🌊 Wave Nature of Light

Part 1 of 7 — Superposition and Interference

Light is a wave — and when waves overlap, they combine. This principle of superposition is the foundation for understanding interference patterns, one of the most beautiful phenomena in physics.

The Principle of Superposition

When two or more waves overlap in the same region of space, the resultant displacement at any point is the algebraic sum of the individual displacements:

$y_{\text{total}} = y_1 + y_2$

This is called the principle of superposition. It applies to all linear waves, including light, sound, and water waves.

Key Requirement: Coherent Sources

For a stable interference pattern, the sources must be coherent:

• Same frequency (wavelength)
• Constant phase relationship

Incoherent sources (like two light bulbs) produce rapidly shifting patterns that average out — no visible interference.

Two Types of Interference

Path Difference: The Key to Interference

The type of interference depends on the path difference $\Delta r$ — the difference in distance each wave travels from its source to the observation point.

Constructive Interference (Bright)

Waves arrive in phase when the path difference is a whole number of wavelengths:

$\Delta r = m\lambda \quad (m = 0, \pm 1, \pm 2, \ldots)$

The waves reinforce each other, producing maximum intensity.

Destructive Interference (Dark)

Waves arrive out of phase when the path difference is a half-integer number of wavelengths:

$\Delta r = \left(m + \frac{1}{2}\right)\lambda \quad (m = 0, \pm 1, \pm 2, \ldots)$

The waves cancel each other, producing zero intensity.

The Central Maximum

At the center of the pattern ($m = 0$), the path difference is zero — both waves travel the same distance. This always gives constructive interference.

Phase and Path Difference Quiz 🎯

Intensity and Interference

For two coherent waves of equal amplitude $E_0$, the resultant intensity depends on the phase difference $\phi$:

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

where $I_0$ is the intensity of each individual wave.

Phase Difference and Path Difference

The phase difference $\phi$ is related to the path difference $\Delta r$:

$\phi = \frac{2\pi}{\lambda} \Delta r$

Energy Conservation

Interference doesn't create or destroy energy — it redistributes it. The extra intensity at bright spots comes at the expense of dark spots. The average intensity over the full pattern is $2I_0$, the same as without interference.

Interference Concept Check 🔬

Exit Quiz — Wave Superposition ✅

🔦 Young's Double-Slit Experiment

Part 2 of 7 — The Experiment That Proved Light Is a Wave

In 1801, Thomas Young performed the most famous experiment in optics. By shining light through two narrow slits, he produced an interference pattern that could only be explained if light is a wave.

The Double-Slit Setup

Geometry

• Two narrow slits separated by distance $d$
• A screen at distance $L$ from the slits ($L \gg d$)
• Monochromatic light of wavelength $\lambda$

The Key Equations

Bright fringes (constructive interference) occur when: $d \sin\theta = m\lambda \quad (m = 0, \pm 1, \pm 2, \ldots)$

Dark fringes (destructive interference) occur when: $d \sin\theta = \left(m + \frac{1}{2}\right)\lambda \quad (m = 0, \pm 1, \pm 2, \ldots)$

where $\theta$ is the angle from the central axis to the point on the screen.

The Small-Angle Approximation

When $\theta$ is small (fringes near the center), $\sin\theta \approx \tan\theta = y/L$, so:

$d \cdot \frac{y}{L} = m\lambda \implies y_m = \frac{m\lambda L}{d}$

This gives the position of the $m$-th bright fringe on the screen.

Fringe Spacing

The distance between consecutive bright fringes is called the fringe spacing:

$\Delta y = \frac{\lambda L}{d}$

What Affects Fringe Spacing?

Intensity Pattern

The intensity at angle $\theta$ is:

$I = 4I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right)$

This produces equally spaced bright and dark fringes — the hallmark of double-slit interference.

Order Number $m$

• $m = 0$: central maximum (always the brightest, at the center)
• $m = \pm 1$: first-order maxima
• $m = \pm 2$: second-order maxima
• Higher orders exist until $\sin\theta > 1$ (impossible), so $m_{\max} = \lfloor d/\lambda \rfloor$

Double-Slit Concept Quiz 🎯

Double-Slit Calculation Drill 🧮

A double-slit experiment uses $\lambda = 550$ nm, slit separation $d = 0.25$ mm, and screen distance $L = 2.0$ m.

1. Fringe spacing $\Delta y$ (in mm)
2. Position of the 3rd bright fringe from center (in mm)
3. Position of the 1st dark fringe from center (in mm)

Round all answers to 3 significant figures.

Advanced Double-Slit Problems 🔬

1. Light of wavelength 480 nm passes through two slits 0.20 mm apart. A screen is 1.5 m away. What is the distance from the central maximum to the 2nd dark fringe? (in mm)

2. In a double-slit experiment, the 5th bright fringe is located 15 mm from the central maximum. The screen is 2.0 m away and $\lambda = 600$ nm. What is the slit separation $d$? (in mm)

Round all answers to 3 significant figures.

Exit Quiz — Double-Slit Mastery ✅

🌅 Single-Slit Diffraction

Part 3 of 7 — When Light Bends Around Edges

Even a single slit produces a pattern of bright and dark bands. This is diffraction — the bending and spreading of waves as they pass through an opening. The narrower the slit, the more the light spreads out.

Single-Slit Diffraction Pattern

The Minima Condition

For a slit of width $a$, dark fringes (minima) occur at:

$a \sin\theta = m\lambda \quad (m = \pm 1, \pm 2, \pm 3, \ldots)$

Note: $m = 0$ is NOT a minimum — it is the bright central maximum!

Central Maximum Width

The central bright fringe extends between the first minima on either side ($m = \pm 1$):

$\text{Angular width} = 2\theta_1 = 2\arcsin\left(\frac{\lambda}{a}\right)$

For small angles:

$\text{Width on screen} = \frac{2\lambda L}{a}$

The central maximum is twice as wide as any other bright fringe.

Key Differences from Double-Slit

Intensity Pattern

The intensity for single-slit diffraction is:

$I = I_0 \left(\frac{\sin\beta}{\beta}\right)^2$

where $\beta = \frac{\pi a \sin\theta}{\lambda}$.

Key Features

• Central maximum: the brightest and widest peak
• Secondary maxima: much dimmer — the first secondary maximum is only about 4.7% of the central peak intensity
• Minima: exactly zero intensity at $a\sin\theta = m\lambda$

Width and Slit Size

The relationship between slit width $a$ and the diffraction pattern:

The narrower the slit, the wider the diffraction pattern — this is an inverse relationship!

Single-Slit Concept Quiz 🎯

Single-Slit Diffraction Drill 🧮

Light of wavelength 600 nm passes through a single slit of width $a = 0.15$ mm. A screen is placed $L = 2.0$ m away.

1. Angular position of the 1st minimum $\theta_1$ (in degrees, to 3 significant figures)
2. Width of the central maximum on the screen (in mm)
3. Position of the 2nd minimum from the center (in mm)

Double-Slit vs Single-Slit 🔍

Exit Quiz — Single-Slit Diffraction ✅

🌈 Diffraction Gratings

Part 4 of 7 — Splitting Light into a Rainbow

A diffraction grating is like a double slit on steroids — thousands of equally spaced slits that produce extremely sharp, well-separated bright fringes. Gratings are the key tool for analyzing the wavelengths present in light.

How Gratings Work

The Grating Equation

A grating with slit spacing $d$ produces bright maxima at:

$d \sin\theta = m\lambda \quad (m = 0, \pm 1, \pm 2, \ldots)$

This looks identical to the double-slit equation — but the pattern is very different.

Slit Spacing from Line Count

If the grating has $N$ lines per millimeter (or per cm), the slit spacing is:

$d = \frac{1}{N}$

Example: A grating with 500 lines/mm has $d = 1/500$ mm $= 2.0 \times 10^{-6}$ m $= 2.0\;\mu\text{m}$.

Why Gratings Are Better Than Double Slits

With $N$ slits, each principal maximum is $N$ times as narrow as the double-slit maximum. More slits → sharper peaks.

Spectral Analysis with Gratings

Separating Colors

When white light hits a grating, each wavelength diffracts at a different angle:

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

Longer wavelengths (red) diffract more than shorter wavelengths (violet). This creates a spectrum at each order $m$.

Rainbow Pattern for Each Order

• $m = 0$: central white maximum (all colors overlap)
• $m = \pm 1$: first-order spectrum (violet closest to center, red farthest)
• $m = \pm 2$: second-order spectrum (wider spread, may overlap with other orders)

Resolving Power

The ability to distinguish two close wavelengths is the resolving power:

$R = \frac{\lambda}{\Delta\lambda} = mN$

where $N$ is the total number of slits and $m$ is the order.

Higher orders and more slits → better resolution of closely spaced spectral lines.

Maximum Order

The highest observable order is limited by $\sin\theta \leq 1$:

$m_{\max} = \left\lfloor \frac{d}{\lambda} \right\rfloor$

Grating Concepts Quiz 🎯

Grating Calculation Drill 🧮

A diffraction grating has 800 lines/mm. Monochromatic light of $\lambda = 550$ nm is incident on it.

1. Slit spacing $d$ (in μm)
2. Angle of the 1st-order maximum (in degrees, to 3 significant figures)
3. Maximum observable order $m_{\max}$

Resolving Power Drill 🔬

A grating has 5000 total slits.

1. What is the resolving power in the 2nd order?
2. In the 2nd order, what is the minimum wavelength difference $\Delta\lambda$ that can be resolved near $\lambda = 589$ nm? (in nm, to 3 significant figures)

Exit Quiz — Diffraction Gratings ✅

🫧 Thin-Film Interference

Part 5 of 7 — The Colors of Soap Bubbles and Oil Slicks

When light reflects off a thin film (like a soap bubble or oil on water), the reflections from the top and bottom surfaces interfere. This produces the beautiful rainbow colors you see in everyday life.

Phase Shifts on Reflection

The key to thin-film problems is understanding phase shifts:

The Rule

• Light reflecting off a surface with a higher index of refraction undergoes a ½λ phase shift (equivalent to a path difference of $\lambda/2$)
• Light reflecting off a surface with a lower index of refraction undergoes no phase shift

Example: Soap Film in Air

A soap film ($n_{\text{film}} > n_{\text{air}}$) in air:

Since only one reflection has a phase shift, the two reflected rays start ½λ out of phase. This changes the conditions!

Wavelength in the Film

Inside the film, light has a shorter wavelength:

$\lambda_{\text{film}} = \frac{\lambda}{n}$

The extra path traveled by ray 2 inside the film is $2t$ (down and back up), where $t$ is the film thickness.

Interference Conditions for Thin Films

Case 1: One Phase Shift (most common — soap bubbles, oil on water)

When there is exactly one ½λ phase shift between the two reflections:

Constructive (bright reflection): $2nt = \left(m + \frac{1}{2}\right)\lambda \quad (m = 0, 1, 2, \ldots)$

Destructive (no reflection): $2nt = m\lambda \quad (m = 0, 1, 2, \ldots)$

Case 2: Zero or Two Phase Shifts

When both reflections have the same phase shift (both shift, or neither shifts):

Constructive: $2nt = m\lambda \quad (m = 0, 1, 2, \ldots)$

Destructive: $2nt = \left(m + \frac{1}{2}\right)\lambda \quad (m = 0, 1, 2, \ldots)$

Memory Aid 🧠

"Odd shifts swap the conditions" — if there's an odd number of ½λ phase shifts, constructive and destructive conditions are swapped compared to ordinary interference.

Anti-Reflection Coatings

A thin coating with $n$ between air and glass can eliminate reflection at one wavelength. The coating thickness is chosen so the two reflected rays destructively interfere. Minimum thickness for destructive reflection:

$t = \frac{\lambda}{4n}$

Thin-Film Concept Quiz 🎯

Thin-Film Calculation Drill 🧮

A soap film ($n = 1.33$) in air is illuminated with white light.

1. What is the minimum thickness for the film to strongly reflect light of wavelength $\lambda = 630$ nm? (in nm, round to nearest integer)

2. An oil film ($n = 1.40$) floats on water ($n = 1.33$). What is the minimum thickness for constructive reflection of $\lambda = 560$ nm? (in nm)

3. An anti-reflection coating ($n = 1.38$) on glass ($n = 1.52$). What minimum thickness eliminates reflection at $\lambda = 550$ nm? (in nm, round to nearest integer)

Thin-Film Applications 🔍

Exit Quiz — Thin-Film Interference ✅

🔭 Resolution and the Rayleigh Criterion

Part 6 of 7 — The Diffraction Limit

Every optical instrument — telescope, microscope, camera, even your eye — has a fundamental limit on how fine a detail it can resolve. This limit comes not from manufacturing defects, but from the wave nature of light itself.

Diffraction Through Circular Apertures

When light passes through a circular opening of diameter $D$, it doesn't form a perfect point — it forms a diffraction pattern called an Airy disk:

• A bright central disk surrounded by faint rings
• The first dark ring occurs at angle:

$\sin\theta = 1.22 \frac{\lambda}{D}$

For small angles ($\sin\theta \approx \theta$ in radians):

$\theta_{\text{min}} = 1.22 \frac{\lambda}{D}$

Why 1.22?

The factor 1.22 comes from the mathematics of diffraction through a circular aperture (it involves Bessel functions). For a rectangular slit, the factor would be 1.00.

The Airy Disk

The central bright disk contains about 84% of the total light. The remaining 16% is spread across the surrounding rings. The angular radius of the central disk is $\theta_{\text{min}} = 1.22\lambda/D$.

The Rayleigh Criterion

The Problem

When two point sources (like two stars) are close together, their Airy disks overlap. At some point, you can no longer tell them apart.

Rayleigh's Rule

Two sources are just resolved when the central maximum of one falls on the first minimum of the other. This gives the minimum resolvable angle:

$\theta_{\text{min}} = 1.22 \frac{\lambda}{D}$

Key Insight

To improve resolution (resolve finer details), you need:

• Larger aperture $D$ → smaller $\theta_{\text{min}}$
• Shorter wavelength $\lambda$ → smaller $\theta_{\text{min}}$

This is why:

• Telescopes are built with large mirrors
• Microscopes use blue/UV light or electron beams for better resolution
• Radio telescopes must be enormous to achieve decent resolution

Resolution Concept Quiz 🎯

Resolution Calculation Drill 🧮

1. A telescope has a mirror diameter of $D = 2.4$ m. What is the minimum resolvable angle for light of $\lambda = 550$ nm? (in arcseconds, 1 rad = 206,265 arcsec; round to 3 significant figures)

2. Two stars are separated by an angle of $0.10$ arcseconds. What minimum telescope diameter is needed to resolve them at $\lambda = 500$ nm? (in meters, to 3 significant figures)

3. The pupil of the human eye has diameter $D \approx 5.0$ mm. What is the angular resolution limit at $\lambda = 550$ nm? (in arcminutes, to 3 significant figures; 1 rad = 3438 arcmin)

Resolution Applications 🔍

Exit Quiz — Rayleigh Criterion ✅

🎯 Synthesis & AP Review

Part 7 of 7 — Mastering Interference and Diffraction for the AP Exam

Let's bring everything together: compare the key phenomena, review common AP mistakes, and practice the types of questions you'll see on the exam.

Interference vs Diffraction — Side by Side

The Big Picture

Key Formulas to Memorize

⚠️ Common AP Mistakes

Mistake 1: Mixing Up Maxima and Minima Formulas

• Double slit: $d\sin\theta = m\lambda$ gives bright fringes
• Single slit: $a\sin\theta = m\lambda$ gives dark fringes
• They look identical but mean opposite things!

Mistake 2: Forgetting Phase Shifts in Thin Films

• Always check both surfaces for phase shifts
• Low $n$ → high $n$ reflection: ½λ shift
• High $n$ → low $n$ reflection: no shift
• Count the total shifts, then choose the right condition

Mistake 3: Central Maximum Width

• Single slit: central max is 2× the width of other maxima
• Double slit: all fringes are the same width

Mistake 4: Effect of Changing Variables

• Increasing wavelength → wider fringes (for both single and double slit)
• Increasing slit separation $d$ → narrower fringes
• Increasing slit width $a$ → narrower central max (less diffraction)
• Students often get the direction wrong!

Mistake 5: Rayleigh Criterion Units

• The formula gives $\theta$ in radians
• Convert to degrees, arcminutes, or arcseconds as needed
• Larger aperture = better (smaller) resolution angle

Formula Sorting Challenge 🧠

Match each situation to the correct formula or result:

AP FRQ Practice 📝

A student performs a double-slit experiment with $\lambda = 632.8$ nm (He-Ne laser), slit separation $d = 0.30$ mm, and screen distance $L = 1.5$ m.

1. Calculate the fringe spacing (in mm, to 3 significant figures)
2. The student then covers one slit. The central maximum width of the resulting single-slit pattern is 12.0 mm. What is the slit width $a$? (in mm, to 3 significant figures)
3. A diffraction grating with 300 lines/mm replaces the double slit. At what angle does the 2nd-order maximum appear? (in degrees, to 3 significant figures)

Final Mastery Quiz 🏆

Final Exit Quiz ✅