1,2,3\dots$

$\text{or }x=\frac{n\lambda D}{d}\text{ (7)}$

This equation gives the distances of the bright fringes from the point C. At C, the path difference is zero and a bright fringe is formed.

When, $n=1,x_1=\frac{\lambda D}{d}$ $n=2,x_2=\frac{2\lambda D}{d}$ $n=3,x_3=\frac{3\lambda D}{d}$ $x_n=\frac{n\lambda D}{d}$

Therefore the distance between any two consecutive bright fringes. $x_2-x_1=\frac{2\lambda D}{d}-\frac{\lambda D}{d}=\frac{\lambda D}{d}\text{ (8)}$

Dark fringes: If the path difference is an odd number multiple of half Wavelength, the point P is dark.

$\frac{xd}{D}=(2n+1)\frac{\lambda }{2}\text{ where }n=0,1,2,3\dots$

$\text{or }x=\frac{(2n+1)\lambda D}{2d}\text{ (9)}$

This equation gives the distances of the dark fringes from the point C.

When, $n=0,x_0=\frac{\lambda D}{2d}$ $n=1,x_1=\frac{3\lambda D}{2d}$ $n=2,x_2=\frac{5\lambda D}{2d}$ $and{x}_n=\frac{(2n+1)\lambda D}{2d}$

The distance between any two consecutive dark fringes. $x_2-x_1=\frac{5\lambda D}{2d}-\frac{3\lambda D}{2d}=\frac{\lambda D}{d}\text{ (10)}$

The distance between any two consecutive bright or dark fringes is known as fringe width. Therefore, alternately bright and dark parallel fringes are formed. The fringes are formed on both sides of C.

The fringes width $\beta =\frac{\lambda D}{d}$

Fresnel’s Biprism

Image/Diagram Note: Diagram illustrating Fresnel’s Biprism setup, showing the virtual sources A and B, the biprism (angle $\alpha$), and the distances $y_1,y_2,$ and D for measurement.

The fringes width $\beta =\frac{\lambda D}{d}$

$\therefore \lambda =\frac{\beta d}{D}\text{ (11)}$

Determination of wavelength of light $d=\sqrt{d_1{d}_2}$ $d=2(\mu -1)\alpha y_1$

Interference in Thin Films

Image/Diagram Note: Two diagrams showing a setup (likely for determining $d$), involving a source S, virtual sources A and B, lens $L_1$ yielding distance $d_1$ and lens $L_2$ yielding distance $d_2$.

Newton and Hooke observed and developed the interference phenomenon due to multiple reflections from the surface of thin transparent materials. Everyone is familiar with the beautiful colors produced by thin film of oil on the surface of water and also by the thin film of a soap-bubble. Hooke observed such colors in thin films of mica and similar thin transparent plates.

Newton was able to show the interference rings when a convex lens was placed on a plane glass-plate. Young was able to explain the phenomenon on the basis of interference between light reflected from the top and the bottom surface of a thin film.

Interference due to Reflected Light

Image/Diagram Note: Diagram showing a ray SA incident on a thin film (thickness $t$, refractive index $\mu$) surrounded by air. The diagram illustrates reflection (AT) and refraction/reflection paths (AB, BC, CQ, and the transmitted ray F).

Consider a transparent film of thickness $t$ and refractive index $\mu$. A ray SA incident on the upper surface of the film is partly reflected along AT and partly reflected along AB. At B part of it is reflected along BC and finally emerges out along CQ.

The optical path difference $x=\mu (AB+BC)-AN$

Here, $\mu =\frac{\sin i}{\sin r}=\frac{AN}{CM}$

$\therefore AN=\mu \cdot CM$

$x=\mu (AB+BC)-\mu \cdot CM$

$\therefore x=\mu (AB+BC-CM)=\mu (PC-CM)$

$=\mu \cdot PM$

In the $\Delta APM$, $\cos r=\frac{PM}{AP}$

$\text{or }PM=AP\cdot \cos r=(AE+EP)\cos r$

$(\therefore AE=EP=t)=2t\cos r$

$\therefore x=\mu \cdot PM=2\mu t\cos r\text{ (12)}$

This equation in the case of reflected light does not represent the correct path difference but only apparent. It has been, established on the basis of electromagnetic theory that, when light is reflected from the surface of an optically denser medium a phase change $\pi$ equivalent to a path difference $\lambda /2$ occurs.

Therefore, the correct path difference in this case, $x=2\mu t\cos r-\frac{\lambda }{2}\text{ (13)}$

If the path difference $x=n\lambda$ where $n=0,1,2,3,4,\text{ etc., constructive interference takes place and the film appears bright}$.

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

$\text{or }2\mu t\cos r=(2n+1)\frac{\lambda }{2}\text{ (14)}$

If the path difference $x=(2n+1)\lambda /2$ where $n=1,2,3,4,\text{ etc., destructive interference takes place and the film appears dark}$.

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

$\text{or }2\mu t\cos r=(n+1)\lambda \text{ (15)}$

Here $n$ is an integer only, therefore $(n+1)$ can also be taken as $n$.

$\therefore 2\mu t\cos r=n\lambda \text{ Where }n=0,1,2,3,4\dots$

The amplitude of the incident ray is $a$. Let $r$ be the reflection coefficient, $t$ the transmission coefficient from rarer to denser medium and $t^{\prime }$ the transmission coefficient from denser to rarer medium.

Image/Diagram Note: Diagram showing multiple internal reflections and refractions within a thin film, labeling the incident ray amplitude $a$ and the resulting reflected ray amplitudes (e.g., $ar,atr{t}^{\prime },at{r}^3{t}^{\prime },\dots$).

The amplitudes of the reflected rays are: $ar,atr{t}^{\prime },at{r}^3{t}^{\prime },at{r}^5{t}^{\prime },\dots \text{ etc}$. The ray 1 is reflected at the surface of a denser medium. It undergoes a phase change $\pi$. The rays 2, 3, 4 etc. are all in phase but out of phase with ray 1 by $\pi$.

The resultant amplitude of 2, 3, 4 etc. is given by $A=atr{t}^{\prime }+at{r}^3{t}^{\prime }+at{r}^5{t}^{\prime }+\dots$

$A=at{t}^{\prime }r[1+r^2+r^4+\dots ]$

As $r$ is less than 1, the terms inside the brackets form a geometric series.

$A=at{t}^{\prime }r\left[\frac{1}{1-r^2}\right]=\left[\frac{at{t}^{\prime }r}{1-r^2}\right]$

According to the principle of reversibility $t{t}^{\prime }=1-r^2$

$\therefore A=\frac{a(1-r^2)r}{(1-r^2)}=ar$

Thus the resultant amplitude of 2, 3, 4, … etc. is equal in magnitude of the amplitude of ray 1 but out of phase with it. Therefore the minima of the reflected system will be of zero intensity.

Intensities of Maxima and Minima in The Interference Pattern of Reflected and Transmitted Beams in Thin Films

Image/Diagram Note: Two graphs showing intensity ($I_r$ and $I_t$) versus phase difference, illustrating the wave patterns for reflected beam intensity ($I_r$, peaking at 14.79%) and transmitted beam intensity ($I_t$, peaking at 100% and dipping to 85.21%).

The intensity of the transmitted beam is given by (vide theory of Fabry-Perot Interferometer) $I_t=\frac{I_0}{1+\frac{4r^2}{(1-r^2{)}^2}{\sin }^2\frac{\delta }{2}}$

Here $\delta$ is the phase difference, $r^2$ is the reflection coefficient and $I_0$ is the maximum intensity.

For values of $\delta =\pi ,3\pi ,5\pi ,\text{ etc.}$. ${\sin }^2\frac{\delta }{2}=1$

For $r^2=0.04$. , $[\text{i.e. Reflectance of }4\%]$

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Now that you have the complete text on Interference, we can begin working through these concepts. What is the first part you’d like to understand or practice? Perhaps Young’s experiment setup or the calculation of fringe width?