Interference 

Interference of Light

• When the waves emitting from the coherent sources propagating in the same direction are superposed over one another, then redistribution of energy takes place in space and this phenomenon is called interference.
• It is based on the energy conservation principle. Total energy of the waves remains constant, only redistribution of energy takes place.

Mathematical Analysis: Let two waves having amplitude $A_1$ and $A_2$ and and same frequency, and constant phase difference superimpose. Let their displacements are:

$y_1=A_1\sin (\omega t)$

$y_1=A_2\sin (\omega t+\varphi )$

$\Delta \varphi =(\omega t+\varphi )-(\omega t)=\varphi$

$A_{res}=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \Delta \varphi }$

$\text{ Phase angle }\theta ={\tan }^{-1}\left(\frac{A_2\sin \varphi }{A_1+A_2\cos \varphi }\right)$

$A_{res}^2=A_1^2+A_2^2+2A_1{A}_2\cos \Delta \varphi$

We know that $I\propto A^2$

$I=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \Delta \varphi$

$2\sqrt{I_1}\sqrt{I_2}\cos \Delta \varphi$ is known as interference factor

• Here intensity varies as the function of position.
• The existence of one wave is not affected due to presence or absence of another wave in the medium, waves propagate independently in the medium.

Conditions for sustained interference

• The two sources should be coherent.
• The separation between two coherent sources should be small.
• The distance of the screen from the two sources should be large.
• For good contrast between maxima and minima, the amplitude of two interfering waves should be as nearly equal as possible.

1.0Young's Double-Slit Experiment

In 1802 Thomas Young devised a method to produce a stationary interference pattern. This was based upon division of a single wave front into two; these two wave fronts acted as if they emanated from two sources having a fixed phase relationship. Hence when they were allowed to interfere, a station ary interference pattern was observed.

Figure: In Young’s interference experiment, light diffracted from pinhole $S_0$ encounters pinholes $S_1$ and $S_2$ in screen B. Light diffracted from these two pinholes overlaps in the region between screen B and viewing screen $C_1$

producing an interference pattern on screen C.

Analysis of Interference Pattern:

We have insured in the above arrangement that the light wave passing through $S_1$ is in phase with that passing through $S_2$ . However, the wave reaching P from $S_2$ may not be in phase with the wave reaching P from $S_1$ , because the latter must travel a longer path to reach P than the former. We have already discussed the phase-difference arising due to path difference.

If the path difference is equal to zero or is an integral multiple of wavelengths, the arriving waves are exactly in phase and undergo constructive interference.

If the path difference is an odd multiple of half a wavelength, the arriving waves are out of phase and undergo fully destructive interference. Thus, it is the path difference $\Delta x$, which determines the intensity at a point P.

 Path difference

$\Delta x=S_{1}P-S_{2}P=\sqrt{{\left(y+\frac{d}{2}\right)}^2+D^2}-\sqrt{{\left(y-\frac{d}{2}\right)}^2+D^2}$

Approximation-1:For $D\gg d$ we can approximate rays 1 and 2 as being approximately parallel at angle $\theta$ to the principal axis.

$S_{1}P-S_{2}P=S_{1}A=S_1{S}_2\sin \theta$

$\text{ path difference }=d\sin \theta \dots \dots \dots .(1)$

Approximation-2: Further if $\theta$ is small ,i.e $y\ll D$

$\sin \theta =\tan \theta =\frac{y}{D}$

And hence$\text{ path difference }=\frac{dy}{D}\dots ..\text{ (2) }$

For Maxima(constructive interference)

$\Delta x=\frac{dy}{D}=n\lambda$

$y=\frac{n\lambda D}{d},n=0,\pm 1,.\pm 2,\pm 3\dots \dots \text{ (3) }$

Here $n=0$ corresponds to the central maxima $n=\pm 1$ corresponds to the $1^{st}$ maxima $n=\pm 2$ corresponds to the $2^{nd}$ maxima and so on

For Minima(destructive interference)

$\Delta x=\pm \frac{\lambda }{2},\pm \frac{3\lambda }{2},\pm \frac{5\lambda }{2}$

$\Delta x=(2n-1)\frac{\lambda }{2}\text{ if }n=1,2,3$

$\Delta x=(2n+1)\frac{\lambda }{2}\text{ if }n=-1,-2,-3$

$y=(2n-1)\frac{\lambda D}{2d}\text{ if }n=1,2,3$

$\Delta x=(2n+1)\frac{\lambda D}{2d}\text{ if }n=-1,-2,-3$

$\Delta x=(2n+1)\frac{\lambda D}{2d}\text{ if }n=-1,-2,-3$

Here $n=\pm 1$ corresponds to first minima

Here $n=\pm 2$ corresponds to second minima and so on

2.0Fringe width:

It is the distance between two maxima of successive order on one side of the central maxima. This is also equal to the distance between two successive minima. 

Fringe width $\beta =\frac{\lambda D}{d}\dots ..(5)$

Notice that it is directly proportional to wavelength and inversely proportional to the distance between the two slits. 

Maximum order of Interference Fringes:

We obtained$,y=\frac{n\lambda D}{d},n=0,\pm 1,\pm 2\dots$

For interference maxima, $n$ cannot take infinitely large values, as that would violate the approximation (2).

i.e. $\theta$ is small or $y\ll D$

$\frac{y}{D}=\frac{n\lambda }{d}\ll 1$

Hence the above formula (3 and 4) for interference maxima/minima are applicable when $n\ll \frac{d}{\lambda }$

when $n$ becomes comparable to $\frac{d}{\lambda }$ path difference can no longer be given by equation (1) but by (2)

Hence for maxima $\Delta x=n\lambda \Rightarrow d\sin \theta =n\lambda \Rightarrow n=\frac{d\sin \theta }{\lambda }$

Hence higher order of interference maxima,

$n_{\max }=\left[\frac{d}{\lambda }\right]$

Where$\text{ [ ] }$represents the greatest integer function

Similarly, highest order of interference maxima

$n_{\min }=\left[\frac{d}{\lambda }+\frac{1}{2}\right]$

Alter:

$\Delta x=S_{1}P-S_{2}P$

$\Delta x\le d\Rightarrow \Delta x_{\max }=d$

($3^{rd}$ side of  a triangle is always greater than the difference in length of the other two sides)

3.0Intensity:

Suppose the electric field components of the light waves arriving at point$\text{ P\$ }$in the Figure) from the two slits $S_1$ and $S_2$ vary with time as

$E_1=E_0\sin \omega t$ and $\varphi =k\Delta x=\frac{2\pi }{\lambda }\Delta x$ and we have assumed that intensity of the two slits $S_1$ and $S_2$ are same (say $I_0$) hence waves have same amplitude $E_0$.

then the resultant electric field at point P is given by:

$E=E_1+E_2=E_0\sin \omega t+E_0\sin (\omega t+\varphi )=E_0^{\prime }\sin \left(\omega t+{\varphi }^{\prime }\right)$

$E_0^{\prime 2}=E_0^2+E_0^2+2E_0\cdot E_0\cos \varphi =4E_0^2{\cos }^2\frac{\varphi }{2}$

Hence the resultant intensity at point P:

$I=4I_0{\cos }^2\frac{\varphi }{2}$

$I_{\max }=4I_0\text{ when }\frac{\varphi }{2}=n\pi ;n=0,\pm 1,\pm 2$

$I_{\min }=0$ when $\frac{\varphi }{2}=\left(n-\frac{1}{2}\right);n=0,\pm 1,\pm 2$

$\varphi =k\Delta x=\frac{2\pi }{\lambda }\Delta x$

If $D\gg d$

$\varphi =\frac{2\pi }{\lambda }d\sin \theta$

If $D\gg d$ and If $y\ll d$

$\varphi =\frac{2\pi }{\lambda }d\frac{y}{D}$

However ,If the two slits have different intensities $I_1$ and $I_2$,say

$E_1=E_{01}\sin \omega t$ and $E_2=E_{02}\sin (\omega t+\varphi )$

Then resultant field at point P

$E=E_1+E_2=E_0\sin (\omega t+\theta )$

$E_0^2=E_{01}^2+E_{02}^2+2E_{01}\cdot E_{02}\cos \varphi$

Hence resultant intensity at point P 

$I=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \Delta \varphi$

Angular Width $(\alpha )$

$\tan \alpha =\frac{w}{D}=\frac{\lambda D}{dD}=\frac{\lambda }{d}$

If $\alpha$ is small than 

$\tan \alpha =\frac{\lambda }{d}(\therefore \alpha =\text{ angular width })$

Angular Fringe Width $\alpha =\frac{\beta }{D},\alpha =\frac{\lambda }{d}\left[\because \alpha =\frac{\beta }{D}=\frac{\lambda }{d}\right]$

YDSE with white light:

The central maxima will be white because all wavelengths will constructively interference here. However slightly below (or above) the position of central maxima fringes will be coloured. For example, if P is a point on the screen such that $S_{2}P-S_{1}P=\frac{{\lambda }_{\text{violet }}}{2}=190\mathrm{nm}$ completely destructive interference will occur for violet light. Hence, we will have a line devoid of violet colour that will appear reddish.

And if $S_{2}P-S_{1}P=\frac{{\lambda }_{\text{Red }}}{2}=350\mathrm{nm}$ completely destructive interference for red light results and the line at this position will be violet. The coloured fringes disappear at points far away from the central white fringe; for these points there are so many wavelengths which interfere constructively, that we obtain a uniform white illumination. For example, if $S_{2}P-S_{1}P=3000\mathrm{nm}$ then constructive interference will occur for wavelengths $\lambda =\frac{3000}{n}\mathrm{nm}$. In the visible region these wavelengths are 750 nm(red), 600 nm(yellow), 500 nm (greenish-yellow), 428.6 nm (violet). Clearly such a light will appear white to the unaided eye. Thus, with white light we get a white central fringe at the point of zero path difference, followed by a few colored fringes on its both sides, the color soon fading off to a uniform white. In the usual interference pattern with a monochromatic source, a large number of identical interference fringes are obtained and it is usually not possible to determine the position of central maxima. Interference with white light is used to determine the position of central maxima in such cases.

Shape of Interference fringes in YDSE:

We discuss the shape of fringes when two pinholes are used instead of the two slits in YDSE. Fringes are locus of points which move in such a way that its path difference from the two slits remains constant.

$S_{2}P-S_{1}P=\Delta =\text{ constant }$

Above equation represents a hyperbola with its two foci at $S_1$ and $S_2$

If $\Delta x=\pm \frac{\lambda }{2}$, the fringe represents $1^{st}$ minima

If $\Delta x=\pm \frac{3\lambda }{2}$, it represents $2^{nd}$ minima

If $\Delta x=0$, it represents central maxima 

If $\Delta x=\pm \lambda$, it represents $1^{st}$ maxima

The interference pattern which we get on screen is the section of the hyperboloid of revolution when we revolve the hyperbola about the axis $S_1{S}_2$.

A. If the screen is perpendicular to the X axis, i.e., in the YZ plane, as is generally the case, fringes are hyperbolic with a straight central section.

B. If the screen is in the XY plane, again fringes are hyperbolic.

C. If screen is perpendicular to $Y$ axis (along $S_1{S}_2$.), i.e., in the XZ plane, fringes are concentric circles with center on the axis $S_1{S}_2$; the central fringe is bright if $S_1{S}_2=n\lambda$ and dark if $S_1{S}_2=(2n-1)\frac{\lambda }{2}$

YDSE with Oblique Incidence

In YDSE , ray is incident on the slit at an inclination of ${\theta }_0$ to the axis of symmetry of the experimental set-up for points above the central point on the screen (say for $P_1$), $\Delta x=d\sin {\theta }_0+\left(S_2{P}_2-S_1{P}_1\right)$

$\Delta x=d\sin {\theta }_0+d\sin {\theta }_1\text{ If }(d\ll D)$

And for points below O on the screen say for $P_2$

$\Delta x=\left|\left(d\sin {\theta }_0+S_2{P}_2\right)-S_1{P}_2\right|=\left|d\sin {\theta }_0-\left(S_1{P}_2-S_2{P}_2\right)\right|\text{ If }(d\ll D)$

$\Delta x=\left|d\sin {\theta }_0-d\sin {\theta }_2\right|=0\text{ If }(d\ll D)$

We obtain central maxima at a point where $\Delta p=0$

$\left(d\sin {\theta }_0-d\sin {\theta }_2\right)=0\Rightarrow {\theta }_2={\theta }_0$

This corresponds to the point $O’$ in the diagram

Hence, we have finally for path difference

$\Delta x=d\left(\sin {\theta }_0+\sin \theta \to \text{ for points above }O\right)$

$\Delta x=d\left(\sin {\theta }_0-\sin \theta \to \text{ for points between }O\text{ and }{O}^{\prime }\right)$

$\Delta x=d\left(\sin {\theta }_0-\sin {\theta }_0\to \text{ for points below }{O}^{\prime }\right)$

Displacement of fringe:

On introduction of the thin glass-slab of thickness t and refractive index $\mu$, the optical path of the ray $S_1$ P increases by $t(\mu -1)$.

Now the path difference between waves coming form $S_1$ and $S_2$ at any point P is 

$\Delta x=S_{2}P-\left(S_{1}P+t(\mu -1)\right)=\left(S_{2}P-S_{1}P\right)-t(\mu -1)$

$\Delta x=d\sin \theta -t(\mu -1)Ifd\ll D$

And $\Delta x=\frac{yd}{D}-t(\mu -1)$

If $y\ll D$ as well

For central bright fringe  $\Delta x=0\Rightarrow \frac{yd}{D}=t(\mu -1)$

$y=O$

$O^{\prime }=(\mu -1)t\frac{D}{d}=(\mu -1)t\cdot \frac{\beta }{\lambda }$

The whole fringe pattern gets shifted by the same distance

$\Delta x=(\mu -1)\frac{D}{d}=(\mu -1)t\cdot \frac{\beta }{\lambda }$

Thin-Film interference:

In YDSE we obtained two coherent sources from a single (incoherent) source by division of the wave-front. Here we do the same by division of Amplitude (into reflected and refracted waves).

When a plane wave (parallel rays) is incident normally on a thin film of uniform thickness $d$ then waves reflected from the upper surface interfere with waves reflected from the lower surface.

Clearly the wave reflected from the lower surface travels an extra optical path of $2\mu d$, where is refractive index of the film.

Further if the film is placed in air the wave reflected from the upper surface (from a denser medium) suffers a sudden phase change of, while the wave reflected from the lower surface (from a rarer medium) suffers no such phase change. Consequently, conditions for constructive and destructive interference in the reflected light is given by,

$2\mu d=n\lambda$ For destructive interference 

$2\mu d=\left(n+\frac{1}{2}\right)\lambda$ For constructive interference ……(1)

Where $n=0,1,2$ And $\lambda =\text{ wavelength in free space }$

Interference will also occur in the transmitted light and here condition of constructive and destructive interference will be the reverse of (1)

$2\mu d=n\lambda$ For constructive interference 

$2\mu d=\left(n+\frac{1}{2}\right)\lambda$ For destructive  interference……….(2)

This can easily be explained by energy conservation (when intensity is maximum in reflected light it has to be minimum in transmitted light). However the amplitude of the directly transmitted wave and the wave transmitted after one reflection differ substantially and hence the fringe contrast in transmitted light is poor. It is for this reason that thin film interference is generally viewed only in the reflected light. In deriving equation (1) we assumed that the medium surrounding the thin film on both sides is rarer compared to the medium of thin film. If the medium on both sides are denser, then there is no sudden phase change in the wave reflected from the upper surface, but there is a sudden phase change of $\pi$ in waves reflected from the lower surface. The conditions for constructive and destructive interference in reflected light would still be given by equation (1). However, if the medium on one side of the film is denser and that on the other side is rarer, then either there is no sudden phase in any reflection, or there is a sudden phase change of in both reflection from upper and lower surfaces. Now the condition for constructive and destructive interference in the reflected light would be given by equation (2) and not equation (1).

4.0Solved JEE-Level Problems

Illustration-1:   In YDSE with D = 1m, d = 1mm, light of wavelength 500nm is incident at an angle of $0.57^{\circ }$ w.r.t. the axis of symmetry of the experimental set up. If the centre of symmetry of the screen is O as shown.

(1) Find the position of central maxima

(2) Intensity at point O in terms of intensity of central maxima $I_0$

(3) Number of maxima lying between O and the central maxima.

Solution:

1. $\theta ={\theta }_0=0.57^{\circ }$

$y=-D\tan \theta =-D\theta =-1\text{ meter }\times \left(\frac{0.57}{57}\mathrm{rad}\right)=-1\mathrm{cm}$

2. For Point $O\theta =0$

Hence $\Delta x=d\sin {\theta }_0;d{\theta }_0=1\mathrm{mm}\times \left(10^{-2}\mathrm{rad}\right)=10,000\mathrm{nm}=20\times (500\mathrm{nm})$

$\Delta x=20\lambda$

Hence point O corresponds to $20^{th}$ maxima

Intensity at $0=I_0$

3. 19 maxima lie between central maxima and O, excluding maxima at O and central maxima.

Illustration-2:White light is used in a YDSE with D = 1m and d = 0.9 mm. Light reaching the screen at position y = 1mm is passed through a prism and its spectrum is obtained. Find the missing lines in the visible region of this spectrum.

Solution:

$\Delta x=\frac{yd}{D}=9\times 10^{-4}\times 1\times 10^{-3}m=900nm$

For minima $\Delta x=(2n-1)\frac{\lambda }{2}$

 $\lambda =\frac{2\Delta x}{(2n-1)}=\frac{1800}{(2n-1)}=\frac{1800}{1},\frac{1800}{3},\frac{1800}{5},\frac{1800}{7}\dots \dots$ of these 600 nm and 360 nm lie in the visible range. Hence these will be missing lines in the visible spectrum.

Illustration-3: A beam of light consisting of wavelengths$\text{ 6000A˚ }$and $\text{4500A˚ }$is used in a YDSE with D = 1m and d = 1 mm. Find the least distance from the central maxima, where bright fringes due to the two wavelengths coincide.

Solution:

${\beta }_1=\frac{{\lambda }_{1}D}{d}=\frac{6000\times 10^{-10}\times 1}{10^{-3}}=0.6\mathrm{mm}$

${\beta }_2=\frac{{\lambda }_{2}D}{d}=0.45\mathrm{mm}$

Let $n_{1}th$ maxima of ${\lambda }_1$ and $n_{2}th$ maxima of ${\lambda }_2$ coincide at a position y.

$y=n_1{\beta }_1=n_2{\beta }_2=LCM\text{ of }{\beta }_1\text{ and }{\beta }_2$

 y = LCM of 0.6 mm and 0.45 mm

y = 1.8 mm

At this point 3rd maxima for$\text{ 6000A˚ }$and 4th maxima for $\text{4500A˚ }$ coincide

Illustration-4: In a YDS$ with d = 1mm and D = 1m, slabs of $(t=\mu m,\mu =3)$and $(t=0.5\mu m,\mu =2)$ are introduced in front of the upper and lower slit respectively. Find the shift in the fringe pattern.

Solution: Optical path for light coming from upper slit $S_1$ is

$S_{1}P+1\mu m(3-1)=S_{1}P+2\mu m$

Similarly, optical path for light coming from $S_2$ is

$S_{2}P+0.5\mu m(2-1)=S_{2}P+0.5\mu m$

Path difference $\Delta x=\left(S_{2}P+0.5\mu m\right)-\left(S_{1}P+2\mu m\right)=\left(S_{2}P-S_{1}P\right)-1.5\mu m=\frac{yd}{D}-1.5\mu m$

For central bright fringe $\Delta x=0$

$\Rightarrow y=\frac{1.5\mu m}{1\mathrm{mm}}\times 1\mathrm{m}=1.5\mathrm{mm}$

The whole pattern is shifted by 1.5mm upwards.

Illustration-5: White light, with a uniform intensity across the visible wavelength range 430 - 690 nm, is perpendicularly incident on a water film, of index of refraction $\mu =1.33$ and thickness d = 320 nm, that is suspended in air. At what wavelength \lambda is the light reflected by the film brightest to an observer?

Solution:

Solving for \lambda and inserting the given data, we obtain

$\lambda =\frac{2\mu d}{m+\frac{1}{2}}=\frac{(2)(1.33)(320nm)}{m+\frac{1}{2}}=\frac{85}{m+\frac{1}{2}}$

For $m=0$, this give use $\lambda =1700\mathrm{nm}$, which is in the infrared region. For $m=1$, we find 

$\lambda =567\mathrm{nm},$ which is yellow-green light, near the middle of the visible spectrum. For $m=2,\lambda =340\mathrm{nm}$, which is in the ultraviolet region. So, the wavelength at which the light seen by the observer is brightest is

$\lambda =567\mathrm{nm}$.

5.0FAQs 

Q1. What do you mean by interference?

Answer: Interference is the phenomenon where two or more waves overlap and combine, resulting in a new wave pattern. This can produce regions of increased intensity (constructive interference) or decreased intensity (destructive interference)..

Q2. What are the types of interference?

Answer: The two main types are:

1. Constructive interference – waves combine to form bright fringes.
2. Destructive interference – waves cancel each other, forming dark fringes.

Q3. What is interference with light?

Answer: Interference with light occurs when two or more coherent light waves combine to produce bright and dark fringes on a screen. This is commonly observed in Young’s Double-Slit Experiment and thin film interference.

Q4. What is the condition for interference?

Answer: Interference occurs only when the sources are coherent, have the same frequency, and maintain a constant phase difference.