AP Physics 2 Unit 6 Study Notes: Waves, Sound, and Physical Optics

Wave fundamentals: describing periodic waves

A wave is a traveling disturbance that transfers energy and momentum from one place to another without transporting matter overall. That last part is important: individual pieces of the medium (like air molecules for sound, or points on a string) oscillate around equilibrium, while the pattern moves.

In AP Physics 2, you’ll constantly move between three levels of thinking about waves:

1. Microscopic motion (what a little piece of the medium does)
2. Macroscopic pattern (the moving shape of the wave)
3. Measurable quantities (frequency, wavelength, speed, intensity)

If you keep those layers separate, many common confusions disappear.

Types of waves and what “transverse” vs “longitudinal” really means

A wave can be:

• Mechanical: requires a material medium (sound in air, waves on a string, water waves).
• Electromagnetic: does not require a medium (light), but still behaves like a wave.

Mechanical waves are often classified by the direction of oscillation compared to the direction the wave travels:

• Transverse wave: the medium oscillates perpendicular to the direction of travel (string waves, electromagnetic waves).
• Longitudinal wave: the medium oscillates parallel to the direction of travel (sound in air).

A classic misconception is thinking that “longitudinal means the wave moves along the medium” and “transverse means it moves across.” Both kinds travel along some direction. The terms describe the direction of the oscillation, not the direction of travel.

Key measurable quantities

For a repeating (periodic) wave, you describe the oscillation in time and in space.

• Amplitude: maximum displacement from equilibrium (or maximum pressure variation for sound). It connects to energy and intensity.
• Period T: time for one complete cycle.
• Frequency f: cycles per second.

f = \frac{1}{T}

• Wavelength \lambda: spatial period, the distance between repeating points (crest-to-crest, compression-to-compression).
• Wave speed v: how fast a point of constant phase (like a crest) moves.

These are related by the most-used wave relationship in the unit:

v = f\lambda

Why it matters: This equation is the bridge between time behavior and spatial behavior. If you know two of v, f, and \lambda, you know the third. Many exam questions are disguised versions of this relationship.

What sets the wave speed?

A deep idea: wave speed is set by the medium and the type of wave, not by how hard you “shake” it.

• For a wave on a particular string, speed depends on string tension and mass per length (you may be given this relationship in some contexts, but AP Physics 2 typically emphasizes the concept that the medium controls speed).
• For sound, speed depends strongly on the medium and temperature.
• For light in vacuum, speed is constant (but this unit focuses on wave behavior of light more than on deriving its speed).

Common misconception: “Higher amplitude waves move faster.” Not for ordinary linear waves. Increasing amplitude increases energy transfer but typically does not change v.

A basic traveling-wave model (what the equation means)

A common mathematical model for a sinusoidal wave traveling in the +x direction is:

y(x,t) = A\sin(kx - \omega t + \phi)

Where:

• A is amplitude
• k is the wave number, related to wavelength
• \omega is the angular frequency, related to frequency
• \phi is a phase constant that shifts the wave

The relationships are:

k = \frac{2\pi}{\lambda}

\omega = 2\pi f

How to interpret it: At a fixed position x, the displacement varies sinusoidally in time. At a fixed time t, the displacement varies sinusoidally in space.

Even if your class doesn’t emphasize this equation, understanding “phase” helps tremendously with interference later.

Example: linking graphs in space and time

Suppose a wave has frequency f = 5\text{ Hz} and wavelength \lambda = 2\text{ m}.

1) Wave speed:

v = f\lambda = (5)(2) = 10\text{ m/s}

2) Period:

T = \frac{1}{f} = \frac{1}{5} = 0.2\text{ s}

Interpretation: every 0.2\text{ s} a given point repeats its motion, and the pattern advances 10\text{ m} each second.

Exam Focus

• Typical question patterns:
  • Given f and \lambda (or a graph), find v or T.
  • Identify whether a wave is transverse or longitudinal from a description.
  • Translate between “how a point moves” (time graph) and “what the shape looks like” (position graph).
• Common mistakes:
  • Treating amplitude as affecting wave speed.
  • Confusing frequency (time cycles) with wavelength (space cycles).
  • Mixing up the direction of travel with direction of oscillation.

Superposition: interference as “adding displacements”

The most powerful idea in this unit is the principle of superposition: when two (or more) waves overlap in a linear medium, the resulting displacement is the algebraic sum of the individual displacements.

This matters because many wave phenomena that look “complicated” are just consequences of addition:

• Interference patterns
• Standing waves
• Beats
• Diffraction (which is interference from many points)

Constructive and destructive interference

When two waves overlap, you look at their phase relationship.

• Constructive interference happens when the waves arrive in phase (crest with crest). The resultant amplitude is larger.
• Destructive interference happens when the waves arrive out of phase by half a cycle (crest with trough). The resultant amplitude is smaller, possibly zero.

A crucial point: destructive interference does not violate energy conservation. Energy is redistributed—often into regions of constructive interference.

Path difference and phase difference (the interference “decision tool”)

For many interference setups, especially with light and sound, the key quantity is the path difference \Delta L between two waves traveling from two sources to the same point.

• If \Delta L is an integer multiple of wavelength, the waves arrive in phase.

\Delta L = m\lambda

• If \Delta L is a half-integer multiple of wavelength, they arrive out of phase.

\Delta L = \left(m + \frac{1}{2}\right)\lambda

Here m = 0, 1, 2, \dots

These conditions show up repeatedly for two-source interference (double slit), diffraction gratings, and sometimes sound-source interference.

Coherence: when stable interference is possible

Stable, lasting interference patterns require coherent sources: sources with a constant phase relationship.

• Two independent flashlights are not coherent, so you don’t see a stable fringe pattern.
• A double-slit experiment uses a single source split into two paths, creating coherence.

Example: two speakers and a “dead spot”

Two in-phase speakers emit sound of wavelength \lambda = 0.85\text{ m}. You are standing at a point where one speaker is 3.40\text{ m} away and the other is 3.83\text{ m} away.

Compute path difference:

\Delta L = 3.83 - 3.40 = 0.43\text{ m}

Compare to wavelength:

\frac{\Delta L}{\lambda} = \frac{0.43}{0.85} \approx 0.51

That is close to \frac{1}{2}, so you are near destructive interference (a quieter spot). On an exam, you’d usually be given cleaner numbers that hit exactly 0.5\lambda.

What can go wrong: Students often try to use frequency conditions here. For a fixed frequency (thus fixed wavelength in a given medium), interference at a point is about path difference, not “how loud the speakers are.”

Exam Focus

• Typical question patterns:
  • Decide constructive vs destructive interference from a path difference.
  • Use diagrams to compute \Delta L and compare to \lambda.
  • Explain qualitatively why energy is not destroyed in destructive interference.
• Common mistakes:
  • Using amplitude to decide constructive vs destructive; it’s phase/path difference.
  • Forgetting that “in phase at the sources” matters.
  • Claiming destructive interference means energy disappears.

Standing waves and resonance: waves trapped by boundaries

A standing wave forms when two waves of the same frequency and amplitude travel in opposite directions and interfere. Instead of a traveling pattern, you get a fixed pattern of nodes and antinodes.

Standing waves matter because they explain:

• Musical instruments (strings and air columns)
• Resonance in pipes, rooms, and cavities
• Why only certain frequencies “fit” in a system

Nodes, antinodes, and why the boundary conditions matter

• A node is a point of zero displacement (always at equilibrium).
• An antinode is a point of maximum displacement (largest oscillation).

Standing waves are controlled by boundary conditions:

• A string fixed at an end must have zero displacement there, so the end is a node.
• An open end of an air column is a displacement antinode (air can move freely).
• A closed end of an air column is a displacement node (air can’t move through the boundary).

A common mistake is mixing displacement patterns with pressure patterns in sound. In air columns:

• Displacement node corresponds to pressure antinode.
• Displacement antinode corresponds to pressure node.

AP questions sometimes test this qualitatively.

Allowed wavelengths and harmonics on a string (or open-open pipe)

For a string of length L fixed at both ends, the standing wave must have nodes at both ends.

The allowed wavelengths are:

\lambda_n = \frac{2L}{n}

and the allowed frequencies are:

f_n = \frac{nv}{2L}

where n = 1, 2, 3, \dots and v is the wave speed in the medium.

• n = 1 is the fundamental (first harmonic).
• Higher n values are harmonics (overtones).

Why it matters: This is the physics behind “only certain notes” coming from a particular string length and tension.

One end closed, one end open (closed-open pipe)

For an air column closed at one end and open at the other:

• Closed end: displacement node
• Open end: displacement antinode

This forces a different set of allowed wavelengths:

\lambda_n = \frac{4L}{2n - 1}

and frequencies:

f_n = \frac{(2n - 1)v}{4L}

where n = 1, 2, 3, \dots

Only odd harmonics appear (1st, 3rd, 5th, …). Students often memorize this but forget it comes from the boundary conditions.

Resonance: large amplitude at the natural frequencies

Resonance occurs when a driving frequency matches one of the system’s natural frequencies. Because the standing-wave pattern is “supported” by the boundaries, energy input each cycle adds efficiently, producing a large amplitude.

Real systems have damping, so resonance peaks are finite rather than infinite. Conceptually:

• Less damping means a sharper, higher resonance peak.
• More damping means a broader, lower peak.

Worked problem: third harmonic on a string

A string of length L = 0.80\text{ m} has wave speed v = 120\text{ m/s}. Find the frequency of the third harmonic.

Use:

f_n = \frac{nv}{2L}

For n = 3:

f_3 = \frac{3(120)}{2(0.80)} = \frac{360}{1.6} = 225\text{ Hz}

Sanity check: higher harmonics have higher frequency, so 225\text{ Hz} should be 3 times the fundamental.

Worked problem: fundamental of a closed-open pipe

A tube is closed at one end and open at the other, with length L = 0.25\text{ m}. Assume speed of sound v = 340\text{ m/s}. Find the fundamental frequency.

For closed-open, the fundamental corresponds to n = 1:

f_1 = \frac{v}{4L} = \frac{340}{4(0.25)} = \frac{340}{1.0} = 340\text{ Hz}

Exam Focus

• Typical question patterns:
  • Determine which harmonics are allowed given end conditions (fixed, open, closed).
  • Compute a resonant frequency from L and v.
  • Interpret node/antinode diagrams and connect them to harmonic number.
• Common mistakes:
  • Using the wrong length-wavelength relationship (mixing string/open-open with closed-open).
  • Confusing displacement nodes with pressure nodes in air columns.
  • Treating “resonance” as a property only of sound; it’s a general wave/oscillation phenomenon.

Sound waves: production, speed, intensity, and perception

Sound in air is a longitudinal mechanical wave: oscillations of pressure and density travel through the medium. You can picture alternating compressions (high pressure) and rarefactions (low pressure).

Sound topics in AP Physics 2 connect wave physics to real measurements: loudness, pitch, hearing, and how sound spreads in space.

Speed of sound: what controls it

The speed of sound depends on properties of the medium (how stiff it is and how much inertia it has).

Qualitative rules you should know:

• Sound travels faster in solids than in liquids, and faster in liquids than in gases (generally) because particles are more strongly coupled in stiffer media.
• In gases, sound speed increases with temperature (warmer gas particles move faster, enabling disturbances to propagate more quickly).

Once the medium and conditions are set, sound speed is essentially fixed—changing frequency changes wavelength via:

v = f\lambda

So in a fixed medium, if f goes up, \lambda must go down.

Intensity: power spread over area

Intensity is the rate of energy transfer per area perpendicular to the wave direction:

I = \frac{P}{A}

where P is power and A is area.

For a point source radiating uniformly in all directions (a common idealization), the wave spreads over a sphere of radius r, so:

I = \frac{P}{4\pi r^2}

This is the inverse-square law behavior.

Why it matters: Many “loudness vs distance” problems are really geometry problems. Doubling distance does not halve intensity—it quarters it.

Decibels: a logarithmic intensity scale

Human hearing spans an enormous range of intensities, so sound level is measured on a logarithmic scale.

The intensity level \beta in decibels (dB) is:

\beta = 10\log_{10}\left(\frac{I}{I_0}\right)

where I_0 is a reference intensity typically taken as:

I_0 = 1.0\times 10^{-12}\text{ W/m}^2

Key implications:

• Increasing intensity by a factor of 10 increases \beta by 10 dB.
• Increasing intensity by a factor of 2 increases \beta by about 3 dB.

Common misconception: Many students assume dB is proportional to intensity. It’s proportional to the logarithm of intensity.

Pitch vs loudness (frequency vs amplitude)

• Pitch is primarily related to frequency: higher f means higher pitch.
• Loudness is related to intensity (and therefore amplitude), but human perception is not linear.

So:

• If you double frequency, you change pitch but do not automatically change loudness.
• If you increase amplitude, you change loudness but not pitch (for a pure tone).

Worked problem: inverse-square intensity change

A siren emits power P = 2.0\text{ W} uniformly. Find intensity at r = 5.0\text{ m}.

Use:

I = \frac{P}{4\pi r^2} = \frac{2.0}{4\pi(5.0)^2} = \frac{2.0}{100\pi}

I \approx \frac{2.0}{314} \approx 6.4\times 10^{-3}\text{ W/m}^2

If you move to r = 10\text{ m}, intensity becomes one quarter of this value (because distance doubled).

Worked problem: decibel comparison

Two sounds have intensities I_1 = 1.0\times 10^{-6}\text{ W/m}^2 and I_2 = 1.0\times 10^{-8}\text{ W/m}^2. Find the difference in intensity level.

Difference in dB:

\Delta \beta = 10\log_{10}\left(\frac{I_1}{I_2}\right) = 10\log_{10}(100) = 10(2) = 20\text{ dB}

Notice you didn’t need I_0 because it cancels when comparing two levels.

Exam Focus

• Typical question patterns:
  • Use inverse-square spreading to relate intensity at two distances.
  • Convert intensity ratios to decibel differences.
  • Concept questions distinguishing pitch vs loudness, or how wavelength changes when frequency changes.
• Common mistakes:
  • Saying sound speed increases when frequency increases (in a fixed medium).
  • Treating dB as linear with intensity.
  • Forgetting the area for spherical spreading is 4\pi r^2.

Doppler effect and beats: when frequency changes (or nearly matches)

Two “everyday weird” sound phenomena—ambulance pitch changes and pulsing loudness—are both explained by interference and timing.

Doppler effect: perceived frequency shifts from relative motion

The Doppler effect is the change in observed frequency due to relative motion between a source and an observer.

Conceptually:

• If the source and observer move toward each other, wavefronts arrive more frequently, so the observed frequency increases.
• If they move away, wavefronts arrive less frequently, so observed frequency decreases.

A useful algebraic expression (for motion along the line connecting source and observer) is:

f' = f\left(\frac{v \pm v_o}{v \mp v_s}\right)

Where:

• f' is observed frequency
• f is emitted frequency
• v is wave speed in the medium (speed of sound)
• v_o is observer speed relative to the medium
• v_s is source speed relative to the medium

Sign logic:

• In the numerator, use +v_o when the observer moves toward the source.
• In the denominator, use -v_s when the source moves toward the observer.

Students often try to memorize signs without meaning. A safer approach is to reason: “toward increases frequency,” then choose signs to make the factor bigger than 1.

Important limit: This formula assumes speeds are less than the wave speed and the medium is at rest in the chosen frame.

Worked problem: moving source toward a stationary observer

A car horn emits f = 500\text{ Hz}. The car moves toward you at v_s = 20\text{ m/s}. You are stationary, and take v = 340\text{ m/s}.

Observer is stationary so v_o = 0. Source moves toward observer so use minus in denominator:

f' = f\left(\frac{v}{v - v_s}\right) = 500\left(\frac{340}{340 - 20}\right) = 500\left(\frac{340}{320}\right)

f' = 500(1.0625) = 531\text{ Hz}

So you hear a higher pitch.

Beats: interference in time from slightly different frequencies

Beats occur when two waves of similar frequency overlap. Sometimes they reinforce and sometimes they cancel, producing a periodic rise and fall in intensity (a “wah-wah” effect).

The beat frequency is the absolute difference of the frequencies:

f_{\text{beat}} = |f_1 - f_2|

Why it matters: Beats are used to tune instruments. When the beat frequency goes to zero, the frequencies match.

Worked problem: tuning with beats

A tuning fork of 440\text{ Hz} is compared with a piano string. You hear beats at 3\text{ Hz}.

Then:

|f_{\text{string}} - 440| = 3

So:

f_{\text{string}} = 443\text{ Hz}

or

f_{\text{string}} = 437\text{ Hz}

You would need additional information (e.g., whether tightening the string increases or decreases beats) to choose which.

Exam Focus

• Typical question patterns:
  • Compute f' for Doppler situations with moving source and or moving observer.
  • Reason qualitatively about when observed frequency is greater or less than emitted.
  • Use f_{\text{beat}} = |f_1 - f_2| for tuning or identifying unknown frequency.
• Common mistakes:
  • Mixing up which speed goes in numerator vs denominator in Doppler calculations.
  • Forgetting beats are a time variation in loudness, not a spatial fringe pattern.
  • Choosing a single answer for beats when two frequencies are possible.

Light as a wave: two-slit interference and fringe geometry

Physical optics focuses on what cannot be explained by rays alone: interference and diffraction. These phenomena reveal that light behaves like a wave with a wavelength small enough that you typically need narrow slits or precise setups to notice wave effects.

Young’s double-slit experiment: the core idea

In a double-slit setup:

• A monochromatic light source illuminates two nearby slits separated by distance d.
• Each slit acts like a coherent source.
• On a screen far away, light from the two slits overlaps.

Where waves arrive in phase, you see bright fringes; where they arrive out of phase, you see dark fringes.

Constructive and destructive conditions (two-slit)

For a point at angle \theta from the central axis, the path difference is approximately d\sin\theta.

Bright fringes (constructive interference):

d\sin\theta = m\lambda

Dark fringes (destructive interference):

d\sin\theta = \left(m + \frac{1}{2}\right)\lambda

Here m = 0, 1, 2, \dots

Why it matters: Many AP questions reduce to “find the angle or position of the mth bright fringe.”

Small-angle screen geometry

If the screen is a distance L away and the fringe is at vertical position y from the center, then:

\tan\theta = \frac{y}{L}

For small angles, \sin\theta \approx \tan\theta \approx \frac{y}{L}, giving:

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

This approximation is used frequently in algebra-based problems.

Common misconception: Students sometimes use y_m = m\lambda L/d even when angles are not small. In AP problems, you’ll usually be in the small-angle regime unless explicitly told otherwise.

Intensity depends on amplitude, not just “bright or dark”

Interference is not only about where maxima and minima occur. The brightness pattern is determined by how amplitudes add.

• Two equal-amplitude waves in phase produce an amplitude of 2A, which corresponds to an intensity proportional to (2A)^2 = 4A^2 . So the bright fringes can be much brighter than one slit alone.
• Destructive interference corresponds to near-zero amplitude.

AP often emphasizes the qualitative idea “intensity depends on amplitude squared” even if you don’t compute exact intensities.

Worked problem: fringe spacing

A double-slit has separation d = 0.25\text{ mm}. A screen is L = 2.0\text{ m} away. Light has wavelength \lambda = 500\text{ nm}. Find the distance between adjacent bright fringes.

Adjacent bright fringes differ by \Delta m = 1, so spacing:

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

Convert units:

• \lambda = 500\times 10^{-9}\text{ m}
• d = 0.25\times 10^{-3}\text{ m}

Compute:

\Delta y = \frac{(500\times 10^{-9})(2.0)}{0.25\times 10^{-3}}

\Delta y = \frac{1000\times 10^{-9}}{0.25\times 10^{-3}} = \frac{1.0\times 10^{-6}}{2.5\times 10^{-4}} = 4.0\times 10^{-3}\text{ m}

So fringes are spaced by about 4\text{ mm}.

Exam Focus

• Typical question patterns:
  • Calculate bright fringe locations using d\sin\theta = m\lambda or y_m = m\lambda L/d.
  • Identify how changing \lambda, d, or L changes fringe spacing.
  • Explain why coherence is required and why two slits create it.
• Common mistakes:
  • Forgetting to convert nm and mm to meters.
  • Using the wrong slit distance (mixing slit width and slit separation).
  • Treating bright fringes as simply “twice as bright” rather than relating intensity to amplitude squared.

Diffraction: interference from a single aperture (and the limits of sharp images)

Diffraction is the spreading and bending of waves when they pass through an opening or around an obstacle. In wave terms, you can think of each point across an opening as generating secondary wavelets; those wavelets interfere to create a pattern.

Diffraction matters because it sets a fundamental limit: even a perfect lens cannot focus light into an infinitely small point. Any finite aperture causes spreading.

When diffraction becomes significant

Diffraction effects are strongest when the opening size is comparable to the wavelength.

• For sound (wavelengths from centimeters to meters), diffraction around doorways and corners is very noticeable.
• For visible light (wavelengths around hundreds of nanometers), everyday objects are huge compared with \lambda, so diffraction is usually subtle unless you use narrow slits.

Single-slit diffraction pattern (minima condition)

For a slit of width a, dark fringes (minima) occur approximately when:

a\sin\theta = m\lambda

where m = 1, 2, 3, \dots

The central bright maximum is broad, and the intensity falls off in side lobes.

Important distinction:

• Double-slit interference: spacing primarily depends on slit separation d.
• Single-slit diffraction: envelope (overall spread) depends on slit width a.

In many real experiments, you see both: fine double-slit fringes modulated by a broad single-slit envelope.

Diffraction gratings (many slits)

A diffraction grating has many equally spaced slits. The condition for principal maxima is similar to the double-slit case:

d\sin\theta = m\lambda

But with many slits, maxima become much sharper and brighter, which is why gratings are used to separate wavelengths in spectroscopy.

What can go wrong: Students sometimes think gratings use a different equation. The angle condition looks the same; the pattern sharpness is what changes.

Resolution and the “diffraction limit” (qualitative)

Even with a perfect optical system, two nearby point sources produce overlapping diffraction patterns. If the patterns overlap too much, the sources can’t be distinguished. This is the idea of diffraction-limited resolution.

Qualitatively, to improve resolution you can:

• Use a smaller wavelength \lambda (bluer light resolves better than red light)
• Use a larger aperture (bigger lens or telescope)

AP Physics 2 typically emphasizes these relationships conceptually rather than requiring a specific resolution formula.

Worked problem: first minimum angle for a single slit

Light of wavelength \lambda = 650\text{ nm} passes through a slit of width a = 0.050\text{ mm}. Find the angle to the first minimum.

Use m = 1:

a\sin\theta = \lambda

Convert:

• \lambda = 650\times 10^{-9}\text{ m}
• a = 0.050\times 10^{-3}\text{ m} = 5.0\times 10^{-5}\text{ m}

So:

\sin\theta = \frac{\lambda}{a} = \frac{650\times 10^{-9}}{5.0\times 10^{-5}} = 1.3\times 10^{-2}

Thus \theta \approx 0.013\text{ rad} (small angle), about 0.75^\circ.

Exam Focus

• Typical question patterns:
  • Use a\sin\theta = m\lambda to find minima angles or relate slit width to pattern spread.
  • Compare diffraction and interference patterns conceptually.
  • Explain why smaller \lambda or larger aperture improves resolution.
• Common mistakes:
  • Using slit separation d when the problem is about slit width a.
  • Starting m at 0 for single-slit minima (it starts at 1).
  • Interpreting diffraction as “light turning corners because it slows down” rather than as wave spreading and interference.

Polarization: a wave property that reveals transverse behavior

Polarization describes the orientation of oscillations in a transverse wave. It is a uniquely transverse phenomenon.

This is a big conceptual lever:

• Light can be polarized, so light must have transverse-wave behavior.
• Sound in air cannot be polarized (it’s longitudinal in air).

What it means for light to be polarized

For an electromagnetic wave, the electric field oscillates in a direction perpendicular to propagation. Unpolarized light has electric field directions that are randomly distributed over time. Linearly polarized light has its electric field confined to one plane.

A polarizing filter only transmits the component of the electric field aligned with its transmission axis.

Malus’s law (intensity through polarizers)

If polarized light of intensity I_0 passes through a polarizer whose axis makes an angle \theta with the incoming polarization direction, the transmitted intensity is:

I = I_0\cos^2\theta

Special cases you should be fluent with:

• \theta = 0^\circ: I = I_0 (maximum transmission)
• \theta = 90^\circ: I = 0 (ideal extinction)
• \theta = 45^\circ: I = I_0/2

If unpolarized light passes through a single ideal polarizer, the transmitted intensity is half:

I = \frac{1}{2}I_0

This happens because the average of \cos^2\theta over all orientations is 1/2.

Worked problem: three polarizers

Unpolarized light with intensity I_0 passes through three ideal polarizers:

1) First polarizer: transmits half.

I_1 = \frac{1}{2}I_0

2) Second polarizer: at 45^\circ relative to the first.

I_2 = I_1\cos^2 45^\circ = \frac{1}{2}I_0\left(\frac{1}{2}\right) = \frac{1}{4}I_0

3) Third polarizer: at 90^\circ relative to the first, so it is 45^\circ relative to the second.

I_3 = I_2\cos^2 45^\circ = \frac{1}{4}I_0\left(\frac{1}{2}\right) = \frac{1}{8}I_0

This result surprises many students because if you used only two polarizers at 90^\circ (crossed), you’d get zero. The middle polarizer “rotates” the polarization in steps, allowing some light through.

Real-world applications

• Polarized sunglasses reduce glare because reflections off surfaces (like water or roads) are often partially polarized.
• Polarization is used in LCD screens and stress analysis in transparent materials.

Exam Focus

• Typical question patterns:
  • Apply Malus’s law to one or more polarizers.
  • Explain why polarization implies transverse waves.
  • Predict qualitative changes in brightness as the polarizer rotates.
• Common mistakes:
  • Using I = I_0\cos\theta instead of I = I_0\cos^2\theta.
  • Forgetting the first polarizer halves unpolarized light.
  • Claiming sound can be polarized in air (it cannot, because it is longitudinal).