Light, Electromagnetic Radiation

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

What Light Actually Is

Must know

Light is a self-sustaining wave of oscillating electric ($\vec{E}$) and magnetic ($\vec{B}$) fields, each generating the other. Unlike sound, EM radiation needs no medium — it crosses the vacuum of space.

Propagation is perpendicular to both $\vec{E}$ and $\vec{B}$, and $\vec{E} \perp \vec{B}$. (Wave along $+z$ → $\vec{E}$ along $x$, $\vec{B}$ along $y$.) This mutual perpendicularity is directly tested.

Speed, Frequency, and Wavelength

Must know

In a vacuum, all EM radiation travels at the same speed:

$c = 3.00 \times 10^8 \ \text{m/s}, \qquad c = f\lambda$

where $f$ is frequency (Hz) and $\lambda$ is wavelength (m). Since $c$ is constant, frequency and wavelength are inversely related.

When light enters a medium, its speed decreases and wavelength shrinks, but its frequency stays the same (frequency is the invariant property of the photon). The index of refraction quantifies the slowdown:

$n = \frac{c}{v}, \qquad \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n}$

Quick check: A yellow photon is 580 nm in vacuum. Its wavelength in glass ($n = 1.5$)?

Answer: $\lambda_{\text{glass}} = 580/1.5 \approx 387 \ \text{nm}$. Frequency is unchanged; only wavelength and speed shrink.

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The Electromagnetic Spectrum and Photon Energy

The Spectrum

Must know

The ordering from lowest to highest energy/frequency (longest to shortest wavelength):

Region	Approx. Wavelength	Context
Radio	$> 1 \ \text{m}$	MRI (NMR)
Microwave	$\sim 1 \ \text{mm}–1 \ \text{m}$	Heating
Infrared (IR)	$\sim 700 \ \text{nm}–1 \ \text{mm}$	Molecular vibration, heat
Visible	$\sim 400–700 \ \text{nm}$	Vision
Ultraviolet (UV)	$\sim 10–400 \ \text{nm}$	DNA damage
X-rays	$\sim 0.01–10 \ \text{nm}$	Imaging, crystallography
Gamma	$< 0.01 \ \text{nm}$	Nuclear decay, highest energy

Mnemonic: Radio, Microwave, IR, Visible, UV, X-ray, Gamma.

Photon Energy: $E = hf$

Must know

Light comes in discrete photons, each carrying:

$E = hf = \frac{hc}{\lambda}$

where $h = 6.626 \times 10^{-34} \ \text{J·s}$ (Planck's constant). Higher frequency = shorter wavelength = higher energy. This is why UV (not visible) light damages DNA — each photon has enough energy to break bonds.

Passage-level

A convenient unit is the electron-volt, $1 \ \text{eV} = 1.6 \times 10^{-19} \ \text{J}$; visible photons carry ~1.8–3.1 eV.

Quick check: Energy of a 400 nm violet photon, in J and eV?

Answer: $E = hc/\lambda = (6.626\times10^{-34})(3.00\times10^8)/(4.00\times10^{-7}) = 4.97\times10^{-19}\ \text{J} \approx 3.1\ \text{eV}$ — right at the violet end of the visible range. A 700 nm red photon is $700/400 = 1.75\times$ less energetic (~1.8 eV).

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How Light Interacts With Matter

The Photoelectric Effect

Must know

When light of high enough frequency strikes a metal, it ejects photoelectrons — the canonical proof that light behaves as particles. Each photon delivers all its energy to one electron. To escape, an electron must overcome the work function $\Phi$; leftover energy becomes kinetic:

$KE_{\max} = hf - \Phi$

• Threshold frequency $f_0 = \Phi/h$: below it, no electrons eject no matter how intense the light.
• More intensity (above threshold) → more electrons, but same max KE.
• Higher frequency → higher KE per electron.

The stopping potential $V_s$ just halts the fastest electrons: $eV_s = KE_{\max}$.

Quick check: Metal with $\Phi = 2.0$ eV, hit by 2.0 eV photons. Electrons ejected? Answer: Just barely — $KE_{\max} = 0$. At threshold; any lower frequency ejects none.

Atomic Line Spectra: Emission and Absorption

Must know

Electrons occupy discrete (quantized) energy levels. A photon interacts only when its energy exactly matches a level gap:

$E_{\text{photon}} = hf = \Delta E = E_{\text{higher}} - E_{\text{lower}}$

• Absorption: cool gas removes matching wavelengths from white light → dark lines.
• Emission: excited atoms drop and emit those same wavelengths → bright lines on dark.

Each element's line spectrum is a unique fingerprint — the basis of flame tests and absorption spectroscopy.

Quick check: A hydrogen electron falls from $-1.5$ eV to $-3.4$ eV. Photon energy? Answer: $\Delta E = (-1.5)-(-3.4) = 1.9$ eV — the red Balmer line.

Fluorescence and Phosphorescence

Know the logic

In fluorescence, a molecule absorbs a high-energy (often UV) photon, loses some energy as heat, then re-emits a lower-energy / longer-wavelength photon — the Stokes shift. Emission is essentially immediate, so it stops the instant the source is removed (underlies fluorescence microscopy, GFP assays). Phosphorescence is the same process but via a long-lived excited state, so emission is delayed (glow-in-the-dark, seconds to minutes).

The Doppler Effect for Light

Know the logic

Relative motion shifts observed frequency:

• Approaching → higher frequency, shorter wavelength → blue shift.
• Receding → lower frequency, longer wavelength → red shift.

The red shift of distant galaxies is the primary evidence the universe is expanding. Only relative velocity matters (no medium).

Quick check: A star's absorption lines appear at longer wavelengths than in the lab. Approaching or receding? Answer: Receding — red shift indicates motion away.

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The Visual Spectrum and Color

Colors of Visible Light

Must know

Within ~400–700 nm, the brain reads wavelength as color, longest to shortest (lowest to highest energy):

$\text{Red} > \text{Orange} > \text{Yellow} > \text{Green} > \text{Blue} > \text{Violet}$

Mnemonic: ROY G BIV. Red is lowest energy; violet is highest.

Perceived Color vs. Absorbed Color

Must know

An object appears the color it reflects/transmits and absorbs the complementary color. Complementary pairs sit opposite on the color wheel — e.g., Red ↔ Cyan, Green ↔ Magenta, Blue ↔ Yellow.

Example: a solution looks blue because it absorbs orange/red and transmits blue — the principle behind colorimetric assays.

Dispersion

Know the logic

White light is all visible wavelengths mixed. In a prism, $n$ depends slightly on $\lambda$ (higher $n$ for shorter $\lambda$), so violet bends most, red bends least. This spreading is dispersion; water droplets cause rainbows the same way.

Quick check: Why does red bend less than violet in glass?

Answer: Glass has a slightly lower $n$ for red (longer $\lambda$), so red slows and bends less (Snell's law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$).

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Wave Nature of Light: Interference

The Core Idea

Must know

Waves superpose: crest + crest → constructive interference (bigger amplitude); crest + trough → destructive (cancels). This is the foundation of all interference and diffraction.

Young's Double-Slit Experiment

Must know

Monochromatic light through two slits separated by $d$ produces alternating bright/dark fringes on a screen distance $L$ away. The path difference between the two rays sets constructive vs. destructive:

$\text{Bright:} \quad d\sin\theta = m\lambda \qquad \text{Dark:} \quad d\sin\theta = \left(m + \tfrac{1}{2}\right)\lambda$

For small angles, $\sin\theta \approx y/L$, so the $m$-th bright fringe sits at:

$y_m = \frac{m\lambda L}{d}, \qquad \Delta y = \frac{\lambda L}{d}$

Larger $\lambda$ → wider spacing; larger $d$ → narrower spacing. Optional the same wave behavior applies to electrons (matter waves), linking to de Broglie wavelength.

Quick check: Replace red light (700 nm) with blue (400 nm). Fringe spacing wider or narrower? Answer: Narrower — $\Delta y \propto \lambda$.

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Diffraction

Must know

Diffraction is the bending of waves around edges or through openings, most pronounced when the opening is comparable to the wavelength.

Single-Slit Diffraction

Must know

A single slit of width $a$ gives a broad central maximum with dimmer side maxima. Minima (dark fringes) occur at:

$a\sin\theta = m\lambda \quad (m = \pm1, \pm2, \ldots)$

For single-slit, $a\sin\theta = m\lambda$ gives dark fringes; for double-slit, $d\sin\theta = m\lambda$ gives bright fringes. A narrower slit → wider central maximum (constraining a wave in space spreads it in angle).

Diffraction Grating

Must know

A diffraction grating has many equally spaced slits (spacing $d$), giving the same condition:

$d\sin\theta = m\lambda$

Because $d$ is tiny, the bright orders are far apart and sharp. Gratings are used in spectroscopy to separate wavelengths — relevant to atomic-structure passages.

Quick check: A grating with 500 lines/mm ($d = 2000 \ \text{nm}$): first-order ($m=1$) angle for red (700 nm)? Answer: $\sin\theta = (1)(700)/(2000) = 0.35$, so $\theta \approx 20.5°$.

Thin-Film Interference

Know the logic

Light hitting a thin film (soap bubble, oil slick, anti-reflective coating) partially reflects off the top and bottom surfaces; the two reflected rays interfere. Two factors set the phase difference:

1. Path length: the bottom ray travels an extra $2t$; in terms of film wavelength ($\lambda/n$), that's $2nt/\lambda$ cycles.
2. Reflection phase shift: reflecting off a higher-$n$ medium adds a $\lambda/2$ (180°) shift; off a lower-$n$ medium, none.

For a film of index $n$ in air, only the top reflection (air → film) gets a shift, so there is one net $\lambda/2$ shift, which flips the conditions:

$\text{Dark (destructive):} \quad 2nt = m\lambda \qquad \text{Bright (constructive):} \quad 2nt = \left(m + \tfrac{1}{2}\right)\lambda$

Key logic: 0 or 2 phase flips → constructive when $2nt = m\lambda$; 1 flip → constructive when $2nt = (m+\tfrac{1}{2})\lambda$.

Anti-reflective coatings exploit thin-film destructive interference to cancel reflected glare.

Worked example: Oil film ($n=1.5$) on water ($n=1.33$); minimum thickness for destructive interference (zero reflection) at $\lambda = 600$ nm? Top (air→oil): phase shift. Bottom (oil→water, high→low $n$): no shift → one net flip → destructive when $2nt = m\lambda$. Minimum at $m=1$: $t = \lambda/(2n) = 600/(2\cdot1.5) = 200 \ \text{nm}$. ✓ (right range for thin-film effects.)

X-Ray Diffraction and Bragg's Law

Must know

X-ray diffraction (XRD) works because crystal plane spacings (~ångströms) match X-ray wavelengths. Constructive interference off parallel atomic planes obeys Bragg's Law:

$2d\sin\theta = m\lambda$

where $d$ is plane spacing and $\theta$ is the glancing angle (measured from the plane, not the normal).

Passage-level

X-ray diffraction produced the key images of DNA and is standard for protein structures — it can appear in biochem passages.

Quick check: For fixed $\lambda$ and $d$, increasing $\theta$ does what to order $m$? Answer: $m = 2d\sin\theta/\lambda$; larger $\theta$ → higher orders at larger angles.

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Polarization of Light

The Concept

Must know

Unpolarized light has $\vec{E}$ oscillating in all transverse directions. Polarization restricts $\vec{E}$ to one orientation.

Linear Polarization

Know the logic

In linearly polarized light, $\vec{E}$ oscillates in a single plane. A polarizer transmits only the $\vec{E}$ component along its axis. Unpolarized light through one polarizer loses half its intensity. When already-polarized light hits a second polarizer (analyzer) at angle $\theta$:

• Aligned ($0°$): full transmission.
• Crossed ($90°$): zero transmission.
• Intermediate: partial.

Passage-level

This qualitative behavior is what matters — no intensity calculation required.

Circular Polarization

Passage-level

In circularly polarized light, $\vec{E}$ rotates as the wave propagates (arising from two perpendicular components with equal amplitude and a 90° phase difference).

MCAT relevance: chiral molecules (amino acids, sugars) rotate the plane of polarized light — optical rotation, measured by a polarimeter.

Brewster's Angle

Know the logic

At Brewster's angle $\theta_B$, reflected light is completely linearly polarized:

$\tan\theta_B = n$

($n$ = index of the second medium). This is why polarized sunglasses (vertical axis) block horizontally polarized glare from roads and water.

Quick check: Why do polarized sunglasses have vertical transmission axes? Answer: Glare reflected off horizontal surfaces is horizontally polarized (Brewster reflection); a vertical polarizer blocks it.

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Common Confusions & Tricks

1. Single-slit minima vs. double-slit maxima — same equation, opposite meaning. $a\sin\theta = m\lambda$ (single) gives dark; $d\sin\theta = m\lambda$ (double) gives bright. Note to self: "single slit: m → dark."

2. Thin film: count the phase flips. A $\lambda/2$ shift happens only at a low-$n$ → high-$n$ boundary. Label $n$ values, count flips: 0 or 2 → $2nt = m\lambda$ for constructive; 1 → $2nt = (m+\tfrac{1}{2})\lambda$ for constructive.

3. Bragg's Law: $\theta$ is the glancing angle, not from the normal. Opposite to Snell's convention. If given the angle from the perpendicular, subtract from 90°.

4. Frequency is invariant when light changes medium; wavelength and speed change. $E = hf$ doesn't change — it's the same photon. Wavelength ($\lambda = v/f$) shrinks.

5. Absorbed color ≠ observed color. A blue solution absorbs orange. Mix-ups are common in spectroscopy passages.

6. Polarizer vs. analyzer. Unpolarized light on the first polarizer always loses half, regardless of orientation. Only already-polarized light on a second polarizer has angle-dependent transmission.

7. Energy increases toward gamma, not radio. $E = hf$ — radio lowest, gamma highest. Don't confuse wavelength ordering with energy ordering.

8. Soap bubbles / oil slicks → thin-film interference. Different thicknesses destructively remove different wavelengths from reflected light.

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Key Equations

Equation	Variables & Use
$c = f\lambda$	$c$ = speed of light in vacuum, $f$ = frequency, $\lambda$ = wavelength
$n = c/v$	index of refraction; higher $n$ → slower speed, shorter $\lambda$
$E = hf = hc/\lambda$	photon energy; $h = 6.626\times10^{-34}$ J·s
$KE_{\max} = hf - \Phi$	photoelectric effect; threshold $f_0 = \Phi/h$
$E_{\text{photon}} = \Delta E = E_{\text{higher}} - E_{\text{lower}}$	atomic line spectra; photon energy = level gap
$d\sin\theta = m\lambda$	double-slit / grating bright fringes
$y_m = m\lambda L/d$	$m$-th bright fringe position (small-angle)
$a\sin\theta = m\lambda$	single-slit dark minima ($m = \pm1, \pm2,...$)
$2nt = m\lambda$	thin film, 0 or 2 flips: constructive
$2nt = (m+\tfrac{1}{2})\lambda$	thin film, 1 flip: constructive (min at $m=0 \Rightarrow t = \lambda/(4n)$)
$2d\sin\theta = m\lambda$	Bragg's Law; $d$ = lattice plane spacing, $\theta$ = glancing angle
$\tan\theta_B = n$	Brewster's angle; $n$ = index of second medium