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Interference - Class 12 Physics

Notes and Question Solutions for Class 12 NEB Physics on Interference

Superposition Principle

When two or more waves meet, they interfere and produce a resultant wave whose properties can be calculated by using the principle of a preposition. According to this principle, when two or more wave motions travelling through a medium superimpose one another, a new wave is formed in which resultant displacement $(\vec y)$ at any instant is equal to the vector sum of the displacements due to individual vectors $(\vec y_1 ,\vec y_2, \dots )$ at that instant,

$$ \text {i.e.} \: \vec y = \vec y_1 + \vec y_2 + \vec y_3 \dots $$

If there are only two waves, then the resultant displacement is given by:

$$ \vec y = \vec y_1 + \vec y_2 $$

Here, $\vec y, \vec y_1 \: \text {and} \: \vec y_2 $ are the functions of time and space.

For example, when the crest of one wave falls on a crest of other, the amplitude of the resultant wave is a sum of the amplitude of two waves, as shown in the figure. When the crest of one wave falls on the trough of the other, the amplitude of the resultant wave, as shown in the figure.

The superposition principles hole good for mechanical waves as well as for electromagnetic or light waves.

Coherent Sources

Two sources of light, which continuously emit light waves of same wavelength or frequency and are always in phase differences are called coherent sources.

Coherent sources can be obtained from a single sources S as shown in the figure. Light waves which are in phase at S reach in phase on slits S₁ and S₂ as both S₁ and S₂ lie on the same wave front and are at equal distance from s. since they are derived from a single wavefront, the waves from S₁ and S₂ have same frequency or wavelength and so they are coherent sources of light.

The coherent sources obtained as above are from the division of a wavefront. Such coherent sources are also obtained from the division of amplitude as in Lloyd’s single mirror method in which a monochromatic light source and its image act as two coherent sources. Similarly, two virtual images obtained from the refraction through Fresnel’s biprism are also two coherent sources.

Two independent sources of light cannot be coherent sources. This is because light emitted by individual atoms of the light sources is the random and so light source is random and so the light from these atoms are not in phase. Thus, two independent sourced of light are incoherent sources.

Interference of Light

When light waves from two coherent sources P and Q as shown in figure superpose, they produce a non-uniform distribution of energy in different directions. At some points in this region where the crest of one wave falls on the crest of the other, resultant amplitude becomes maximum. Hence, the intensity of light is maximum at these points. At certain other points, a crest of one wave falls on the trough of the other and the resultant amplitude becomes zero. So, the intensity of the light is zero. This non-uniform distribution of energy is called interference of light. Thus, interference of light is the phenomenon of non-uniform distribution of light energy in a medium due to a superposition of light waves from two coherent sources.

There are two types of interference i.e. constructive and destructive interference.

Constructive Interference

When light waves from two coherent sources P and Q reach at a point Y in phase as shown in the figure, the amplitude of a resultant wave is equal to the sum of the amplitude of the two waves. The intensity of light at such point then is maximum that is constructive interference. In the constructive interference, a crest of one wave falls on the crest of the other and the trough of one fall on the trough of the other. If PX is greater than QX by a whole number of wavelength λ, the wave from P will be in phase with the wave form Q. So, for constructive interference, the path difference between the waves from two sources is

$$ \text {path difference,} \: QX – PX = n\lambda $$

where n is the integer number.

Destructive Interference

If waves from P and Q reaching Y are out of phase or differ in phase by 180^o as shown in a figure, the resultant amplitude is the difference in amplitudes of two waves which becomes zero. So, no light energy is obtained at Y and this is called destructive interference. In destructive interference, the crest of one wave falls on the through of the other wave and vice versa. If the distance PY is greater than QY by a half wavelength, then waves at Y will be in out of phase, and so for destructive interference, we have

$$ \text {path difference,} \: QY – PY = (n + \frac 12) \lambda $$

where n is the integer number.

Path Difference and Phase Difference

When a wave passes through a medium, the particles of the medium vibrate. When the particles completer one to and fro motion, the wave advances by a distance equal to its wavelength λ. For a complete wave, the wavelength varies in λ and the phase is changed through 2p.

Let there be two waves with a path difference of λ. Then, the phase difference them will be 2p. If the path difference is x, then path difference $ = \frac {2\pi }{\lambda } \times x. $

$$ \text {hence, phase difference} = \frac {2\pi }{\lambda } \times \text {path difference} $$

Optical Path

The product of the distance travelled by the light in a medium and the refractive index of that medium is called the optical path. If d be the distance travelled by the light in a medium of refractive index µ, then by definition,

$$ \text {Optical Path} = \mu d $$

Let us consider an optically denser medium in which the light travels with a velocity v. if the distance travelled by the light in the medium is d, then the time taken to ravel this distance is given by

$$ t = \frac dv \dots (i) $$

In this time, the light travels a distance L in free space. Then,

$$ L = ct \dots (ii) $$

Now, the refractive index of the medium is given by

$$ \mu = \frac cv \dots (iii) $$

where c is the velocity of the light in a free space. \ equation (i) in equation (ii), we get

$$ \begin{align*} L = c \times \frac dv = \frac {cv}{v} \\ \text {or,} \: L = \mu d \\ \end{align*} $$

This is called the optical path.

Conditions for Sustained Interference of Light

The two sources of light must be coherent.
The amplitude of waves from the two sources should be equal.
Two sources should be monochromatic. Otherwise, the fringes of different colours will overlap.
The coherent sources must be very close to each other.
The two sources should be point sources or very narrow sources.
The wave from the sources be in same phase or maintain a constant phase difference with time.
The interfering beams should be of same wavelength and frequency.

Young’s Double Slit Experiment

S is a narrow vertical slit (of width about 1 mm) illuminated by a monochromatic source of light. At a suitable distance (about 10 cm ) from S, there are two fine slits S₁ and S₂ about 0.5 mm apart at equidistant from S. when a screen is placed at a larger (about 2m) from the slits S₁ and S₂, alternate bright and dark bands appear on the screen. The appearance of bright and dark bands are called the fringes.

Theory of Interference of Light

Suppose S₁ and S₂ be two fine slits at a small distance d apart in the figure. Let slits are illuminated by monochromatic light from a strong source S of wavelength λ and MN is a screen at a distance D from the double slits. The two waves starting from S₁ and S₂ superimpose upon each other resulting an interference pattern on the screen placed parallel to the double slit as in the figure.

Let O be the centre between the slits S₁ and S₂. Draw S₁P, S₂P and OC perpendicular to MN. The intensity of light at a point on the screen will depend upon the path difference between the two waves arriving at the point. The point C on the screen lies on the perpendicular bisector of S₁ and S₂. Therefore, the path difference between two waves reaching C is zero and hence, they are in phase. So, the point C is the position of maximum intensity. It is called central maximum.

Consider a point P at a distance x from C. The path difference between two waves arriving at P is given by

$$\text {path difference} = S_2P – S_1P $$

From the geometry in figure, it is found that

$$ \begin{align*} PQ = x-\frac d2 ; PR = x + \frac d2 \\ \text {And} (S_2P)^2 –(S_1P)^2 = \left [ D^2 + \left ( x + \frac d2 \right )^2 \right ] - \left [ D^2 + \left ( x - \frac d2 \right )^2 \right ]\\ \text {or,} \: (S_2P – S_P)(S_2P + S_1P) = 2xd \\ \text {or,} \: (S_2P – S_P)= \frac {2xd}{BP + AP} \\ \text {In practice, point P lies very close to C.} \\ \text {So} \: S_2P \approx S_1P \approx D \\ S_2P + S_1P = D + D = 2D \\ \text {Path difference} = S_2P – S_1P = \frac {2xd}{2D} = \frac {xd}{D} \\ \end{align*} $$

The waves from S₁ and S₂ arriving at a point on the screen will interfere constructively or destructively depending upon this path difference. The phase difference for this path difference is given by

$$ \text {Phase difference,} \: \phi = \frac {2\pi}{\lambda } \left (\frac {xd}{D} \right )$$

Bright fringes
If the path difference is an integral is an integral multiple of wavelength λ, then point P is bright. Therefore, for bright fringes
$$ \begin{align*} \frac {xd}{D} &= n\lambda \\ x &= n\lambda \frac Dd \dots (i) \\ \end{align*} $$
where n = 0, 1, 2, 3, … The distance of the various bright fringes from the central maximum at C can be found as follows:
$$ \begin{align*} \text {For} \: n=0, x_0 = 0 \dots \text {central bright fringes} \\ \text {For} \: n=1, x_1 = \frac {\lambda D}{d} \dots \text {first bright fringes}\\ \text {For} \: n=2, x_2 = \frac {2\lambda D}{d} \dots \text {second bright fringes} \\ \text {For} \: n=n, x_n = n \lambda \frac {D}{d} \dots n^{th} \text { bright fringes} \\ \end{align*} $$
The distance between any two consecutive bright fringes is called fringe width, denoted by β.
$$ \begin{align*} \text {Fringe width,} \: \beta = x_2 – x_1 = \frac {2\lambda D}{d} - \frac {\lambda D}{d} = \frac {\lambda D}{d} \\ \therefore \beta = \frac {\lambda D}{d} \dots (ii) \end{align*} $$
Dark fringes
If the path difference is an odd integral multiple of half wavelength λ, then point P is dark. Therefore, for dark fringes;
$$ \begin{align*} \frac {xd}{D} &= (2n-1)\frac {\lambda }{2} \: \text {where} \: n = 1,2,3, \dots \\ \text {or,}\: x &= (2n-1)\frac {\lambda D}{2d} \dots (iii) \\ \end{align*} $$
Equation (iii) gives the distance of the dark fringes from point C. the distance of the various dark fringes from point C can be calculated as below:
$$ \begin{align*} \text {For} \: n=1, x_1 = \frac {\lambda D}{2d} \dots \text {first dark fringe}\\ \text {For} \: n=2, x_2 = \frac {3\lambda D}{2d} \dots \text {second dark fringes} \\ \text {For} \: n=n, x_n = (2n-1)\lambda \frac {\lambda D}{2d} \dots n^{th} \text { dark fringes} \\ \end{align*} $$
The distance between any two consecutive dark fringes is called fringe width β, given as
$$ \begin{align*} \text {Fringe width,} \: \beta = x_2 – x_1 = \frac {3\lambda D}{2d} - \frac {\lambda D}{2d} = \frac {\lambda D}{d} \\ \therefore \beta = \frac {\lambda D}{d} \dots (iv)\\ \end{align*} $$
from equation (ii) and (iv), it is clear that width of the bright is equal to the width of the dark fringe. From these two equations it is clear that fringe width increases as the
1. Wavelength increases.
2. Distance D of the screen from the sources increases
3. Distance between the sources decreases.