Wave Optics

Class 12 PCM . Physics . Physics Part II . Wave Optics

Chapter 2: Wave Optics

Ch 2

Physics

Class 12 PCM

Lecture 1

Introduction

Reference video · Lecture 1Introduction

Corpuscular and Wave Models of Light

Evidence for the Wave Nature of Light

Geometrical Optics

Maxwell’s Electromagnetic Theory

Topics in This Chapter

Quick recap

Descartes’ corpuscular model and Huygens’ wave model made opposite predictions about light’s speed when it bends towards the normal.
Foucault’s 1850 result that light is slower in water than in air supported the wave prediction.
Young’s 1801 interference experiment and later diffraction observations established strong evidence for light’s wave nature.
Geometrical optics neglects wavelength; a ray represents energy propagation as wavelength tends to zero.
Maxwell identified light as an electromagnetic wave that propagates through vacuum.
Interference uses superposition, diffraction uses the Huygens–Fresnel principle, and polarisation reflects light’s transverse nature.

Lecture 2

Huygens Principle

Reference video · Lecture 2Huygens Principle

Wavefronts and Their Types

A point source produces diverging spherical wavefronts.

A small part of a distant spherical wavefront is approximated as a plane wavefront.

Huygens’ Principle

Secondary wavelets form the forward-moving spherical wavefront; no backward wavefront forms.

Huygens’ construction for a plane wavefront moving forward.

Refraction of a Plane Wave

Wavefront Construction: Angle Relations

$$\sin i=\frac{BC}{AC}=\frac{v_1\tau}{AC},\qquad \sin r=\frac{AE}{AC}=\frac{v_2\tau}{AC}$$

From the geometry of the construction:

Refraction Relation

$$\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$$

Dividing the angle relations gives:

A plane wavefront refracts at a boundary and bends towards the normal when it enters a slower medium.

Refractive Index and Snell’s Law

Refractive index: For light in a medium, refractive index is the speed of light in vacuum divided by its speed in that medium.
Wave-theory prediction: If light bends towards the normal, its speed in the second medium is lower. This prediction is opposite to the corpuscular model’s prediction; experiments support wave theory.
Snell’s law: The speed relation expressed using refractive indices gives the law of refraction.

Refractive Indices

$$n_1=\frac{c}{v_1},\qquad n_2=\frac{c}{v_2}$$

For media 1 and 2, where $c$ is the speed of light in vacuum:

Snell’s Law

$$n_1\sin i=n_2\sin r$$

For light passing between two media:

Wavelength and Frequency During Refraction

Speed, Wavelength and Frequency

$$\frac{v_1}{\lambda_1}=\frac{v_2}{\lambda_2}$$

Since frequency equals wave speed divided by wavelength:

⚠ Common mistake

Frequency does not change during refraction: speed and wavelength change together, while frequency stays constant.

State Huygens’ principle.

Answer: Every point on a wavefront acts as a source of secondary wavelets; the forward common envelope of these wavelets forms the later wavefront.

State Snell’s law and the relation between the angles and wave speeds.

Answer: $n_1\sin i=n_2\sin r$ and $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$.

What happens to light’s speed, wavelength and frequency when it enters a slower medium?

Answer: Its speed and wavelength decrease, while its frequency remains unchanged.

Quick recap

A wavefront is a surface of constant phase; wave energy travels perpendicular to it.
Point sources produce spherical wavefronts; a small distant region can be treated as plane.
Secondary wavelets spread at wave speed, and their forward common envelope gives the later wavefront.
Refraction involves a speed change across a boundary; direction changes only when the incidence is not normal.
Refraction obeys $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$ and $n_1\sin i=n_2\sin r$.
In a slower medium, light’s speed and wavelength decrease but its frequency remains constant.

Lecture 3

Refraction at a rarer medium

Reference video · Lecture 3Refraction at a rarer medium

Refraction into a Rarer Medium

Critical angle

$$\sin i_c=\frac{n_2}{n_1}$$

For light travelling from a denser medium with refractive index $n_1$ into a rarer medium with refractive index $n_2$:

A plane wavefront bends away from the normal as it enters a faster, rarer medium.

Reflection of a Plane Wave

Law of reflection

$$i=r$$

The incident and reflected angles are measured from the normal.

Incident and reflected plane wavefronts at a reflecting surface.

Wavefronts in Optical Devices

★ Remember

Equal travel time: For an object point and its corresponding image point, the travel time is the same along every ray.
Convex lens example: A central ray travels a shorter path but moves more slowly in glass; its travel time matches that of rays near the lens edge.

A prism tilts a plane wavefront; a convex lens and concave mirror form converging spherical wavefronts.

Frequency and Intensity of Light

Wave intensity

$$I\propto A^2$$

Intensity is proportional to the square of the wave amplitude.

Why do reflected and refracted light retain the incident frequency?

Answer: Atoms in the material undergo forced oscillations at the incident frequency and emit light at that same frequency.

Does lower wave speed necessarily mean lower light energy?

Answer: No. Wave energy depends on amplitude, not on the speed of propagation.

What determines light intensity in the wave and photon pictures?

Answer: In the wave picture, intensity is proportional to amplitude squared. For a given frequency in the photon picture, it depends on the number of photons crossing unit area per unit time.

Quick recap

Light entering a rarer medium bends away from the normal and obeys Snell’s law.
At the critical angle, refraction is at $90^\circ$; above it, total internal reflection occurs.
Wavefront construction gives the law of reflection, $i=r$.
Prisms, lenses and mirrors shape wavefronts by delaying or redirecting portions of them.
For an object-image pair, travel time is the same along every ray.
Frequency is unchanged at a boundary; wave intensity is proportional to amplitude squared, while photon intensity depends on photon rate per unit area.

Lecture 4

Coherent and Incoherent Addition of Waves

Reference video · Lecture 4Coherent and Incoherent Addition of Waves

Superposition and Coherent Sources

Two in-phase sources produce nodal and antinodal lines.

Constructive and Destructive Interference

Interference from Two In-Phase Sources

Constructive interference: Waves arrive in phase and reinforce; for equal-amplitude waves the resultant intensity is $4I_0$.
Destructive interference: Waves arrive out of phase by an odd multiple of $\pi$ and cancel; for equal amplitudes the intensity is zero.
Equal-path example: At a point equidistant from the sources, in-phase waves arrive in phase, giving maximum intensity.
Path difference: A difference of $2\lambda$ corresponds to phase difference $4\pi$, so the waves again arrive in phase.
Amplitude assumption: For points far from the sources compared with their separation, the two wave amplitudes are approximately equal.

Constructive Interference Condition

$$\left|S_1P-S_2P\right|=n\lambda,\quad n=0,1,2,\ldots$$

The path difference is an integer multiple of the wavelength.

Destructive Interference Condition

$$\left|S_1P-S_2P\right|=\left(n+\frac{1}{2}\right)\lambda,\quad n=0,1,2,\ldots$$

The path difference is an odd half-wavelength.

Wave paths show constructive interference at Q and destructive interference at R.

Resultant Displacement and Intensity

Two Equal-Amplitude Waves

Phase difference: At a general point, the waves can differ in phase by $\phi$.
Amplitude: The resultant amplitude is $2a\cos(\phi/2)$; intensity depends on the square of the amplitude.
Maximum intensity: For $\phi=0,\pm2\pi,\pm4\pi,\ldots$, the waves reinforce and $I=4I_0$.
Zero intensity: For $\phi=\pm\pi,\pm3\pi,\ldots$, the waves cancel.

Resultant Displacement

$$y=2a\cos\left(\frac{\phi}{2}\right)\cos\left(\omega t+\frac{\phi}{2}\right)$$

For waves with displacements $y_1=a\cos\omega t$ and $y_2=a\cos(\omega t+\phi)$:

Intensity for Coherent Waves

$$I=4I_0\cos^2\left(\frac{\phi}{2}\right)$$

$I_0$ is the intensity due to either one of the equal-amplitude sources.

Path Difference and Phase Difference

Relation: A path difference of $\Delta r$ produces phase difference $\phi=2\pi\Delta r/\lambda$.
Maxima: $\Delta r=n\lambda$ corresponds to phase difference $2n\pi$.
Minima: $\Delta r=(n+\tfrac12)\lambda$ corresponds to phase difference $(2n+1)\pi$.

Loci of points with integer-multiple wavelength path differences.

Incoherent Addition of Waves

Time-Averaged Intensity

Incoherent sources: Their phase difference changes rapidly with time, so maxima and minima shift rapidly.
Observed result: The rapidly changing pattern is seen as a time-averaged intensity; the individual intensities add.
Example: Two separate light sources illuminating a wall are incoherent.

Average Intensity for Two Incoherent Equal Sources

$$I=2I_0$$

For sources each producing intensity $I_0$, the average intensity is the sum of their intensities.

Quick recap

Superposition: Resultant displacement is the vector sum of the individual wave displacements.
Coherence: A constant phase difference gives a stable interference pattern.
Constructive interference: Path difference $n\lambda$ gives maximum intensity $4I_0$ for equal sources.
Destructive interference: Path difference $(n+\tfrac12)\lambda$ gives zero intensity for equal sources.
Resultant displacement: $y=2a\cos(\phi/2)\cos(\omega t+\phi/2)$; the amplitude is $2a\cos(\phi/2)$.
Coherent-wave intensity: $I=4I_0\cos^2(\phi/2)$.
Incoherent sources: Rapidly varying phase difference gives average intensity $2I_0$ for two equal sources.

Lecture 5

Interference of Light Waves and Young’s Experiment

Reference video · Lecture 5Interference of Light Waves and Young’s Experiment

Interference and Coherence

Young’s Double-Slit Experiment

Definition

Path difference — The difference between the distances travelled by the two waves from their sources to a point on the screen.

Symbols and screen geometry

$d$ is the separation between the two pinholes; $D$ is their distance from the screen.
$x$ is the displacement of a fringe from the central point; $\lambda$ is the wavelength of light.
Path difference At displacement $x$, the path difference is approximately $\Delta = xd/D$.

Bright and Dark Fringes

Conditions and positions

Bright fringe Constructive interference occurs when the path difference is an integer multiple of the wavelength: $\Delta=n\lambda$.
Dark fringe Destructive interference occurs when the path difference is a half-integer multiple of the wavelength: $\Delta=(n+\frac{1}{2})\lambda$.
Order Here, $n$ is an integer; the central bright fringe corresponds to $n=0$.

Bright-fringe positions

$$x_n=\frac{n\lambda D}{d}$$

Use for constructive interference on the screen.

Dark-fringe positions

$$x_n=\frac{(n+\frac{1}{2})\lambda D}{d}$$

Use for destructive interference on the screen.

Fringe width

$$\beta=\frac{\lambda D}{d}$$

The equal spacing between successive bright fringes, or successive dark fringes.

Why do two independent sodium lamps not produce stable interference fringes?

Answer: Their phase relationship changes abruptly and is not fixed. The light is incoherent, so the intensities add instead of forming stable fringes.

How does Young obtain coherent sources in the double-slit experiment?

Answer: He illuminates two nearby pinholes with light from the same source, so the emerging waves share phase changes and remain coherent.

Quick recap

Stable interference requires coherent sources with a fixed phase relationship.
Independent ordinary sources are incoherent, so their intensities add without stable fringes.
Young’s two nearby pinholes act as coherent sources when illuminated by the same original source.
The path difference at screen displacement $x$ is approximately $xd/D$.
Bright fringes occur at $x_n=n\lambda D/d$; dark fringes occur at $x_n=(n+\frac{1}{2})\lambda D/d$.
Bright and dark fringes are equally spaced, with fringe width $\beta=\lambda D/d$.

Lecture 6

Diffraction

Reference video · Lecture 6Diffraction

Definition

Diffraction — The spreading of waves around obstacles or through narrow openings into regions of geometrical shadow.

When diffraction matters

All waves can diffract, including sound, light, water and matter waves.
Visibility: Light’s wavelength is much smaller than most everyday obstacles, so its diffraction is usually hard to notice.
Applications and effects: Diffraction limits the resolution of eyes and optical instruments; the colours seen on a CD arise from diffraction.

Diffraction at a Single Slit

Setup and wavefront idea

Setup: A parallel monochromatic beam falls normally on a slit of width $a$; the diffracted light reaches a screen.
Secondary sources: Treat each small part of the slit as a source of waves. They are initially in phase because the incoming wavefront is parallel to the slit.
At an angle: For a screen point viewed at angle $\theta$ from the normal, waves from different parts of the slit travel different distances and acquire phase differences.

Single-slit geometry showing the paths from different slit points to a screen point.

Path difference across the slit

$$a\sin\theta$$

The path difference between waves from the two edges of the slit is:

Minima

$$\theta\approx\frac{n\lambda}{a},\qquad n=\pm1,\pm2,\pm3,\ldots$$

Destructive cancellation gives zero intensity at the minima. For small angles, their positions are approximately:

Intensity maxima

$$\theta\approx\left(n+\frac{1}{2}\right)\frac{\lambda}{a}$$

There is a central maximum at zero angle; approximate positions of the weaker secondary maxima are:

Single-slit pattern

Central bright region: The central maximum is broad and lies at $\theta=0$.
Side fringes: Dark minima and bright secondary maxima alternate on both sides.
Intensity: Secondary maxima become progressively weaker farther from the centre.

A single slit produces a broad central bright band with weaker bright and dark fringes on either side.

Interference vs Diffraction

Point	Interference	Diffraction
Usual description	Often used when a small number of sources, such as two, combine.	Often used when contributions from many parts of a wavefront combine.
Physical basis	Superposition of waves.	Superposition of waves; there is no sharp fundamental physical distinction from interference.
Double-slit pattern	The two slits produce an interference pattern.	Each slit also produces a single-slit diffraction pattern; the observed pattern is their superposition.

Wavelength and fringe width

Wavelength dependence: Fringe positions depend on wavelength, so white light can produce coloured bands.
Red and blue light: Red light has a longer wavelength than blue light, so its fringes are wider.

Energy in Diffraction and Interference

Redistribution: Light energy shifts between regions: reduced intensity at dark fringes is balanced by increased intensity in bright regions.
Conservation: Interference and diffraction redistribute energy; they do not create or destroy it.

State the condition for minima in single-slit diffraction.

Answer: For small angles, minima occur at $\theta\approx n\lambda/a$, where $n=\pm1,\pm2,\pm3,\ldots$.

Describe the single-slit diffraction pattern.

Answer: It has a broad central maximum, with alternating dark minima and weaker secondary maxima on both sides. The secondary maxima decrease in intensity away from the centre.

How does a double-slit pattern combine interference and diffraction?

Answer: The observed pattern is the superposition of the single-slit diffraction pattern from each slit and the interference pattern from the two slits.

Quick recap

Diffraction is the spreading of waves around obstacles or through narrow openings.
A single slit acts as many in-phase secondary sources whose waves combine at the screen.
Single-slit minima occur approximately at $\theta=n\lambda/a$ for nonzero integer $n$.
The pattern has a broad central maximum and weaker secondary maxima on either side.
A double-slit pattern combines single-slit diffraction with two-slit interference.
Diffraction and interference redistribute light energy without changing the total.

Lecture 7

Polarisation

Reference video · Lecture 7Polarisation

Transverse and Polarised Waves

Definition

Transverse wave — A wave is transverse when its displacement is perpendicular to its direction of propagation.

Wave Description

String wave: If a string moves up and down while the wave travels along $+x$, displacement is along $y$ and is perpendicular to propagation.
Wave equation: A sinusoidal wave travelling along $+x$ is described by $y(x,t)=a\sin(kx-\omega t)$.
Parameters: $a$ is amplitude, $\omega=2\pi\nu$ is angular frequency, $\nu$ is frequency, and $\lambda=\frac{2\pi}{k}$ is wavelength.
Linear polarisation: A wave whose vibrations are restricted to one fixed direction is linearly polarised.
Plane polarisation: A wave whose vibrations remain confined to a single plane is plane-polarised.

Wave Vibration Directions

Wave type	Vibration	Example
Y-polarised	Along $y$	$y(x,t)=a\sin(kx-\omega t)$
Z-polarised	Along $z$	$z(x,t)=a\sin(kx-\omega t)$
Unpolarised	Direction changes randomly, but remains perpendicular to propagation	Ordinary light

Sinusoidal string-wave displacement at two times and at a fixed point.

Polarisation of Light

Definition

Polarisation — Polarisation is the restriction of a transverse wave’s vibrations to a particular direction or plane.

Light and Polaroids

Light waves: Light is transverse; its electric field oscillates perpendicular to the direction of propagation.
Polaroid structure: A polaroid contains long-chain molecules aligned in one direction.
Selective absorption: Electric-field oscillations parallel to the aligned molecules are absorbed; the transmitted field oscillates perpendicular to them.
Pass-axis: The direction of the transmitted electric-field oscillation is the polaroid’s pass-axis.

Transmission through Polaroids

Unpolarised input: A polaroid transmits half the incident intensity and produces linearly polarised light.
Rotation of one polaroid: Rotating it does not change the intensity transmitted from unpolarised light, because the input has no preferred vibration direction.
Two polaroids: The first polarises the light; the second controls transmission according to the angle between their pass-axes.
Crossed polaroids: Pass-axes at $90^\circ$ ideally block the polarised beam; parallel axes transmit the maximum intensity.

Malus’ Law

$$I=I_0\cos^2\theta$$

For polarised light incident on a second polaroid, transmitted intensity depends on the angle between the pass-axes.

Using Malus’ Law

Symbols: $I_0$ is the intensity of polarised light incident on the second polaroid; $\theta$ is the angle between the pass-axes.
Reason: The transmitted electric-field amplitude is proportional to $E\cos\theta$; intensity is proportional to amplitude squared.
Parallel axes: At $\theta=0^\circ$, $I=I_0$.
Crossed axes: At $\theta=90^\circ$, $I=0$ ideally.

Transmission through two polaroids as the angle between their pass-axes changes.

A Polaroid between Crossed Polaroids

Setup: The first and final polaroids are crossed. A third polaroid has its pass-axis at angle $\theta$ to the first.
After the middle polaroid: The intensity is $I_0\cos^2\theta$, where $I_0$ is the intensity after the first polaroid.
After the final polaroid: The angle between the middle and final axes is $\frac{\pi}{2}-\theta$; hence $I=I_0\cos^2\theta\cos^2(\frac{\pi}{2}-\theta)=\frac{I_0}{4}\sin^2(2\theta)$.
Maximum transmission: The intensity is maximum at $\theta=\frac{\pi}{4}$ (or $45^\circ$), giving $I_{\max}=\frac{I_0}{4}$.
Zero transmission: The final intensity is zero when the middle polaroid is parallel to either crossed outer polaroid.

Applications

Intensity control: Polaroids vary transmitted light by changing the angle between their pass-axes.
Uses: Polaroids are used in sunglasses, windowpanes, photographic cameras and 3D movie cameras.

⚠ Common mistake

Do not apply Malus’ law directly to unpolarised light: account first for the initial polaroid, which reduces its intensity to half.

State Malus’ law and explain its symbols.

Answer: For polarised light incident on a second polaroid, $I=I_0\cos^2\theta$, where $I_0$ is the incident polarised intensity and $\theta$ is the angle between the pass-axes.

What is the transmitted intensity when a polaroid is placed between two crossed polaroids?

Answer: If the middle polaroid is at angle $\theta$ to the first and the intensity after the first is $I_0$, then $I=\frac{I_0}{4}\sin^2(2\theta)$. It is maximum, $\frac{I_0}{4}$, at $\theta=45^\circ$.

Quick recap

A transverse wave vibrates perpendicular to propagation; a linearly or plane-polarised wave has vibrations restricted to a fixed direction or plane.
Light is transverse, and its electric field oscillates perpendicular to propagation.
A polaroid absorbs the electric-field component parallel to its aligned molecules and transmits the component along its pass-axis.
An ideal polaroid transmits half the intensity of unpolarised light.
For polarised light, transmission through a second polaroid follows Malus’ law: $I=I_0\cos^2\theta$.
A middle polaroid between crossed polaroids gives maximum transmission at $45^\circ$, with intensity $I_0/4$.
Polaroids control light intensity and are used in sunglasses, windowpanes and cameras.

Lecture 8

Summary and Points to Ponder

Reference video · Lecture 8Summary and Points to Ponder

Huygens’ Principle and Wavefronts

Superposition and Interference

Single-Slit Diffraction

Polarisation

Why Wave Optics Matters

Quick recap

Huygens’ principle builds a later wavefront from secondary wavelets.
Reflection and refraction follow from Huygens’ construction.
Stable interference requires equal frequencies and a stable phase difference.
Young’s double slit produces equally spaced fringes; a single slit produces a central diffraction maximum and minima.
Diffraction limits optical resolution, while polarisation identifies transverse-wave behaviour.

Physics — ALL CHAPTERS

Electric Charges And Fields

View

Electrostatic Potential And Capacitance

View

Current Electricity

View

Moving Charges And Magnetism

View

Magnetism And Matter

View

Electromagnetic Induction

View

Alternating Current

View

Electromagnetic Waves

View

Ray Optics And Optical Instruments

View

Wave Optics

View

Dual Nature Of Radiation And Matter

View

Atoms

View

Nuclei

View

Semiconductor Electronics: Materials, Devices And Simple Circuits

View

Wave type	Vibration	Example
Y-polarised	Along \(y\)	\(y(x,t)=a\sin(kx-\omega t)\)
Z-polarised	Along \(z\)	\(z(x,t)=a\sin(kx-\omega t)\)
Unpolarised	Direction changes randomly, but remains perpendicular to propagation	Ordinary light