Module 7: The Nature of Light | Year 12 Physics HSC | StudyCave

Pillar 1 of 3

Electromagnetic Waves & The Wave Model

Explore Maxwell's unification of electricity and magnetism, Hertz's experimental validation and the wave properties of light through interference, diffraction and polarisation.

Maxwell's Unification

James Clerk Maxwell unified the theories of electricity and magnetism into a single framework, predicting that electromagnetic waves could propagate through space without requiring a medium. This revolutionary insight came from his equations showing that changing electric fields create magnetic fields and changing magnetic fields create electric fields.

Maxwell's Prediction

Self-Propagating Electromagnetic Waves:

Maxwell's equations predicted that a changing electric field generates a changing magnetic field, which in turn generates a changing electric field, creating a self-sustaining wave that propagates through space.

Speed of Light Calculation:

c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}

Where:

$c$ = speed of light (m/s)
$\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}$ (permeability of free space)
$\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$ (permittivity of free space)

Key Insight: When Maxwell calculated this value, he got approximately $3 \times 10^8 \text{ m/s}$ , which exactly matched the known speed of light! This suggested that light itself was an electromagnetic wave.

The Four Maxwell Equations:

Gauss's Law (Electric)

Electric field lines originate from positive charges and terminate on negative charges. The total electric flux through a closed surface is proportional to the enclosed charge.

Gauss's Law (Magnetic)

There are no magnetic monopoles. Magnetic field lines always form closed loops. The total magnetic flux through any closed surface is zero.

Faraday's Law

A changing magnetic field induces an electric field. This is the principle behind electromagnetic induction and electric generators.

Ampère-Maxwell Law

Magnetic fields are produced by electric currents AND by changing electric fields (Maxwell's addition). This was the key to predicting electromagnetic waves.

Worked Example

Calculate the speed of electromagnetic waves using Maxwell's equation $c = 1/\sqrt{\mu_0 \epsilon_0}$ . Compare this to the accepted speed of light.

Given: $\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}, \quad \epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$

Solution:

$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$

$c = \frac{1}{\sqrt{(4\pi \times 10^{-7})(8.85 \times 10^{-12})}}$

$c = \frac{1}{\sqrt{1.11 \times 10^{-17}}}$

$c = \frac{1}{3.33 \times 10^{-9}}$

$c = \mathbf{3.00 \times 10^8 \text{ m/s}}$

This matches the experimentally measured speed of light exactly! This remarkable agreement led Maxwell to conclude that light must be an electromagnetic wave.

Practice Question

Explain how Maxwell's equations predict self-propagating electromagnetic waves. What is the significance of the calculated wave speed matching the speed of light?

Show Answer

Explanation:

Maxwell's equations show that a changing electric field creates a changing magnetic field (Ampère-Maxwell Law) and a changing magnetic field creates a changing electric field (Faraday's Law). Once started, this process becomes self-sustaining:

An oscillating electric field generates an oscillating magnetic field perpendicular to it
This oscillating magnetic field generates another oscillating electric field perpendicular to both
The wave propagates through space without needing a medium
The E and B fields are perpendicular to each other and to the direction of propagation (transverse wave)

Significance of matching speed: When Maxwell calculated the speed of these predicted electromagnetic waves using fundamental constants from electricity and magnetism, he got $3 \times 10^8 \text{ m/s}$ - exactly the measured speed of light. This was too precise to be coincidence, leading Maxwell to conclude that light IS an electromagnetic wave. This unified optics with electromagnetism.

Hertz's Experiment

Heinrich Hertz experimentally validated Maxwell's predictions in 1887 by producing and detecting electromagnetic waves (radio waves). His experiments proved that these waves could reflect, refract, interfere and polarise - behaving exactly like light.

Hertz's Apparatus

Production (Spark Gap Transmitter):

Design: Two metal spheres connected to a high-voltage AC source (induction coil) with a small gap between them.

Operation: When voltage was high enough, sparks jumped across the gap. These sparks represented rapidly oscillating currents, which according to Maxwell should produce electromagnetic waves.

Result: The oscillating charges in the spark gap radiated electromagnetic waves into the surrounding space.

Detection (Receiver Loop):

Design: A circular wire loop with a small gap, placed several meters away from the transmitter.

Operation: When electromagnetic waves from the transmitter reached the receiver, the changing magnetic field induced an EMF in the loop (Faraday's Law).

Result: Small sparks appeared in the receiver gap, proving that electromagnetic energy had traveled through space without wires.

Wave Properties Demonstrated:

Reflection

Hertz placed metal sheets behind the transmitter and observed that the reflected waves could interfere with direct waves, creating standing wave patterns. This proved electromagnetic waves obey the law of reflection.

Refraction

By passing waves through large prisms made of pitch (asphalt), Hertz showed that electromagnetic waves change direction when entering a different medium, just like light.

Interference

Using metal reflectors, Hertz created interference patterns where waves from the transmitter and reflected waves combined. Nodes (weak signal) and antinodes (strong signal) appeared at regular intervals.

Polarization

Hertz oriented the receiver loop at different angles. Maximum detection occurred when the loop was parallel to the transmitter's orientation. This proved electromagnetic waves are transverse (not longitudinal).

Why Polarization Proves Transverse Nature

Key Concept: Only transverse waves can be polarised. Longitudinal waves (like sound) cannot.

Transverse Waves: Oscillations are perpendicular to the direction of propagation. The wave can oscillate in any plane perpendicular to the direction of travel. Polarization restricts oscillation to one specific plane.

Longitudinal Waves: Oscillations are parallel to the direction of propagation. There's only one direction to oscillate (forward/backward), so the concept of polarization doesn't apply.

Hertz's Evidence: When the receiver loop was perpendicular to the transmitter orientation, no sparks appeared. This directional dependence is only possible if the waves are transverse, with the electric field oscillating in a specific plane.

Practice Question

Describe Hertz's experimental setup and explain how his observation of polarization provided evidence that electromagnetic waves are transverse, not longitudinal.

Show Answer

Experimental Setup:

Hertz used a spark gap transmitter (two metal spheres with high voltage AC creating oscillating sparks) to produce electromagnetic waves. A receiver loop (circular wire with a small gap) was placed several meters away to detect the waves through induced sparks.

Polarization Evidence:

When Hertz rotated the receiver loop, he observed that:

Maximum sparks (strongest signal) when receiver loop was parallel to transmitter orientation
No sparks (no signal) when receiver loop was perpendicular to transmitter orientation
Intermediate signal at intermediate angles

Why This Proves Transverse Nature:

This angular dependence can only occur if the electromagnetic waves are transverse. In a transverse wave, the electric field oscillates in a specific plane perpendicular to propagation. The receiver detects the wave best when aligned with this oscillation plane. Longitudinal waves (like sound) oscillate parallel to propagation and have no preferred orientation - they cannot be polarised. The fact that Hertz could "block" the signal by rotating the receiver 90° proved electromagnetic waves must be transverse, exactly like light.

Spectroscopy

Spectroscopy is the analysis of light by separating it into its component wavelengths (spectrum). By analyzing stellar spectra, astronomers can determine a star's temperature, chemical composition and velocity without ever visiting it.

Three Key Stellar Properties from Spectra:

1. Surface Temperature (Wien's Law)

Method: Analyze the peak wavelength (dominant color) of the continuous spectrum using Wien's Displacement Law.

\lambda_{max} T = b

where $b = 2.898 \times 10^{-3} \text{ mK}$ (Wien's constant)

Example: A blue star has peak wavelength ~400 nm → Temperature ~7000 K. A red star has peak wavelength ~700 nm → Temperature ~4000 K.

2. Translational Velocity (Doppler Shift)

Method: Analyze the shift in spectral line positions compared to laboratory reference wavelengths.

Red Shift: Spectral lines shifted toward longer wavelengths (red end) → Star moving away from Earth

Blue Shift: Spectral lines shifted toward shorter wavelengths (blue end) → Star moving toward Earth

Formula:

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

For velocities much less than $c$

3. Chemical Composition (Absorption Lines)

Method: Match dark absorption lines in the spectrum to known element "fingerprints."

How it works: Each element absorbs specific wavelengths of light. When starlight passes through the star's cooler outer atmosphere, atoms absorb their characteristic wavelengths, creating dark lines.

Hydrogen lines: Balmer series (visible) - 656 nm (red), 486 nm (blue-green), 434 nm (violet)

Helium lines: 587.6 nm (yellow), 501.6 nm (green)

Analysis: By comparing the pattern of absorption lines to laboratory spectra, astronomers identify which elements are present in the star's atmosphere.

Worked Example

A star's spectrum shows peak intensity at wavelength 500 nm. A hydrogen absorption line, normally at 656 nm in the laboratory, appears at 658 nm in the star's spectrum. Calculate: (a) the surface temperature of the star, (b) the radial velocity of the star.

Given:

$\lambda_{max} = 500 \text{ nm} = 5.0 \times 10^{-7} \text{ m}$

$\lambda_{lab} = 656 \text{ nm}, \quad \lambda_{obs} = 658 \text{ nm}$

$b = 2.898 \times 10^{-3} \text{ mK}$

(a) Surface temperature:

Using Wien's Law: $\lambda_{max} T = b$

$T = \frac{b}{\lambda_{max}}$

$T = \frac{2.898 \times 10^{-3}}{5.0 \times 10^{-7}}$

$T = \mathbf{5796 \text{ K}}$

Similar to our Sun's temperature (~5800 K)

(b) Radial velocity:

Wavelength shift: $\Delta\lambda = 658 - 656 = 2 \text{ nm}$

Since shift is toward red (longer wavelength), star is moving away.

$\frac{\Delta\lambda}{\lambda} = \frac{v}{c}$

$v = c \times \frac{\Delta\lambda}{\lambda}$

$v = (3.0 \times 10^8) \times \frac{2}{656}$

$v = \mathbf{9.15 \times 10^5 \text{ m/s}}$

Star is moving away at approximately 915 km/s

Practice Question

A distant galaxy shows a hydrogen line that should be at 486 nm (laboratory value) but appears at 510 nm. Calculate the galaxy's recession velocity and explain what this tells us about the universe.

Show Answer

Calculation:

Wavelength shift: $\Delta\lambda = 510 - 486 = 24 \text{ nm}$

This is a red shift (moving away).

$v = c \times \frac{\Delta\lambda}{\lambda} = (3.0 \times 10^8) \times \frac{24}{486}$

$v = \mathbf{1.48 \times 10^7 \text{ m/s} = 14,800 \text{ km/s}}$

Cosmic Significance:

This significant red shift indicates the galaxy is receding at about 5% the speed of light. Nearly all distant galaxies show red shift, with more distant galaxies having larger red shifts (Hubble's Law).

Conclusion: This is evidence for an expanding universe. Space itself is expanding, carrying galaxies away from each other. This discovery supports the Big Bang theory.

Interference: Young's Double Slit

When light from a coherent source passes through two narrow slits, an interference pattern of alternating bright and dark fringes appears on a screen. This phenomenon can only be explained by the wave nature of light and was crucial evidence against the particle theory.

Conditions for Interference

Coherent Sources:

The two sources must have constant phase difference. In Young's experiment, this is achieved by using a single source split into two paths, ensuring the waves maintain a fixed phase relationship.

Monochromatic Light:

Single wavelength (single color) light produces clear, sharp fringes. White light produces colored fringes that overlap, making the pattern less distinct.

Similar Amplitudes:

The two waves should have similar intensities for maximum contrast between bright and dark fringes.

Path Difference and Interference

Constructive Interference (Bright Fringes):

\text{Path Difference (pd)} = m\lambda

where $m = 0, 1, 2, 3, \ldots$ (order number)

Waves arrive in phase. Crests meet crests, troughs meet troughs. Amplitudes add, producing maximum intensity.

Destructive Interference (Dark Fringes):

\text{Path Difference (pd)} = (m + 0.5)\lambda

where $m = 0, 1, 2, 3, \ldots$

Waves arrive out of phase by half a wavelength. Crests meet troughs. Amplitudes cancel, producing minimum intensity (darkness).

Double Slit Formula:

d\sin\theta = m\lambda

Where:

$d$ = slit separation (m)
$\theta$ = angle from center to the fringe
$m$ = order number (0 = central, 1 = first, 2 = second, etc.)
$\lambda$ = wavelength (m)

For Small Angles (typical setup):

x = \frac{m\lambda L}{d}

Where:

$x$ = distance from central bright fringe to mth bright fringe
$L$ = distance from slits to screen (m)

Worked Example

In a double-slit experiment, red light (λ = 650 nm) passes through two slits separated by 0.30 mm. The screen is 2.0 m away. Calculate: (a) the angle to the first bright fringe, (b) the distance from the central maximum to the third bright fringe.

Given:

$\lambda = 650 \text{ nm} = 6.5 \times 10^{-7} \text{ m}$

$d = 0.30 \text{ mm} = 3.0 \times 10^{-4} \text{ m}$

$L = 2.0 \text{ m}$

(a) Angle to first bright fringe (m = 1):

$d\sin\theta = m\lambda$

$\sin\theta = \frac{m\lambda}{d} = \frac{(1)(6.5 \times 10^{-7})}{3.0 \times 10^{-4}}$

$\sin\theta = 2.167 \times 10^{-3}$

$\theta = \mathbf{0.124°}$

Very small angle - typical for interference experiments

(b) Distance to third bright fringe (m = 3):

For small angles: $x = \frac{m\lambda L}{d}$

$x = \frac{(3)(6.5 \times 10^{-7})(2.0)}{3.0 \times 10^{-4}}$

$x = \frac{3.9 \times 10^{-6}}{3.0 \times 10^{-4}}$

$x = \mathbf{0.013 \text{ m} = 13 \text{ mm}}$

Practice Question

In a double-slit experiment, the third bright fringe is observed at a distance of 7.5 mm from the central maximum. The screen is 1.5 m from the slits, which are separated by 0.50 mm. Calculate the wavelength of light used.

Show Answer

Solution:

Using $x = \frac{m\lambda L}{d}$

Rearrange for wavelength: $\lambda = \frac{xd}{mL}$

Substitute values:

$\lambda = \frac{(7.5 \times 10^{-3})(5.0 \times 10^{-4})}{(3)(1.5)}$

$\lambda = \frac{3.75 \times 10^{-6}}{4.5}$

$\lambda = \mathbf{8.33 \times 10^{-7} \text{ m} = 833 \text{ nm}}$

This is in the near-infrared range, just beyond visible red light.

Diffraction

Diffraction is the bending of waves around obstacles or through openings. It's a characteristic behavior of waves and provides strong evidence for the wave nature of light. The amount of diffraction depends on the relationship between wavelength and obstacle/opening size.

Condition for Significant Diffraction

Significant diffraction occurs when:

\lambda \approx w

Wavelength is comparable to slit width/obstacle size

λ ≪ w

Wavelength much smaller than opening

Minimal diffraction → Sharp shadow

λ ≈ w

Wavelength comparable to opening

Maximum diffraction → Waves spread

λ ≫ w

Wavelength much larger than opening

Complete spreading → Circular wavefronts

Single-Slit Diffraction Pattern:

When light passes through a single narrow slit, it produces a characteristic diffraction pattern:

Central maximum: Bright, wide band in the center
Secondary maxima: Fainter bright bands on either side
Minima: Dark bands between bright regions

Position of Minima (Dark Bands):

w\sin\theta = m\lambda

Where $m = 1, 2, 3, \ldots$ (NOT m = 0)

( $w$ = slit width, $\theta$ = angle to minimum)

Note: The central maximum is about twice as wide as the secondary maxima. Intensity decreases rapidly for higher-order maxima.

Why Light Doesn't Diffract Around Large Objects:

Visible light has wavelengths of 400-700 nm (extremely small). Most everyday objects are millions of times larger.

Example:

A doorway (w ≈ 1 m) compared to visible light (λ ≈ 500 nm):

$\frac{w}{\lambda} = \frac{1}{5 \times 10^{-7}} = 2,000,000$

Since λ ≪ w, diffraction is negligible. Light travels in straight lines, creating sharp shadows.

But sound DOES diffract around doorways:

Sound waves have wavelengths of ~0.1-10 m (comparable to room dimensions), so $\lambda \approx w$ . This is why you can hear someone talking in the next room even if you can't see them.

Worked Example

Light of wavelength 600 nm passes through a single slit of width 0.12 mm. Calculate the angle to the first minimum in the diffraction pattern.

Given:

$\lambda = 600 \text{ nm} = 6.0 \times 10^{-7} \text{ m}$

$w = 0.12 \text{ mm} = 1.2 \times 10^{-4} \text{ m}$

$m = 1$ (first minimum)

Solution:

Using $w\sin\theta = m\lambda$ :

$\sin\theta = \frac{m\lambda}{w}$

$\sin\theta = \frac{(1)(6.0 \times 10^{-7})}{1.2 \times 10^{-4}}$

$\sin\theta = 5.0 \times 10^{-3}$

$\theta = \mathbf{0.29°}$

The first dark fringe appears at a very small angle, creating a wide central bright band. This is typical for single-slit diffraction.

Practice Question

Explain why radio waves (λ ≈ 1 m) can diffract around buildings and hills, allowing you to receive signals even without direct line of sight, but light waves cannot.

Show Answer

Answer:

Diffraction is significant when the wavelength is comparable to the size of the obstacle or opening ( $\lambda \approx w$ ).

Radio waves:

Wavelength ~1 m, which is comparable to buildings (~10-100 m) and hills (~100-1000 m). Since $\lambda \approx w$ , significant diffraction occurs. Radio waves bend around obstacles, spreading into shadow regions.

Light waves:

Wavelength ~500 nm (5 × 10⁻⁷ m), which is millions of times smaller than buildings. Since $\lambda \ll w$ , diffraction is negligible. Light travels in straight lines and creates sharp shadows.

Conclusion:

The enormous difference in wavelength (factor of ~2,000,000) explains why radio waves behave so differently from light waves when encountering obstacles, even though both are electromagnetic waves.

Diffraction Gratings

A diffraction grating consists of many parallel slits (typically thousands per centimeter). It produces much sharper and more widely separated interference maxima than a double slit, making it ideal for precise spectroscopy.

Diffraction Grating Equation

d\sin\theta = m\lambda

Where:

$d$ = grating spacing (distance between adjacent slits)
$\theta$ = angle to the mth order maximum
$m$ = order number (0, 1, 2, 3, ...)
$\lambda$ = wavelength

Important: Calculating d from lines per mm

If a grating has N lines per millimeter:

d = \frac{1}{N} \text{ mm} = \frac{10^{-3}}{N} \text{ m}

Example: 600 lines/mm → $d = 1/600 \text{ mm} = 1.67 \times 10^{-6} \text{ m}$

Why Gratings Produce Sharper Maxima:

Double Slit (N = 2)

Only 2 waves interfere. Maxima are relatively broad with gradual intensity falloff. Good for demonstrating interference but poor for precise measurements.

Diffraction Grating (N = 1000+)

Thousands of waves interfere. Constructive interference occurs only at very specific angles. Maxima are extremely sharp and bright. Ideal for spectroscopy and wavelength measurement.

Physical Explanation:

With many slits, even a tiny deviation from the exact angle causes destructive interference from multiple sources. The more slits, the narrower the angular range where constructive interference survives. This creates sharp, well-defined peaks separated by wide dark regions.

Maximum Order Number:

Since $\sin\theta$ cannot exceed 1, the maximum possible order is limited:

d\sin\theta = m\lambda \implies m_{max} = \frac{d}{\lambda}

At $\theta = 90°$ , $\sin\theta = 1$ , so:

m_{max} = \frac{d}{\lambda}

Round down to the nearest integer since m must be a whole number.

Worked Example

A diffraction grating has 400 lines per millimeter. Green light (λ = 550 nm) is incident normally on the grating. Calculate: (a) the grating spacing d, (b) the angle to the second-order maximum, (c) the maximum order visible.

Given:

$N = 400 \text{ lines/mm}, \quad \lambda = 550 \text{ nm} = 5.5 \times 10^{-7} \text{ m}$

(a) Grating spacing:

$d = \frac{1}{N} = \frac{1}{400} \text{ mm} = \frac{10^{-3}}{400}$

$d = \mathbf{2.5 \times 10^{-6} \text{ m}}$

(b) Angle to second-order (m = 2):

$d\sin\theta = m\lambda$

$\sin\theta = \frac{m\lambda}{d} = \frac{(2)(5.5 \times 10^{-7})}{2.5 \times 10^{-6}}$

$\sin\theta = 0.44$

$\theta = \mathbf{26.1°}$

$m_{max} = \frac{d}{\lambda} = \frac{2.5 \times 10^{-6}}{5.5 \times 10^{-7}}$

$m_{max} = 4.545$

Round down: $m_{max} = \mathbf{4}$

Orders visible: 0, ±1, ±2, ±3, ±4 (9 maxima total)

Practice Question

White light is incident on a diffraction grating with 500 lines/mm. The second-order spectrum shows red light (700 nm) at 44.4° and violet light (400 nm) at 23.6°. Explain why the grating separates colors better than a prism and calculate the angular separation between red and violet in the second order.

Show Answer

Angular separation:

$\Delta\theta = \theta_{red} - \theta_{violet} = 44.4° - 23.6°$

$\Delta\theta = \mathbf{20.8°}$

Why gratings are better than prisms:

1. Linear wavelength dependence: From $d\sin\theta = m\lambda$ , angle is directly proportional to wavelength. This makes calibration simple and accurate.

2. Multiple orders: Higher orders (m = 2, 3, ...) spread colors even further apart, increasing resolution. Prisms only have one spectrum.

3. Sharpness: Gratings produce extremely narrow spectral lines (thousands of slits interfering), allowing resolution of closely-spaced wavelengths. Prisms produce broader bands.

4. No absorption: Gratings work by interference (no material needed to be transparent). Prisms absorb some wavelengths, especially UV and IR.

Polarisation & Malus' Law

Plane polarisation restricts the oscillation of a transverse wave to a single plane. For electromagnetic waves, this means the electric field vector oscillates in only one direction. Polarization is definitive evidence that light is a transverse wave.

Polarization Basics

Unpolarised Light:

The electric field oscillates in all possible directions perpendicular to the direction of propagation. Examples: sunlight, light bulb, candle flame.

Plane Polarized Light:

The electric field oscillates in only ONE plane. Can be produced by passing unpolarised light through a polarizing filter (polaroid).

Polarizing Filter (Analyzer):

A material that allows only the component of the electric field parallel to its transmission axis to pass through. The perpendicular component is absorbed.

Malus' Law

When polarised light passes through an analyzer (polarizing filter) at angle $\theta$ to the plane of polarization:

I = I_{\max}\cos^2\theta

Where:

$I$ = transmitted intensity
$I_{\max}$ = incident intensity (maximum when θ = 0°)
$\theta$ = angle between polarization direction and analyzer axis

θ = 0°

$I = I_{\max}$

Parallel alignment: Maximum transmission

θ = 45°

$I = 0.5I_{\max}$

50% intensity transmitted

θ = 90°

$I = 0$

Perpendicular: Complete blocking

Calculator Mode Warning:

Malus' Law uses DEGREES in most physics problems. Make sure your calculator is in degree mode, not radians!

$\cos^2(45°) = 0.5$ ✓ (degrees)

$\cos^2(45 \text{ rad}) = 0.259$ ✗ (wrong!)

Evidence for Transverse Nature:

Why polarization proves light is transverse:

Transverse waves: Oscillations perpendicular to direction of travel. Can oscillate in many perpendicular directions (up-down, left-right, diagonal, etc.). A filter can select one direction, blocking others → POLARIZATION POSSIBLE.

Longitudinal waves: Oscillations parallel to direction of travel. Only one possible direction of oscillation (forward-backward). No way to "filter" direction → POLARIZATION IMPOSSIBLE.

Since light CAN be polarised (demonstrated by polaroid filters reducing intensity), light MUST be a transverse wave. This was crucial evidence against Newton's corpuscular theory and in favor of the wave theory of light.

Worked Example

Polarized light of intensity 100 W/m² passes through a polarizing filter. Calculate the transmitted intensity when the filter is oriented at: (a) 0°, (b) 30°, (c) 60°, (d) 90° to the plane of polarization.

Given:

$I_{\max} = 100 \text{ W/m}^2$

(a) θ = 0° (parallel):

$I = I_{\max}\cos^2(0°) = 100 \times (1)^2$

$I = \mathbf{100 \text{ W/m}^2}$

(b) θ = 30°:

$I = 100\cos^2(30°) = 100 \times (0.866)^2$

$I = \mathbf{75 \text{ W/m}^2}$

$I = 100\cos^2(60°) = 100 \times (0.5)^2$

$I = \mathbf{25 \text{ W/m}^2}$

(d) θ = 90° (perpendicular):

$I = 100\cos^2(90°) = 100 \times (0)^2$

$I = \mathbf{0 \text{ W/m}^2}$

Complete blocking - "crossed polarisers"

Practice Question

Unpolarised light of intensity $I_0$ passes through two polarizing filters. The first filter (polariser) reduces the intensity by half. The second filter (analyzer) is oriented at 60° to the first. Calculate the final transmitted intensity in terms of $I_0$ .

Show Answer

Step 1: First polariser

Unpolarised light passing through a polariser is reduced to half intensity (average of $\cos^2\theta$ over all angles):

$I_1 = \frac{I_0}{2}$

Step 2: Second analyzer at 60°

Apply Malus' Law with $\theta = 60°$ :

$I_2 = I_1\cos^2(60°)$

$I_2 = \frac{I_0}{2} \times (0.5)^2$

$I_2 = \frac{I_0}{2} \times 0.25$

$I_2 = \mathbf{\frac{I_0}{8}}$

The final intensity is 12.5% of the original. The first polariser cuts intensity in half, then the 60° analyzer further reduces it by 75%.

Pillar 2 of 3

The Quantum Model of Light

Explore the particle nature of light through black body radiation, the photoelectric effect, and Planck's revolutionary quantum hypothesis that energy comes in discrete packets.

Black Body Radiation & Ultraviolet Catastrophe

A black body is an idealized object that absorbs all incident radiation and re-emits it based solely on its temperature. The spectrum of this radiation revealed a fundamental failure of classical physics, leading to the birth of quantum mechanics.

Classical Prediction vs Reality

Classical Physics Prediction (Rayleigh-Jeans Law):

According to classical electromagnetic theory, the intensity of radiation should increase continuously as wavelength decreases. At very short wavelengths (UV and beyond), intensity should approach infinity.

This is the "Ultraviolet Catastrophe" - a spectacular failure of classical physics!

Experimental Reality:

Actual black body spectra show:

Intensity peaks at a specific wavelength (depends on temperature)
Intensity drops to zero at both very short and very long wavelengths
Total energy radiated is finite, not infinite

Classical physics could not explain this curve at all.

Planck's Revolutionary Solution (1900)

Planck's Quantum Hypothesis:

Energy is not continuous but comes in discrete packets called "quanta." The energy of each quantum is proportional to its frequency:

E = hf

Where:

$E$ = energy of one photon/quantum (J)
$h = 6.63 \times 10^{-34} \text{ J·s}$ (Planck's constant)
$f$ = frequency (Hz)

Alternative form using wavelength:

Since $c = f\lambda$ , we can write:

E = \frac{hc}{\lambda}

Why this solved the UV catastrophe:

At very short wavelengths (high frequencies), each quantum has enormous energy ( $E = hf$ is huge). It becomes statistically improbable for atoms to have enough energy to emit such high-energy quanta. This naturally suppresses UV and X-ray emission, preventing the catastrophe.

Key Implications:

Energy Quantization

Energy is not infinitely divisible. An atom can only emit or absorb energy in discrete amounts $E = hf$ . This was revolutionary - like saying you can only buy eggs in dozens, never 3.7 eggs.

Birth of Quantum Mechanics

Planck's hypothesis was the first quantum theory. Though initially proposed as a "mathematical trick," it revealed a fundamental truth about nature and opened the door to modern physics.

Planck's Constant

$h = 6.63 \times 10^{-34} \text{ J·s}$ is one of the fundamental constants of nature. Its tiny value explains why quantum effects aren't obvious in everyday life (macroscopic objects involve huge numbers of quanta).

Perfect Fit to Data

Planck's formula matched experimental black body curves perfectly across all temperatures and wavelengths. This agreement was too precise to be coincidence.

Worked Example

Calculate the energy of a single photon of: (a) red light (λ = 700 nm), (b) blue light (λ = 450 nm), (c) X-rays (λ = 1.0 nm). Compare the results and explain the significance.

Given:

$h = 6.63 \times 10^{-34} \text{ J·s}, \quad c = 3.0 \times 10^8 \text{ m/s}$

(a) Red light (λ = 700 nm):

$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{700 \times 10^{-9}}$

$E = \frac{1.989 \times 10^{-25}}{7.0 \times 10^{-7}}$

$E = \mathbf{2.84 \times 10^{-19} \text{ J}}$

Converting to eV: $E = 2.84 \times 10^{-19} / 1.6 \times 10^{-19} = \mathbf{1.78 \text{ eV}}$

(b) Blue light (λ = 450 nm):

$E = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{450 \times 10^{-9}}$

$E = \mathbf{4.42 \times 10^{-19} \text{ J} = 2.76 \text{ eV}}$

$E = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{1.0 \times 10^{-9}}$

$E = \mathbf{1.99 \times 10^{-16} \text{ J} = 1240 \text{ eV}}$

Significance: X-rays have ~700 times more energy per photon than visible light. This is why short-wavelength radiation is dangerous (ionizing) - individual quanta carry enough energy to break chemical bonds and damage DNA.

Practice Question

Explain the ultraviolet catastrophe and how Planck's quantum hypothesis resolved it. Why was this solution so revolutionary?

Show Answer

The Ultraviolet Catastrophe:

Classical physics (Rayleigh-Jeans Law) predicted that a black body should emit infinite energy at short wavelengths (UV and beyond). This absurd result arose because classical theory allowed atoms to oscillate at any frequency with any energy - the higher the frequency, the more modes available, leading to runaway intensity.

Planck's Resolution:

Planck proposed that energy is quantized: $E = hf$ . Atoms can only emit or absorb energy in discrete packets, not continuously. At very high frequencies, each quantum has enormous energy. Statistically, very few atoms have enough thermal energy to emit such high-energy quanta, naturally suppressing UV emission and preventing the catastrophe.

Why Revolutionary:

Broke classical continuity: Energy was always assumed continuous. Quantization was completely unexpected.
Perfect experimental fit: Planck's formula matched black body curves exactly at all temperatures.
Founded quantum mechanics: Though initially seen as a mathematical trick, $E = hf$ revealed a fundamental truth about nature and spawned an entire new physics.
Introduced fundamental constant: Planck's constant $h$ became as important as the speed of light $c$ .

Wien's Displacement Law

Wien's Displacement Law relates the temperature of a black body to the wavelength at which it emits most intensely. Hotter objects emit peak radiation at shorter wavelengths (bluer colors).

Wien's Law Formula

\lambda_{max} T = b

Where:

$\lambda_{max}$ = peak wavelength (m)
$T$ = absolute temperature (K)
$b = 2.898 \times 10^{-3} \text{ m·K}$ (Wien's displacement constant)

Key Insight: $\lambda_{max}$ is inversely proportional to temperature. Double the temperature → half the peak wavelength (shift toward blue/UV).

Applications & Examples:

The Sun (T ≈ 5800 K)

$\lambda_{max} = \frac{2.898 \times 10^{-3}}{5800} = 500 \text{ nm}$

Peak emission in green-yellow (middle of visible spectrum). This is why the Sun appears white - it emits across all visible wavelengths with peak in middle.

Hot Stars (T ≈ 20,000 K)

$\lambda_{max} = \frac{2.898 \times 10^{-3}}{20000} = 145 \text{ nm}$

Peak in UV (invisible), but significant blue emission. These stars appear blue-white because they emit more blue than red in the visible range.

Cool Stars (T ≈ 3000 K)

$\lambda_{max} = \frac{2.898 \times 10^{-3}}{3000} = 966 \text{ nm}$

Peak in near-infrared, but still emit some red light. These stars appear red (red giants, red dwarfs).

Human Body (T ≈ 310 K)

$\lambda_{max} = \frac{2.898 \times 10^{-3}}{310} = 9350 \text{ nm}$

Peak in mid-infrared (~10 $\mu$ m). This is why thermal cameras can detect people - we glow brightly in IR!

Worked Example

A distant star has peak emission at wavelength 400 nm (violet). Calculate: (a) the surface temperature of the star, (b) the peak wavelength if the star's temperature doubled.

Given:

$\lambda_{max} = 400 \text{ nm} = 4.0 \times 10^{-7} \text{ m}$

$b = 2.898 \times 10^{-3} \text{ m·K}$

(a) Surface temperature:

$\lambda_{max} T = b$

$T = \frac{b}{\lambda_{max}}$

$T = \frac{2.898 \times 10^{-3}}{4.0 \times 10^{-7}}$

$T = \mathbf{7245 \text{ K}}$

This is a hot blue-white star, hotter than our Sun (5800 K).

(b) Peak wavelength if temperature doubled:

New temperature: $T' = 2T = 14490 \text{ K}$

$\lambda'_{max} = \frac{b}{T'} = \frac{2.898 \times 10^{-3}}{14490}$

$\lambda'_{max} = \mathbf{2.0 \times 10^{-7} \text{ m} = 200 \text{ nm}}$

Doubling temperature halves peak wavelength. The star would now emit peak radiation in the UV, appearing even bluer.

Practice Question

The filament of an incandescent light bulb operates at 2900 K. Calculate its peak wavelength and explain why incandescent bulbs are inefficient for producing visible light.

Show Answer

Peak wavelength:

$\lambda_{max} = \frac{b}{T} = \frac{2.898 \times 10^{-3}}{2900}$

$\lambda_{max} = \mathbf{1.0 \times 10^{-6} \text{ m} = 1000 \text{ nm}}$

This is in the near-infrared, beyond the visible spectrum (400-700 nm).

Why inefficient:

The peak emission is in the infrared (heat), not visible light. Most of the radiated energy is wasted as heat rather than light.

Rough estimate: At 2900 K, only about 5-10% of emitted radiation is in the visible range (400-700 nm). The rest is infrared, which we feel as heat but can't see.

Comparison: LED bulbs and fluorescent lights produce light through different mechanisms that don't rely on heating, making them 5-10× more efficient. This is why incandescent bulbs have been phased out.

The Photoelectric Effect

The photoelectric effect is the emission of electrons from a metal surface when light shines on it. Classical wave theory failed completely to explain the observed behavior, providing crucial evidence for the particle (photon) nature of light.

Experimental Observations

Observation 1: Threshold Frequency

Below a certain frequency (threshold), NO electrons are emitted regardless of intensity. Above threshold, electrons are emitted immediately.

Observation 2: No Time Delay

Emission is instantaneous (< 10⁻⁹ s) even with very dim light, as long as frequency exceeds threshold.

Observation 3: KE vs Frequency

Maximum kinetic energy of emitted electrons increases linearly with frequency, independent of intensity.

Observation 4: Current vs Intensity

Number of electrons (current) is proportional to light intensity, but their maximum energy is not affected.

Classical Physics Failures

Wave theory predicted (incorrectly):

❌ Any frequency should work if intense enough

Wave energy depends on amplitude (intensity), not frequency. Bright red light should eventually eject electrons.

❌ Time delay expected with dim light

Electrons need to accumulate energy from the wave. Dim light should require seconds/minutes before emission starts.

❌ KE should increase with intensity

More intense light delivers more energy per second, so electrons should be ejected faster (higher KE).

None of these predictions matched experiments. Classical physics was fundamentally wrong about light-matter interactions.

Einstein's Photon Explanation (1905)

Key Idea: Light consists of discrete particles (photons), each carrying energy $E = hf$ . One photon interacts with one electron in an all-or-nothing event.

How This Explains Everything:

Threshold frequency: If $hf < \phi$ (work function), a single photon lacks sufficient energy to free an electron. No amount of dim light helps because you never get 2+ photons cooperating.
Instantaneous emission: A single photon-electron collision transfers energy immediately. No gradual buildup needed.
KE proportional to frequency: $K_{max} = hf - \phi$ . Higher frequency photons have more energy, so liberated electrons are faster.
Current proportional to intensity: More photons per second = more electrons ejected per second, but each electron still gets energy from just one photon.

Einstein won the Nobel Prize (1921) for this explanation, not for relativity!

Practice Question

Explain why bright red light (λ = 650 nm) cannot eject electrons from sodium (work function φ = 2.3 eV), but dim UV light (λ = 300 nm) can. Use calculations to support your answer.

Show Answer

Red light (650 nm):

Energy per photon: $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{650 \times 10^{-9}}$

$E = 3.06 \times 10^{-19} \text{ J} = 1.91 \text{ eV}$

Since $E < \phi$ (1.91 eV < 2.3 eV), each photon lacks sufficient energy to overcome the work function. No electrons can be emitted, regardless of how bright the light is.

UV light (300 nm):

Energy per photon: $E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{300 \times 10^{-9}}$

$E = 6.63 \times 10^{-19} \text{ J} = 4.14 \text{ eV}$

Since $E > \phi$ (4.14 eV > 2.3 eV), each photon has enough energy to eject an electron. Emission occurs immediately, even if the UV light is dim (few photons per second).

Conclusion:

In the photon model, brightness (intensity) only affects the NUMBER of photons, not the energy of each photon. Since electron emission requires a minimum energy per photon (the work function), only light with sufficiently high frequency (short wavelength) can cause the photoelectric effect, regardless of intensity.

Photoelectric Equation

Einstein's photoelectric equation quantitatively describes the energy transformation when a photon ejects an electron from a metal surface. It's a direct application of energy conservation in the quantum world.

The Photoelectric Equation

K_{max} = hf - \phi

Alternative form using wavelength:

K_{max} = \frac{hc}{\lambda} - \phi

Where:

$K_{max}$ = maximum kinetic energy of ejected electrons (J or eV)
$h$ = Planck's constant = $6.63 \times 10^{-34} \text{ J·s}$ or $4.14 \times 10^{-15} \text{ eV·s}$
$f$ = frequency of incident light (Hz)
$\phi$ = work function of the metal (J or eV)
$c$ = speed of light = $3.0 \times 10^8 \text{ m/s}$
$\lambda$ = wavelength (m)

Energy Conservation Interpretation

Energy In = Energy Used + Energy Out

Energy In: Photon energy = $hf$

Energy Used: Work function = $\phi$ (minimum energy to free electron)

Energy Out: Kinetic energy of electron = $K_{max}$

Physical Meaning:

If $hf = \phi$ : Electron barely escapes with zero KE (threshold condition)
If $hf > \phi$ : Excess energy becomes kinetic energy of the electron
If $hf < \phi$ : Electron cannot escape (no photoelectric effect)

⚠️ Why "Maximum" KE?

Not all electrons start at the surface. Electrons deeper in the metal lose some energy through collisions on their way out. The maximum KE is for electrons right at the surface with the shortest escape path.

Worked Example

Ultraviolet light of wavelength 250 nm shines on a potassium surface (work function φ = 2.0 eV). Calculate: (a) the photon energy in eV, (b) the maximum kinetic energy of ejected electrons, (c) the maximum speed of ejected electrons.

Given:

$\lambda = 250 \text{ nm} = 2.5 \times 10^{-7} \text{ m}, \quad \phi = 2.0 \text{ eV}$

$h = 4.14 \times 10^{-15} \text{ eV·s}, \quad c = 3.0 \times 10^8 \text{ m/s}, \quad m_e = 9.1 \times 10^{-31} \text{ kg}$

(a) Photon energy in eV:

Using the eV form of h to avoid unit conversion:

$E = \frac{hc}{\lambda} = \frac{(4.14 \times 10^{-15})(3.0 \times 10^8)}{2.5 \times 10^{-7}}$

$E = \frac{1.242 \times 10^{-6}}{2.5 \times 10^{-7}}$

$E = \mathbf{4.97 \text{ eV}}$

(b) Maximum kinetic energy:

$K_{max} = hf - \phi = E - \phi$

$K_{max} = 4.97 - 2.0$

$K_{max} = \mathbf{2.97 \text{ eV}}$

First convert KE to Joules: $K_{max} = 2.97 \times 1.6 \times 10^{-19} = 4.75 \times 10^{-19} \text{ J}$

Using $K = \frac{1}{2}mv^2$ :

$v = \sqrt{\frac{2K_{max}}{m}} = \sqrt{\frac{2(4.75 \times 10^{-19})}{9.1 \times 10^{-31}}}$

$v = \sqrt{1.04 \times 10^{12}}$

$v = \mathbf{1.02 \times 10^6 \text{ m/s}}$

About 0.34% the speed of light - very fast!

Practice Question

Light of frequency $8.0 \times 10^{14} \text{ Hz}$ ejects electrons from a metal with maximum kinetic energy of 1.5 eV. Calculate the work function of the metal in eV.

Show Answer

Solution:

Using $K_{max} = hf - \phi$ , rearrange for work function:

$\phi = hf - K_{max}$

Calculate photon energy:

$E = hf = (4.14 \times 10^{-15})(8.0 \times 10^{14})$

$E = 3.31 \text{ eV}$

Therefore:

$\phi = 3.31 - 1.5 = \mathbf{1.81 \text{ eV}}$

This work function is typical for alkali metals like sodium or potassium, which have relatively low work functions and are commonly used in photoelectric experiments.

Work Function & Threshold Frequency

The work function (φ) is the minimum energy required to liberate an electron from a metal surface. The threshold frequency (f₀) is the minimum frequency of light that can cause photoemission.

Work Function (φ)

Definition: The minimum energy needed to remove an electron from the surface of a metal to a point just outside the surface (with zero kinetic energy).

Physical Origin: Electrons in a metal are attracted to the positive ion lattice. Energy is required to overcome this electrostatic attraction.

Material Dependent: Different metals have different work functions because they have different atomic structures and electron binding strengths.

Cesium

φ = 2.1 eV

(Easiest)

Sodium

φ = 2.3 eV

(Common)

Zinc

φ = 4.3 eV

(Medium)

Platinum

φ = 5.6 eV

(Hardest)

Threshold Frequency (f₀)

At threshold, the photon energy exactly equals the work function, with zero energy left over for kinetic energy:

hf_0 = \phi

Rearranging:

f_0 = \frac{\phi}{h}

Equivalently, using wavelength (threshold wavelength λ₀):

\lambda_0 = \frac{hc}{\phi}

Critical Conditions:

$f < f_0$ (or $\lambda > \lambda_0$ ): NO photoelectric effect, regardless of intensity
$f = f_0$ (or $\lambda = \lambda_0$ ): Electrons just barely escape with $K_{max} = 0$
$f > f_0$ (or $\lambda < \lambda_0$ ): Photoelectric effect occurs, $K_{max} = hf - \phi$

Worked Example

Aluminum has a work function of 4.1 eV. Calculate: (a) the threshold frequency, (b) the threshold wavelength, (c) what color light (if visible) corresponds to threshold.

Given:

$\phi = 4.1 \text{ eV}$

$h = 4.14 \times 10^{-15} \text{ eV·s}, \quad c = 3.0 \times 10^8 \text{ m/s}$

(a) Threshold frequency:

$f_0 = \frac{\phi}{h} = \frac{4.1}{4.14 \times 10^{-15}}$

$f_0 = \mathbf{9.90 \times 10^{14} \text{ Hz}}$

(b) Threshold wavelength:

$\lambda_0 = \frac{c}{f_0} = \frac{3.0 \times 10^8}{9.90 \times 10^{14}}$

$\lambda_0 = \mathbf{3.03 \times 10^{-7} \text{ m} = 303 \text{ nm}}$

303 nm is in the ultraviolet range (not visible).

Visible light ranges from ~400-700 nm. Since the threshold is below 400 nm, aluminum requires UV light for the photoelectric effect. No visible light can eject electrons from aluminum!

Practice Question

A metal has a threshold wavelength of 540 nm (green light). Will red light (λ = 650 nm) cause photoemission? Will blue light (λ = 450 nm)? Justify your answers.

Show Answer

Red light (λ = 650 nm):

Since $\lambda_{red} > \lambda_0$ (650 nm > 540 nm), red light has a longer wavelength than threshold.

Longer wavelength means lower frequency and lower photon energy.

Result: NO photoemission. Each red photon has less energy than the work function.

Blue light (λ = 450 nm):

Since $\lambda_{blue} < \lambda_0$ (450 nm < 540 nm), blue light has a shorter wavelength than threshold.

Shorter wavelength means higher frequency and higher photon energy.

Result: YES, photoemission occurs. Each blue photon has more than enough energy to overcome the work function, with excess energy becoming kinetic energy of ejected electrons.

Key Rule:

For photoemission to occur: $\lambda < \lambda_0$ (shorter wavelength) OR equivalently $f > f_0$ (higher frequency).

Stopping Voltage

Stopping voltage (V_s) is the reverse potential difference required to stop the most energetic photoelectrons from reaching the collector. It provides an experimental method to measure the maximum kinetic energy of ejected electrons.

Experimental Setup

Standard Configuration:

Emitter (cathode): Metal surface illuminated by light, emitting photoelectrons
Collector (anode): Plate that collects electrons, creating measurable current
Variable voltage source: Can be adjusted to oppose electron flow
Ammeter: Measures photocurrent

How It Works:

When a reverse (retarding) voltage is applied, it creates an electric field that opposes the motion of electrons. As voltage increases, fewer electrons have enough KE to overcome the electric potential energy barrier. At the stopping voltage $V_s$ , even the fastest electrons are stopped, and current drops to zero.

Stopping Voltage Equation

The maximum kinetic energy of photoelectrons is related to stopping voltage by:

K_{max} = eV_s

Where:

$K_{max}$ = maximum kinetic energy (J)
$e = 1.6 \times 10^{-19} \text{ C}$ (elementary charge)
$V_s$ = stopping voltage (V)

Energy Conservation:

An electron with kinetic energy $K_{max}$ must do work against the electric potential to reach the collector. At stopping voltage:

$K_{max} = \text{Work done against electric field} = eV_s$

Combining with Photoelectric Equation:

eV_s = hf - \phi

This is the complete equation connecting stopping voltage, frequency, and work function.

Graph Analysis: V_s vs Frequency

Plotting stopping voltage against frequency gives a straight line:

Rearrange $eV_s = hf - \phi$ into linear form $y = mx + c$ :

V_s = \frac{h}{e}f - \frac{\phi}{e}

Slope: $m = h/e$ (Planck's constant can be determined!)
y-intercept: $c = -\phi/e$ (work function can be found)
x-intercept: When $V_s = 0$ , we get $f = f_0$ (threshold frequency)

Historical Significance:

This linear relationship was crucial experimental evidence for Einstein's photon theory. The fact that the slope $h/e$ is the same for all metals (universal constant) strongly supported the quantum model.

Worked Example

When light of frequency $1.2 \times 10^{15} \text{ Hz}$ shines on a metal surface, a stopping voltage of 2.5 V is required to stop all photoelectrons. Calculate: (a) the maximum kinetic energy of the electrons in eV, (b) the work function of the metal.

Given:

$f = 1.2 \times 10^{15} \text{ Hz}, \quad V_s = 2.5 \text{ V}$

$h = 4.14 \times 10^{-15} \text{ eV·s}, \quad e = 1.6 \times 10^{-19} \text{ C}$

(a) Maximum kinetic energy:

$K_{max} = eV_s$

Since we want the answer in eV, and $V_s$ is already in volts:

$K_{max} = \mathbf{2.5 \text{ eV}}$

(When using eV as energy unit, $K_{max}$ numerically equals $V_s$ )

(b) Work function:

Using $K_{max} = hf - \phi$ :

$\phi = hf - K_{max}$

$\phi = (4.14 \times 10^{-15})(1.2 \times 10^{15}) - 2.5$

$\phi = 4.97 - 2.5$

$\phi = \mathbf{2.47 \text{ eV}}$

This is consistent with sodium (φ ≈ 2.3 eV) or similar alkali metal.

Practice Question

In a photoelectric experiment, two different frequencies of light are used. Light of frequency f₁ = 8.0 × 10¹⁴ Hz requires a stopping voltage of 1.2 V. Light of frequency f₂ = 1.0 × 10¹⁵ Hz requires a stopping voltage of 2.0 V. Use these data to calculate Planck's constant (in eV·s).

Show Answer

Solution:

For each frequency, we have: $V_s = \frac{h}{e}f - \frac{\phi}{e}$

For f₁:

$1.2 = \frac{h}{e}(8.0 \times 10^{14}) - \frac{\phi}{e}$ ... (1)

For f₂:

$2.0 = \frac{h}{e}(1.0 \times 10^{15}) - \frac{\phi}{e}$ ... (2)

Subtract equation (1) from equation (2):

$2.0 - 1.2 = \frac{h}{e}(1.0 \times 10^{15} - 8.0 \times 10^{14})$

$0.8 = \frac{h}{e}(2.0 \times 10^{14})$

$\frac{h}{e} = \frac{0.8}{2.0 \times 10^{14}} = 4.0 \times 10^{-15} \text{ V·s}$

Since we want h in eV·s, and $1 \text{ eV} = e \times 1\text{ V}$ :

$h = \mathbf{4.0 \times 10^{-15} \text{ eV·s}}$

This is very close to the accepted value of $h = 4.14 \times 10^{-15} \text{ eV·s}$ . The small difference is due to rounding in the problem data.

Pillar 3 of 3

Special Relativity

Explore Einstein's revolutionary theory of spacetime, time dilation, length contraction, and mass-energy equivalence at speeds approaching the speed of light.

Michelson-Morley Experiment

The Michelson-Morley experiment (1887) attempted to detect Earth's motion through the hypothetical "luminiferous aether" - a medium believed necessary for light wave propagation. Its null result became one of the most important negative results in physics history.

In the 19th century, all known waves (sound, water, seismic) required a medium to propagate. Since light exhibited wave behavior through interference and diffraction, scientists assumed it must also propagate through a medium called the luminiferous aether. This aether was thought to fill all of space, be completely transparent and massless, yet rigid enough to support high-speed electromagnetic waves. As Earth orbits the Sun at approximately 30 km/s, it should experience an "aether wind" - similar to feeling wind when driving through still air. Light traveling parallel versus perpendicular to Earth's motion should therefore have different measured speeds.

Experimental Design

Michelson and Morley constructed an interferometer with two perpendicular arms, each approximately 11 meters long. A half-silvered mirror split a monochromatic light beam into two perpendicular beams. Mirrors at the end of each arm reflected the light back to recombine at a detector, creating an interference pattern.

If the aether existed, one beam would travel parallel to Earth's motion (with and against the aether wind) while the other traveled perpendicular (across the wind). The different effective light speeds would cause different travel times. When the beams recombined, this phase difference would create a measurable shift in the interference pattern. Rotating the apparatus 90° should shift the pattern by approximately 0.4 fringes. The apparatus was sensitive enough to detect shifts as small as 0.01 fringes.

The Null Result

Observation: No fringe shift was detected, regardless of apparatus orientation, time of day, or time of year. The experiment was repeated many times over several decades - always with the same null result.

Conclusion: Light speed is the same in all directions, independent of Earth's motion. There is no aether wind, and the luminiferous aether does not exist. Light does not require a medium and can propagate through vacuum. This "failed" experiment destroyed the aether theory and paved the way for Einstein's special relativity, proving that sometimes the most important discoveries come from experiments that don't find what they're looking for.

Worked Example

If the aether existed and Earth moved through it at 30 km/s, calculate the expected time difference for light traveling 11 m parallel versus perpendicular to Earth's motion. (Use the approximation that the time difference Δt ≈ Lv²/c³ for small velocities.)

Given:

$L = 11 \text{ m}, \quad v = 30 \text{ km/s} = 3.0 \times 10^4 \text{ m/s}$

$c = 3.0 \times 10^8 \text{ m/s}$

Time difference:

$\Delta t \approx \frac{Lv^2}{c^3}$

$\Delta t \approx \frac{(11)(3.0 \times 10^4)^2}{(3.0 \times 10^8)^3}$

$\Delta t \approx \frac{(11)(9.0 \times 10^8)}{2.7 \times 10^{25}}$

$\Delta t \approx \mathbf{3.7 \times 10^{-16} \text{ s}}$

This tiny time difference would cause a measurable fringe shift of about 0.4 fringes with the wavelength of light used (≈500 nm). The apparatus was sensitive enough to detect this, yet no shift was observed.

Practice Question

Explain the aim of the Michelson-Morley experiment, what result was expected, and what was actually observed. Why was this null result so significant for physics?

Show Answer

The aim was to detect Earth's motion through the luminiferous aether by measuring differences in the speed of light traveling parallel versus perpendicular to Earth's orbital motion. Scientists expected a shift in the interference pattern (approximately 0.4 fringes) when the apparatus was rotated, due to different light travel times in the two perpendicular arms caused by the "aether wind."

However, no fringe shift was detected. The interference pattern remained constant regardless of apparatus orientation, time of day, or season, indicating that light speed appeared identical in all directions.

This null result was profoundly significant. It proved the luminiferous aether doesn't exist and that light doesn't need a medium to propagate. It demonstrated that light travels at c in all directions, independent of the observer's motion. This experimental result provided crucial evidence supporting Einstein's postulate that the speed of light is the same for all observers, forming the foundation for special relativity and forcing physicists to abandon classical assumptions about absolute space and time.

Einstein's Postulates & Inertial Frames

In 1905, Einstein published his theory of special relativity, built on two deceptively simple postulates that revolutionized our understanding of space and time.

Inertial Reference Frames

An inertial reference frame is a frame of reference that is not accelerating. In an inertial frame, Newton's first law holds: an object at rest stays at rest, and an object in motion continues at constant velocity unless acted upon by a force. Examples include a spaceship drifting at constant velocity, a train moving in a straight line at constant speed, or Earth's surface (approximately, ignoring rotation and orbit). Any frame moving at constant velocity relative to an inertial frame is also inertial.

Non-inertial frames are those that are accelerating, such as an accelerating rocket, a car going around a corner, a rotating merry-go-round, or an elevator starting or stopping. Special relativity applies only to inertial frames. In non-inertial (accelerating) frames, you feel fictitious forces like being pushed back in an accelerating car, and the laws of physics appear different.

Einstein's Two Postulates

Postulate 1: Principle of Relativity

"The laws of physics are the same in all inertial reference frames."

You cannot perform any experiment (mechanical, electrical, optical, or otherwise) to determine whether you are at rest or moving at constant velocity. All physics laws work identically in all inertial frames. For example, on a smoothly moving train, dropping a ball looks identical to dropping it while standing still. No experiment can tell you which frame is "really" moving.

Postulate 2: Constancy of Light Speed

"The speed of light in vacuum is the same for all observers, regardless of the motion of the light source or observer."

If you measure the speed of light, you always get $c = 3.0 \times 10^8 \text{ m/s}$ , whether you're stationary, moving toward the light source, or moving away from it. This is completely different from everyday experience with speeds. For example, a spaceship traveling at 0.9c shoots a laser forward. The light travels at c, not 1.9c. An observer on Earth also measures the light at c, not 0.1c.

The second postulate is shocking because it violates classical velocity addition. In everyday experience, if you're on a train moving at 30 m/s and throw a ball forward at 20 m/s, the ball moves at 50 m/s relative to the ground (velocities simply add). But light doesn't follow this rule. If you're on a spaceship moving at 0.5c and shine a flashlight forward, classical physics would predict the light should move at 1.5c relative to Earth. In reality, light moves at exactly c relative to Earth and also c relative to the ship.

For the speed of light to be constant for all observers, something else must change. Einstein realized that time and space must be relative - different observers measure different times and distances. This leads to time dilation and length contraction.

Worked Example

A spaceship moves at 0.6c relative to Earth. The spaceship fires a missile forward at 0.8c relative to the ship. (a) What speed would classical physics predict for the missile relative to Earth? (b) Why is this impossible according to special relativity?

Given:

$v_{ship} = 0.6c, \quad v_{missile/ship} = 0.8c$

(a) Classical prediction:

Using Galilean velocity addition: $v_{missile/Earth} = v_{ship} + v_{missile/ship}$

$v_{missile/Earth} = 0.6c + 0.8c = \mathbf{1.4c}$

(b) Why this is impossible:

According to Einstein's second postulate, nothing can travel faster than the speed of light c. The speed limit is absolute and universal.

The actual relativistic velocity addition formula gives: $v = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}$

$v = \frac{0.6c + 0.8c}{1 + \frac{(0.6c)(0.8c)}{c^2}} = \frac{1.4c}{1 + 0.48} = \frac{1.4c}{1.48} = 0.946c$

The missile travels at 0.946c relative to Earth, safely below c. Velocities don't simply add at relativistic speeds.

Practice Question

State Einstein's two postulates of special relativity. Explain why the second postulate conflicts with classical (Galilean) velocity addition, and what must change to accommodate it.

Show Answer

Einstein's first postulate, the Principle of Relativity, states that the laws of physics are the same in all inertial reference frames, meaning there is no preferred "absolute" frame of rest. His second postulate states that the speed of light in vacuum is the same for all observers (c = 3.0 × 10⁸ m/s), regardless of the motion of the light source or the observer.

This second postulate conflicts with classical Galilean velocity addition, which states that velocities simply add. If object A moves at speed v₁ relative to B, and B moves at v₂ relative to C, then classically A should move at v₁ + v₂ relative to C. This would predict that light from a moving source should have different speeds for different observers (source velocity plus c), directly contradicting the second postulate.

For all observers to measure the same light speed despite different relative motions, the classical assumptions about absolute time and space must be wrong. Time becomes relative, with moving clocks running slow (time dilation). Space becomes relative, with moving objects appearing contracted (length contraction). Simultaneity also becomes relative, with events simultaneous in one frame not being simultaneous in another. These effects become significant only at speeds approaching c, which explains why we don't notice them in daily life.

Relativity of Simultaneity

In classical physics, if two events occur simultaneously for one observer, they occur simultaneously for all observers. Special relativity destroys this assumption: events that are simultaneous in one frame of reference are not necessarily simultaneous in another frame moving relative to the first.

Consider a thought experiment with a train moving at high speed past a platform. Two lightning bolts strike the front and back of the train simultaneously according to an observer on the platform (equidistant from both strikes). However, an observer on the train sees the front lightning strike first because they are moving toward that light signal and away from the rear signal. Both observers are correct in their own reference frames - simultaneity is relative.

Why Simultaneity is Relative

The relativity of simultaneity is a direct consequence of the constancy of light speed. If light speed is the same for all observers, then observers in relative motion must disagree about when distant events occur. The key insight is that there is no absolute "now" that extends across all of space. Different observers moving at different velocities slice through spacetime differently, creating different definitions of simultaneous events.

This effect becomes noticeable only when objects move at significant fractions of the speed of light and events are separated by large distances. For everyday speeds and distances, the time differences are immeasurably small, which is why we never notice this effect in daily life.

Worked Example

Two explosions occur 1000 m apart (as measured on Earth) and are simultaneous according to Earth observers. A spaceship moving at 0.8c passes Earth exactly when the explosions occur. The spaceship is moving from explosion A toward explosion B. Calculate which explosion the spaceship observer sees first and the time difference between observations. (Use the approximation that time difference Δt ≈ vd/c² for this example.)

Given:

$d = 1000 \text{ m}, \quad v = 0.8c$

Analysis:

The spaceship is moving toward explosion B, so light from B reaches the ship first. The ship observer sees B occur before A, even though they're simultaneous for Earth observers.

Using the approximation $\Delta t \approx \frac{vd}{c^2}$ :

$\Delta t \approx \frac{(0.8c)(1000)}{c^2} = \frac{800c}{c^2} = \frac{800}{c}$

$\Delta t \approx \frac{800}{3.0 \times 10^8} = \mathbf{2.67 \times 10^{-6} \text{ s} = 2.67 \text{ }\mu\text{s}}$

The spaceship observer sees explosion B occur 2.67 microseconds before explosion A, even though Earth observers see them as perfectly simultaneous.

Practice Question

Explain what is meant by "the relativity of simultaneity" and why it is a necessary consequence of the constancy of the speed of light. Why don't we observe this effect in everyday life?

Show Answer

The relativity of simultaneity means that two events that are simultaneous (occur at the same time) in one reference frame are not necessarily simultaneous in another reference frame moving relative to the first. There is no absolute, universal "now" - what constitutes simultaneous events depends on the observer's state of motion.

This is a necessary consequence of the constancy of light speed because different observers in relative motion must receive light signals from distant events at different times, yet light travels at the same speed for all of them. If light speed is constant for all observers, then observers moving at different velocities must disagree about when events occur. The only way to maintain constant light speed for all observers is if time itself is relative, including which events are simultaneous.

We don't observe this effect in everyday life because the time differences are proportional to vd/c², where v is relative velocity and d is spatial separation. For everyday speeds (much less than c) and distances, this factor is incredibly tiny. For example, two observers moving at 100 km/h observing events 1 km apart would disagree about simultaneity by only about 10⁻¹³ seconds - far too small to measure. The effect only becomes noticeable at speeds approaching the speed of light.

Time Dilation

Time dilation is the phenomenon where a moving clock runs slower than a stationary clock. An observer watching a moving clock sees it tick more slowly than their own identical clock. This is not a mechanical defect - time itself passes more slowly for the moving observer.

Time Dilation Formula

t = \frac{t_0}{\sqrt{1 - v^2/c^2}}

Where:

$t$ = dilated time (measured by observer watching the moving clock)
$t_0$ = proper time (measured by observer moving with the clock)
$v$ = relative velocity between observers
$c$ = speed of light

⚠️ Critical Frame Identification:

Ask: "Who is holding the clock?" That observer measures proper time $t_0$ . Everyone else sees the clock run slow and measures dilated time $t > t_0$ .

Proper time is always the shortest time interval - it's the time measured by a clock that is present at both the start and end of the event being timed. For any other observer (for whom the clock is moving), more time passes. This seems paradoxical but is a fundamental property of spacetime.

Evidence: Muon Decay

Muons are unstable particles created when cosmic rays strike Earth's upper atmosphere (about 10 km high). They have a proper lifetime of only 2.2 microseconds. Even traveling at 0.98c, they should only travel about 660 meters before decaying - nowhere near enough to reach Earth's surface.

However, muons are detected at sea level in large numbers. From Earth's perspective, the muons' internal clocks run slow due to time dilation. Their dilated lifetime is $t = t_0/\sqrt{1-v^2/c^2} \approx 11 \text{ }\mu\text{s}$ , giving them enough time to travel the 10 km to Earth's surface. This experimental observation perfectly confirms Einstein's time dilation prediction.

Worked Example

An astronaut travels to a star 10 light-years away at 0.8c. According to the astronaut's clock, how long does the journey take? How long does it take according to Earth clocks?

Given:

$d = 10 \text{ light-years}, \quad v = 0.8c$

Earth's perspective (dilated time):

From Earth's view, the astronaut travels 10 light-years at 0.8c:

$t = \frac{d}{v} = \frac{10 \text{ ly}}{0.8c} = 12.5 \text{ years}$

Earth observers see the journey take 12.5 years.

Astronaut's perspective (proper time):

The astronaut is "holding the clock" (traveling with it), so measures proper time:

$t_0 = t\sqrt{1 - v^2/c^2} = 12.5\sqrt{1 - 0.64}$

$t_0 = 12.5\sqrt{0.36} = 12.5 \times 0.6$

$t_0 = \mathbf{7.5 \text{ years}}$

The astronaut ages only 7.5 years during the journey, while Earth experiences 12.5 years!

Practice Question

A spacecraft travels at 0.6c. An astronaut on board celebrates their 30th birthday. How old are they according to Earth observers? Explain your reasoning.

Show Answer

The astronaut is holding the clock (their own body is the "clock" measuring their age), so 30 years is the proper time. From Earth's perspective, the astronaut's biological processes run slow due to time dilation, so more time has passed on Earth.

Using the time dilation formula where the astronaut measures proper time and Earth measures dilated time:

$t = \frac{t_0}{\sqrt{1 - v^2/c^2}} = \frac{30}{\sqrt{1 - 0.36}} = \frac{30}{\sqrt{0.64}} = \frac{30}{0.8} = 37.5 \text{ years}$

According to Earth observers, the astronaut is 37.5 years old when they celebrate their 30th birthday. From Earth's perspective, 37.5 years have passed, but the astronaut has only aged 30 years due to time dilation. This is real aging - if the astronaut returned to Earth, they would indeed be younger than their twin who stayed behind.

Length Contraction

Length contraction (also called Lorentz contraction) is the phenomenon where a moving object appears shortened in the direction of motion. An observer watching a moving object measures it to be shorter than an observer moving with the object.

Length Contraction Formula

l = l_0\sqrt{1 - v^2/c^2}

Where:

$l$ = contracted length (measured by observer watching the moving object)
$l_0$ = proper length (measured by observer at rest relative to the object)
$v$ = relative velocity between observers
$c$ = speed of light

⚠️ Critical Frame Identification:

Ask: "Who is holding the ruler?" That observer measures proper length $l_0$ . Everyone else sees the ruler contracted and measures $l < l_0$ .

⚠️ Direction Matters:

Contraction occurs ONLY in the direction parallel to motion. Perpendicular dimensions are unchanged. A moving meter stick contracts lengthwise but maintains its width and thickness.

Proper length is always the longest measurement - it's the length measured by an observer at rest relative to the object. For any observer watching the object move, the length appears contracted. Like time dilation, this is a real physical effect of relative motion through spacetime, not an optical illusion or measurement error.

Worked Example

A spaceship is 100 m long when measured at rest. It flies past Earth at 0.8c. (a) What length do Earth observers measure? (b) What is the width of the spaceship (perpendicular to motion) if it's 20 m when at rest?

Given:

$l_0 = 100 \text{ m}, \quad v = 0.8c, \quad w_0 = 20 \text{ m}$

(a) Contracted length:

The spaceship crew are "holding the ruler" (at rest relative to the ship), so 100 m is proper length.

Earth observers see the moving ship contracted:

$l = l_0\sqrt{1 - v^2/c^2} = 100\sqrt{1 - 0.64}$

$l = 100\sqrt{0.36} = 100 \times 0.6$

$l = \mathbf{60 \text{ m}}$

Earth observers measure the spaceship as only 60 m long.

(b) Width perpendicular to motion:

Contraction only occurs parallel to the direction of motion.

Perpendicular dimensions are unchanged: $w = w_0 = \mathbf{20 \text{ m}}$

The width remains 20 m for all observers.

Practice Question

From an astronaut's perspective traveling at 0.6c, the distance from Earth to a distant star is measured as 8 light-years. What is the proper distance (distance measured by Earth observers at rest relative to both Earth and the star)?

Show Answer

Earth observers are at rest relative to both Earth and the star, so they are "holding the ruler" measuring the distance. Earth measures proper length. The astronaut sees this distance contracted because they are moving relative to it.

From the astronaut's perspective, the contracted length is 8 light-years. Using the length contraction formula:

$l = l_0\sqrt{1 - v^2/c^2}$

$8 = l_0\sqrt{1 - 0.36}$

$8 = l_0\sqrt{0.64} = l_0(0.8)$

$l_0 = \frac{8}{0.8} = 10 \text{ light-years}$

The proper distance (measured by Earth) is 10 light-years. From the astronaut's moving perspective, this distance appears contracted to 8 light-years, which explains how they can complete the journey in less time on their clock - the distance is literally shorter for them.

Mass-Energy Equivalence

Einstein's most famous equation, $E = mc^2$ , reveals that mass and energy are interchangeable. Mass is a concentrated form of energy, and energy has mass. This equivalence explains the enormous energies released in nuclear reactions.

Mass-Energy Equation

E = mc^2

Where:

$E$ = energy equivalent (J)
$m$ = mass (kg)
$c = 3.0 \times 10^8 \text{ m/s}$ = speed of light

Since c² is enormous (9 × 10¹⁶), even tiny masses correspond to huge energies. Converting just 1 kg of mass to energy yields 9 × 10¹⁶ J - equivalent to about 20 megatons of TNT!

In nuclear reactions (both fission and fusion), a small amount of mass is converted to energy. The total mass of the products is slightly less than the reactants - this "missing mass" (mass defect) has been converted to energy according to E = mc². This is the source of energy in stars, nuclear power plants, and nuclear weapons.

Why Nothing Can Reach c

As an object with mass approaches the speed of light, its kinetic energy approaches infinity. The energy required to accelerate a massive object to exactly c would be infinite, which is impossible. This is a fundamental speed limit of the universe - only massless particles like photons can travel at c.

The relativistic kinetic energy formula is $K = mc^2(\gamma - 1)$ where $\gamma = 1/\sqrt{1-v^2/c^2}$ . As v approaches c, γ approaches infinity, meaning infinite energy would be needed. This explains why particle accelerators can get protons to 0.9999c but never to c itself.

Worked Example

In nuclear fusion, four hydrogen nuclei (total mass 6.693 × 10⁻²⁷ kg) combine to form one helium nucleus (mass 6.645 × 10⁻²⁷ kg). Calculate: (a) the mass defect, (b) the energy released.

Given:

$m_{initial} = 6.693 \times 10^{-27} \text{ kg}$

$m_{final} = 6.645 \times 10^{-27} \text{ kg}$

$c = 3.0 \times 10^8 \text{ m/s}$

(a) Mass defect:

$\Delta m = m_{initial} - m_{final}$

$\Delta m = 6.693 \times 10^{-27} - 6.645 \times 10^{-27}$

$\Delta m = \mathbf{4.8 \times 10^{-29} \text{ kg}}$

(b) Energy released:

$E = \Delta m c^2$

$E = (4.8 \times 10^{-29})(3.0 \times 10^8)^2$

$E = (4.8 \times 10^{-29})(9.0 \times 10^{16})$

$E = \mathbf{4.32 \times 10^{-12} \text{ J}}$

This is the energy released per fusion event. The Sun performs about 10³⁸ of these reactions per second, releasing about 4 × 10²⁶ watts!

Practice Question

Explain how E = mc² explains the source of the Sun's energy and why it's impossible for a proton to be accelerated to the speed of light.

Show Answer

The Sun generates energy through nuclear fusion, where hydrogen nuclei combine to form helium. The total mass of the helium product is slightly less than the total mass of the hydrogen reactants. This "missing mass" (mass defect) has been converted to energy according to E = mc². Because c² is so large, even the tiny mass defect (about 0.7% of the original mass) produces enormous energy. The Sun converts about 4 million tonnes of mass to energy every second, which has sustained it for 4.6 billion years.

A proton cannot be accelerated to the speed of light because doing so would require infinite energy. As a massive object approaches c, its relativistic kinetic energy increases without limit. The energy required is proportional to 1/√(1-v²/c²), which approaches infinity as v approaches c. No finite amount of energy can accelerate a massive particle to exactly c. Only massless particles (like photons) can travel at the speed of light. This is a fundamental consequence of the structure of spacetime, not a technological limitation.

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Electromagnetic Waves & The Wave Model

Maxwell's Unification

Maxwell's Prediction

Gauss's Law (Electric)

Gauss's Law (Magnetic)

Faraday's Law

Ampère-Maxwell Law

Worked Example

Practice Question

Hertz's Experiment

Hertz's Apparatus

Reflection

Refraction

Interference

Polarization

Why Polarization Proves Transverse Nature

Practice Question

Spectroscopy

1. Surface Temperature (Wien's Law)

2. Translational Velocity (Doppler Shift)

3. Chemical Composition (Absorption Lines)

Worked Example

Practice Question

Interference: Young's Double Slit

Conditions for Interference

Path Difference and Interference

Worked Example

Practice Question

Diffraction

Condition for Significant Diffraction

Worked Example

Practice Question

Diffraction Gratings

Diffraction Grating Equation

Double Slit (N = 2)

Diffraction Grating (N = 1000+)

Worked Example

Practice Question

Polarisation & Malus' Law

Polarization Basics

Malus' Law

Worked Example

Practice Question

The Quantum Model of Light

Black Body Radiation & Ultraviolet Catastrophe

Classical Prediction vs Reality

Planck's Revolutionary Solution (1900)

Energy Quantization

Birth of Quantum Mechanics

Planck's Constant

Perfect Fit to Data

Worked Example

Practice Question

Wien's Displacement Law

Wien's Law Formula

The Sun (T ≈ 5800 K)

Hot Stars (T ≈ 20,000 K)

Cool Stars (T ≈ 3000 K)

Human Body (T ≈ 310 K)

Worked Example

Practice Question

The Photoelectric Effect

Experimental Observations

Classical Physics Failures

Einstein's Photon Explanation (1905)

Practice Question

Photoelectric Equation

The Photoelectric Equation

Energy Conservation Interpretation

Worked Example

Practice Question

Work Function & Threshold Frequency

Work Function (φ)

Threshold Frequency (f₀)

Worked Example

Practice Question

Stopping Voltage

Experimental Setup

Stopping Voltage Equation

Graph Analysis: V_s vs Frequency