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Wave Optics Class 12 Physics Notes

Introduction to Wave Optics

For a long time, there were two main ideas about the nature of light: the corpuscular model (light is made of tiny particles) and the wave theory (light is a wave).

Corpuscular Model: Proposed by Descartes in 1637 and developed by Isaac Newton, this model suggested light consists of particles called corpuscles. It could explain reflection and refraction. However, it incorrectly predicted that light would travel faster in a denser medium (like water) than in a rarer medium (like air).
Wave Theory: Proposed by Christiaan Huygens in 1678, this model suggested light propagates as a wave. It also explained reflection and refraction but correctly predicted that light would travel slower in a denser medium. This was later confirmed by Foucault's experiment in 1850, which showed light is slower in water than in air.

Initially, the wave theory wasn't widely accepted because of Newton's immense influence and the fact that waves were thought to need a medium to travel, yet light could travel through a vacuum.

The debate was largely settled in 1801 when Thomas Young's famous interference experiment provided strong evidence that light behaves as a wave. Later, James Clerk Maxwell's electromagnetic theory explained how light waves could travel through a vacuum. He showed that light is an electromagnetic wave, consisting of changing electric and magnetic fields that can sustain each other and propagate through empty space.

This chapter explores the wave nature of light, covering Huygens' principle, interference, diffraction, and polarisation.

Huygens Principle

To understand how light waves travel, we first need to define a wavefront.

A wavefront is a surface where all points on the wave are in the same phase of oscillation. Imagine dropping a stone in a calm pond; the circular ripples are wavefronts because all points on a single ripple are moving up and down together.

Spherical Wavefront: A point source of light emitting waves uniformly in all directions creates spherical wavefronts.
Plane Wavefront: At a very large distance from a source, a small section of a spherical wavefront can be considered flat, forming a plane wavefront.

Huygens' Principle is a geometrical method to find the position and shape of a wavefront at a later time if we know its current position and shape. The principle states:

Every point on a wavefront acts as a source of new disturbances, called secondary wavelets.
These secondary wavelets spread out in all directions with the speed of the wave in that medium.
The new wavefront at a later time is the surface tangent to all these secondary wavelets (the "forward envelope").

Example

Consider a spherical wavefront

F_1F_2

at time

t=0

. To find the wavefront at a later time

t=\tau

, we can draw small spheres (secondary wavelets) of radius

v\tau

from every point on

F_1F_2

, where

v

is the speed of the wave. The new wavefront,

G_1G_2

, is the surface that is tangent to the front of all these wavelets.

Note

A shortcoming of this simple model is that it also predicts a "backwave," which is not observed in reality. Huygens suggested that the amplitude of the wavelets is maximum in the forward direction and zero backward. A more rigorous wave theory confirms the absence of a backwave.

Refraction and Reflection of Plane Waves using Huygens Principle

Huygens' principle can be used to derive the laws of reflection and refraction.

Refraction of a plane wave

Let's consider a plane wavefront AB incident on a surface PP' that separates two media. Let the speed of light be $v_1$ in medium 1 and $v_2$ in medium 2. The angle of incidence is $i$ .

In the time $\tau$ it takes for the wavefront at point B to travel to point C on the surface, the distance covered is $\mathrm{BC}=v_{1} \tau$ . During this same time, the secondary wavelet from point A travels into medium 2, covering a distance $\mathrm{AE}=v_{2} \tau$ . The new, refracted wavefront is the plane CE, which is tangent to the wavelet from A.

From the triangles ABC and AEC, we have: $\sin i=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{v_{1} \tau}{\mathrm{AC}}$ $\sin r=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{v_{2} \tau}{\mathrm{AC}}$

Dividing these two equations gives the relationship between the angles and the speeds: $\frac{\sin i}{\sin r}=\frac{v_{1}}{v_{2}}$

This important result shows that if a ray bends towards the normal ( $r<i$ ), then the speed of light in the second medium is less than in the first ( $v_2 < v_1$ ). This matches experimental observations and supports the wave theory.

Using the refractive index, defined as $n = c/v$ (where $c$ is the speed of light in vacuum), we can rewrite the equation as: $n_{1} \sin i=n_{2} \sin r$ This is Snell's law of refraction.

When light is refracted, its speed and wavelength change, but its frequency remains the same. The relationship is: $\frac{v_{1}}{\lambda_{1}}=\frac{v_{2}}{\lambda_{2}} = \nu$ So, when light enters a denser medium ( $v_1 > v_2$ ), its wavelength decreases ( $\lambda_1 > \lambda_2$ ).

Refraction at a rarer medium

If light travels from a denser medium to a rarer medium ( $v_2 > v_1$ ), the ray bends away from the normal ( $r > i$ ). As the angle of incidence $i$ increases, the angle of refraction $r$ also increases. There is a specific angle of incidence called the critical angle ( $i_c$ ), for which the angle of refraction is $90^{\circ}$ .

The critical angle is defined by: $\sin i_{c}=\frac{n_{2}}{n_{1}}$

If the angle of incidence is greater than the critical angle ( $i > i_c$ ), no refraction occurs. The wave is completely reflected back into the first medium. This phenomenon is called total internal reflection.

Reflection of a plane wave by a plane surface

Now, consider a plane wavefront AB incident on a reflecting surface MN at an angle $i$ . The speed of the wave in the medium is $v$ .

In the time $\tau$ it takes for the wavefront at B to reach C, the distance covered is $\mathrm{BC}=v \tau$ . In the same time, the secondary wavelet from A travels the same distance, $\mathrm{AE}=v \tau$ , because it is in the same medium. The reflected wavefront is the plane CE.

By comparing the triangles EAC and BAC, we can see they are congruent. Therefore, the angle of incidence $i$ is equal to the angle of reflection $r$ . $i=r$ This is the law of reflection.

Coherent and Incoherent Addition of Waves

The phenomenon of interference arises from the superposition principle, which states that when two or more waves overlap at a point, the resultant displacement is the vector sum of the individual displacements.

For a stable interference pattern to be observed, the light sources must be coherent.

Coherent Sources: Two sources are coherent if they emit waves of the same frequency and have a constant phase difference between them.
Incoherent Sources: Two sources are incoherent if the phase difference between their waves changes rapidly and randomly with time. Ordinary light sources, like two separate light bulbs, are incoherent.

Constructive and Destructive Interference

Consider two coherent sources, $S_1$ and $S_2$ , vibrating in phase.

Constructive Interference: At a point P where the waves arrive in phase, they reinforce each other. This happens when the path difference is an integer multiple of the wavelength. The intensity is maximum.
- Condition: Path difference $\mathrm{S}_{1} \mathrm{P} \sim \mathrm{~S}_{2} \mathrm{P}=n \lambda$ , where $n=0,1,2,3, \ldots$
- Resultant Intensity: If each source has intensity $I_0$ , the resultant intensity is $I = 4I_0$ .
Destructive Interference: At a point P where the waves arrive out of phase, they cancel each other out. This happens when the path difference is an odd-integer multiple of half a wavelength. The intensity is minimum (zero).
- Condition: Path difference $\mathrm{S}_{1} \mathrm{P} \sim \mathrm{~S}_{2} \mathrm{P}=\left(n+\frac{1}{2}\right) \lambda$ , where $n=0,1,2,3, \ldots$
- Resultant Intensity: $I = 0$ .

For any arbitrary point, if the phase difference between the two waves is $\phi$ , the resultant intensity is given by: $I=4 I_{0} \cos ^{2}(\phi / 2)$

If the sources are incoherent, the phase difference $\phi$ changes randomly. We observe an average intensity, which is simply the sum of the individual intensities: $I = I_0 + I_0 = 2I_0$ . No interference pattern is seen.

Interference of Light Waves and Young's Experiment

To observe interference with light, we need coherent sources. Since light from ordinary sources undergoes abrupt phase changes in about $10^{-10}$ seconds, two independent sources will be incoherent.

In 1801, Thomas Young devised a clever method to create two coherent sources from one.

Young's Double-Slit Experiment:

Light from a single source (S) illuminates an opaque screen with two very narrow, closely spaced slits ( $S_1$ and $S_2$ ).
Since the light reaching $S_1$ and $S_2$ comes from the same source S, any phase change in S occurs identically at both slits. Therefore, $S_1$ and $S_2$ act as two perfectly coherent sources.
The waves from $S_1$ and $S_2$ spread out and interfere, creating a pattern of alternating bright and dark bands, called fringes, on a viewing screen.

The positions of these fringes can be calculated. Let $d$ be the distance between the slits, $D$ be the distance from the slits to the screen, and $\lambda$ be the wavelength of light.

Bright Fringes (Constructive Interference): The position $x$ of the n-th bright fringe from the center is given by: $x=x_{n}=\frac{n \lambda D}{d} ; \quad n=0, \pm 1, \pm 2, \ldots$
Dark Fringes (Destructive Interference): The position $x$ of the n-th dark fringe from the center is given by: $x=x_{\mathrm{n}}=\left(n+\frac{1}{2}\right) \frac{\lambda D}{d} ; \quad n=0, \pm 1, \pm 2, \ldots$

These equations show that the bright and dark fringes are equally spaced.

Diffraction

Diffraction is the bending or spreading of waves as they pass through a narrow opening or around an obstacle. This phenomenon is a characteristic of all waves, including light. We don't notice light diffraction in everyday life because its wavelength is extremely small compared to the size of most objects. However, diffraction limits the resolving power of optical instruments like microscopes and telescopes.

The single slit

When monochromatic light passes through a single narrow slit, it doesn't just create a sharp image of the slit. Instead, it produces a diffraction pattern on a screen. This pattern consists of:

A very bright and wide central maximum.
A series of much weaker secondary maxima on either side.
Minima (dark fringes) located between the maxima.

This pattern is explained by treating different parts of the wavefront within the slit as secondary sources that interfere with each other.

For a slit of width $a$ , the positions of the minima are given by the angle $\theta$ :

Condition for Minima (dark fringes): $\theta \approx \frac{n \lambda}{a}, \quad n= \pm 1, \pm 2, \pm 3, \ldots$

The secondary maxima are located approximately halfway between the minima:

Condition for Secondary Maxima (bright fringes): $\theta \approx \left(n+\frac{1}{2}\right) \frac{\lambda}{a}, \quad n= \pm 1, \pm 2, \pm 3, \ldots$

Note

Physicist Richard Feynman noted that there is no fundamental physical difference between interference and diffraction. Generally, "interference" is used when there are a few sources (like two slits), and "diffraction" is used when there is a large number of sources (like the many points across a single slit).

Polarisation

Polarisation is a property of transverse waves that specifies the direction of oscillation. Light is a transverse wave, meaning the electric field oscillates perpendicular to the direction the wave is traveling.

Unpolarised Light: In light from a source like the sun or a bulb, the electric field oscillates in all possible directions perpendicular to the direction of propagation. The direction of oscillation changes randomly and rapidly.
Linearly Polarised Light (or Plane Polarised Light): In this type of light, the electric field oscillates along a single, fixed direction.

Unpolarised light can be polarised using a filter called a polaroid. A polaroid sheet contains long-chain molecules aligned in a specific direction. It only allows the component of the electric field perpendicular to the molecular chains to pass through. This direction is called the pass-axis of the polaroid.

When unpolarised light passes through a polaroid, the transmitted light is linearly polarised, and its intensity is reduced by half.

Malus's Law

If already polarised light with intensity $I_0$ is passed through a second polaroid (called an analyser), the intensity of the light that emerges, $I$ , depends on the angle $\theta$ between the pass-axis of the first polaroid and the second one. This relationship is known as Malus's Law.

$I=I_{0} \cos ^{2} \theta$

When the pass-axes are parallel ( $\theta = 0^{\circ}$ ), the transmitted intensity is maximum ( $I = I_0$ ).
When the pass-axes are perpendicular ( $\theta = 90^{\circ}$ ), they are said to be "crossed," and the transmitted intensity is zero ( $I = 0$ ).

Polarisation is a key proof that light waves are transverse. Longitudinal waves, like sound, cannot be polarised.

Example

Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids?

Given

The first polariser ( $P_1$ ) and third polariser ( $P_3$ ) are crossed. This means the angle between their pass-axes is $90^{\circ}$ .
A second polariser ( $P_2$ ) is placed between them and rotated.
Let the angle between the pass-axes of $P_1$ and $P_2$ be $\theta$ .
Let the intensity of polarised light after passing through $P_1$ be $I_0$ .

To Find

The intensity of light emerging from the third polariser ( $P_3$ ) as a function of $\theta$ .

Formula

Malus's Law: $I = I_{incident} \cos^2(\text{angle})$

Solution

Light passing through P2: The intensity of light after passing through the second polariser $P_2$ is given by Malus's law: $I_2 = I_0 \cos^2 \theta$
Light passing through P3: This light, with intensity $I_2$ , now falls on the third polariser $P_3$ . Since $P_1$ and $P_3$ are crossed, the angle between their axes is $90^{\circ}$ . If the angle between $P_1$ and $P_2$ is $\theta$ , then the angle between the axes of $P_2$ and $P_3$ will be $(90^{\circ} - \theta)$ or $(\pi/2 - \theta)$ .
The final intensity $I$ emerging from $P_3$ is: $I = I_2 \cos^2(\frac{\pi}{2} - \theta)$ Substituting the expression for $I_2$ : $I = (I_0 \cos^2 \theta) \cos^2(\frac{\pi}{2} - \theta)$ Since $\cos(\frac{\pi}{2} - \theta) = \sin \theta$ , the equation becomes: $I = I_0 \cos^2 \theta \sin^2 \theta$ Using the trigonometric identity $\sin(2\theta) = 2\sin\theta\cos\theta$ , we can rewrite this as: $I = I_0 (\frac{\sin(2\theta)}{2})^2 = \frac{I_0}{4} \sin^2(2\theta)$

Final Answer The transmitted intensity is given by $I = (I_0/4) \sin^2(2\theta)$ . The intensity will be maximum when $\sin^2(2\theta)$ is maximum (i.e., equal to 1). This occurs when $2\theta = \pi/2$ or $\theta = \pi/4$ (which is $45^{\circ}$ ).

Wave OpticsClass 12 Physics Notes