← All Topics
PHYS 1120 · Chapter 23 · Tutorial

Light, Refraction & Optical Properties

18 slides · inline SVG diagrams · ← → arrow keys to navigate
Slide 1 · Overview

What Chapter 23 is really about

Light is an electromagnetic wave that travels at c = 3×10⁸ m/s in a vacuum. The moment it crosses into a new material — glass, water, diamond — something remarkable happens: it slows down and changes direction. Chapter 23 asks three tightly linked questions:

  • HOW MUCH does it bend? — Snell's Law (n₁ sinθ₁ = n₂ sinθ₂)
  • WHEN does it totally reflect? — Total Internal Reflection and the critical angle
  • HOW do wave properties (speed, wavelength, frequency) change? — index of refraction and the invariance of frequency
Medium 1 (air, n=1.00) Medium 2 (glass, n=1.52) normal incident ray θ₁ refracted ray θ₂ reflected Key facts: • n = c/v (index of refraction) • θ₂ < θ₁ → bending toward normal • Frequency: CONSTANT across boundary • Wavelength: λ = λ₀/n (shrinks) Snell's Law: n₁ sinθ₁ = n₂ sinθ₂
Light entering glass bends toward the normal because it slows down. θ₁ and θ₂ are both measured from the dashed normal line.

Roadmap of this deck

  • Slides 2–3: Index of refraction, speed, wavelength, and frequency in media
  • Slides 4–5: Snell's Law — formula, application, and reading diagrams
  • Slides 6–7: Speed and wavelength comparisons from diagram geometry
  • Slides 8–9: Total Internal Reflection — critical angle, why air→water TIR is impossible
  • Slides 10–12: TIR applications, dispersion, prism problems
  • Slides 13–16: Polarization and worked exam problem types
  • Slides 17–18: Common exam traps and the complete cheat sheet
Why this chapter matters for Exam 3

Exam 3 heavily tests diagram-reading: can you look at a bent ray and immediately say which medium is denser, faster, and has the longer wavelength? Slides 5–7 drill exactly this skill. Total Internal Reflection (slides 8–9) is a consistent source of exam questions — especially the air-to-water case that students almost always get wrong.

Slide 2 · Index of Refraction

Speed of Light in Media: the Index of Refraction

Light travels at c = 3×10⁸ m/s only in a perfect vacuum. In any material, it slows down. The index of refraction n quantifies how much:

n = c / v     ⟹     v = c / n

Since n ≥ 1 always (light never goes faster than c in vacuum), and n is dimensionless. The larger n is, the slower light travels in that material.

Table of Common Indices of Refraction

Material n v = c/n (m/s)
Vacuum 1.000 3.00 × 10⁸
Air 1.0003 ≈ 3.00 × 10⁸ (essentially same)
Water 1.33 2.26 × 10⁸
Glass (typical) 1.52 1.97 × 10⁸
Diamond 2.42 1.24 × 10⁸
Air (n = 1.00) — fast Glass (n = 1.52) — slower direction of propagation → λ₀ (in air) λ = λ₀/n
Wave fronts compress as light enters glass — same frequency, shorter wavelength, slower speed. The spacing between fronts is λ = λ₀/n.
Key insight to memorize

Higher n = slower light = shorter wavelength. Diamond (n=2.42) slows light to 41% of c. Air (n≈1.00) is essentially the same as vacuum — in all exam problems, treat air as n=1.00 unless told otherwise.

Slide 3 · Wave Properties in Media

Wavelength and Frequency in a Medium

When light crosses a boundary into a new medium, two of its properties change, and one stays the same. Understanding which is which is half the battle of Chapter 23 exam problems.

What changes, what doesn't

  • Frequency: NEVER changes. The source (e.g., a laser) oscillates at a fixed rate. The boundary cannot create or destroy oscillations — every wave crest that arrives at the surface must also leave the surface. f is set by the source alone.
  • Speed: DOES change. v = c/n. In a denser medium, the electromagnetic wave interacts more with atoms and slows.
  • Wavelength: DOES change. Since v = fλ and f is constant, if v decreases, λ must decrease proportionally.
v = f λ     (always true) f = constant across boundary v = c/n → λ = v/f = (c/n)/f = (c/f)/n = λ₀/n ∴   λ_medium = λ₀ / n

Worked Example (likely on your exam)

λ₀ = 256 nm in vacuum. Glass has n = 1.6. Find the frequency and wavelength inside the glass.

Step 1 — Frequency (same in glass as in vacuum): f = c/λ₀ = (3×10⁸ m/s) / (256×10⁻⁹ m) = 1.172 × 10¹⁵ Hz ← same in glass Step 2 — Wavelength in glass: λ_glass = λ₀/n = 256 nm / 1.6 = 160 nm Step 3 — Speed in glass (optional check): v_glass = c/n = 3×10⁸/1.6 = 1.875×10⁸ m/s Check: λ = v/f = 1.875×10⁸ / 1.172×10¹⁵ = 160 nm ✓
VACUUM / AIR GLASS (n=1.6) λ₀ = 256 nm λ = 160 nm f = 1.172×10¹⁵ Hz f = 1.172×10¹⁵ Hz (same!) v = c = 3×10⁸ m/s v = 1.875×10⁸ m/s
The wave compresses as it enters glass — frequency stays identical, but wavelength shrinks from 256 nm to 160 nm because the wave moves more slowly.
Exam trap: Which changes?

Questions often ask "what is the frequency inside glass?" The answer is the same as in vacuum. If a question asks for the wavelength, divide by n. Never multiply. The mnemonic: entering a denser medium — wavelength gets shorter (divided by n > 1).

Slide 4 · Snell's Law

Snell's Law: Quantifying the Bend

When light hits an interface between two media at an angle, it bends. The exact relationship between the angle coming in and the angle going out is given by Snell's Law:

n₁ sinθ₁ = n₂ sinθ₂

Critical rule: All angles (θ₁ and θ₂) are measured from the normal — the perpendicular to the surface. NEVER from the surface itself. This is the single most common source of errors on optics exams.

The two cases

  • Air → Glass (n₁ < n₂): Since n₂ > n₁, we need sinθ₂ < sinθ₁, so θ₂ < θ₁. The ray bends toward the normal.
  • Glass → Air (n₁ > n₂): Since n₂ < n₁, we need sinθ₂ > sinθ₁, so θ₂ > θ₁. The ray bends away from the normal.
Medium 1: n₁ Medium 2: n₂ NORMAL θ₁ incident ray refracted ray θ₂ reflected (partial) If n₁ < n₂ (air→glass): sinθ₂ = (n₁/n₂)sinθ₁ n₁/n₂ < 1 → sinθ₂ < sinθ₁ θ₂ < θ₁ (bends toward normal) If n₁ > n₂ (glass→air): sinθ₂ = (n₁/n₂)sinθ₁ n₁/n₂ > 1 → sinθ₂ > sinθ₁ θ₂ > θ₁ (bends away from normal)
Labeled Snell's Law diagram. θ₁ is in medium 1, θ₂ is in medium 2, both measured from the red dashed normal. Here n₁ < n₂ (going denser), so θ₂ < θ₁.
Quick numerical example

Light in air (n₁=1.00) hits glass (n₂=1.52) at θ₁ = 45°.
sinθ₂ = (n₁/n₂)sinθ₁ = (1.00/1.52)sin45° = (0.658)(0.707) = 0.465
θ₂ = sin⁻¹(0.465) = 27.7° — significantly smaller than 45°, confirming it bends toward the normal.

Slide 5 · Exam Skill: Reading Diagrams

How to Read a Refraction Diagram (Critical Exam Skill)

Exam questions frequently show a diagram of a ray crossing a boundary and ask: "which medium has larger n?" or "in which medium is the speed greater?" You do NOT need to know the actual angle values — just compare whether the ray bent toward or away from the normal.

The two key scenarios

Case A: bends TOWARD normal θ₂ < θ₁ → n₂ > n₁ θ₁ θ₂ Medium 1 Medium 2 n₂ > n₁, v₂ < v₁, λ₂ < λ₁ Case B: bends AWAY from normal θ₂ > θ₁ → n₂ < n₁ θ₁ θ₂ Medium 1 Medium 2 n₂ < n₁, v₂ > v₁, λ₂ > λ₁
Case A: ray bends toward normal → denser medium (larger n, slower v, shorter λ). Case B: ray bends away from normal → less dense medium (smaller n, faster v, longer λ). No numbers needed — just the direction of the bend.

Decision tree for exam diagrams

  • Draw the normal (mentally). Is the angle in medium 2 smaller than in medium 1? → n₂ > n₁ (medium 2 is denser, slower, shorter λ).
  • Is the angle in medium 2 larger than in medium 1? → n₂ < n₁ (medium 2 is less dense, faster, longer λ).
  • Are the angles equal? → n₁ = n₂ (same medium on both sides).
The one sentence that solves 80% of diagram questions

"The bent-toward side is the denser side." If the ray bends toward the normal when crossing the boundary, the medium it's entering is denser (higher n). Write this on your scratch paper at the start of the exam.

Slide 6 · Speed from Geometry

Determining Speed Ratio from a Diagram

Once you know how to read a refraction diagram, you can extract quantitative information about the speeds. Snell's Law can be rewritten in terms of wave speeds, which gives a direct connection between angles and speeds.

Snell's Law: n₁ sinθ₁ = n₂ sinθ₂ Since n = c/v: (c/v₁) sinθ₁ = (c/v₂) sinθ₂ Cancel c: sinθ₁/v₁ = sinθ₂/v₂ Rearrange: v₁/v₂ = sinθ₁/sinθ₂ = n₂/n₁

Reading speed from angle comparison

  • If θ₂ > θ₁ (bends away): sinθ₂ > sinθ₁, so v₂ > v₁ — light is faster in medium 2.
  • If θ₂ < θ₁ (bends toward): sinθ₂ < sinθ₁, so v₂ < v₁ — light is slower in medium 2.
  • The speed RATIO: v₂/v₁ = sinθ₂/sinθ₁ (plug in actual angle values if given).
θ₁ v₁ (slower) θ₂ v₂ (faster) θ₂ > θ₁ sinθ₂ > sinθ₁ v₁/v₂ = sinθ₁/sinθ₂ < 1 ∴ v₂ > v₁ Example: θ₁=30°, θ₂=48° v₂/v₁ = sin48°/sin30° = 0.743/0.500 = 1.49 Light is 49% faster in medium 2 Medium 1 (n₁ larger, v₁ smaller) Medium 2 (n₂ smaller, v₂ larger)
Ray bends away from normal crossing into medium 2 → medium 2 is less dense → light is faster in medium 2. The ratio v₁/v₂ = sinθ₁/sinθ₂ directly from angle values.
The fundamental ratio chain

All the ratios link together: sinθ₁/sinθ₂ = v₁/v₂ = n₂/n₁ = λ₁/λ₂. Know this chain and you can extract any quantity from any other. Larger angle → faster speed → smaller n → longer wavelength. They all move together.

Slide 7 · Wavelength from Geometry

Determining Wavelength Ratio from a Diagram

Since λ = λ₀/n (or equivalently λ ∝ v, since f is constant), the wavelength relationship follows directly from the angle comparison. The medium with the smaller refraction angle is the medium with the shorter wavelength.

λ₁/λ₂ = v₁/v₂ = n₂/n₁ = sinθ₁/sinθ₂ All four ratios are equal!

Rules from diagram geometry

  • Ray bends toward normal entering medium 2 (θ₂ < θ₁): wavelength is shorter in medium 2 (λ₂ < λ₁).
  • Ray bends away from normal entering medium 2 (θ₂ > θ₁): wavelength is longer in medium 2 (λ₂ > λ₁).
  • Angles equal: wavelengths are equal.
λ₁ (longer) λ₂ (shorter) AIR / VACUUM GLASS (denser) The Big Ratio Chain λ₁/λ₂ = v₁/v₂ = n₂/n₁ = sinθ₁/sinθ₂ Denser medium: larger n, smaller λ, slower v
The wave compresses in the denser glass medium. Same color of light, but shorter wavelength. The ratio λ₁/λ₂ equals the ratio of sines of the angles in the respective media.
Summary of all wave property comparisons

For any two media, larger angle ↔ larger n? No — it's the opposite! Larger angle ↔ smaller n ↔ faster speed ↔ longer wavelength. The angle and n go in opposite directions. If θ₂ > θ₁, then n₂ < n₁, v₂ > v₁, λ₂ > λ₁. All consistent with the ratio chain.

Slide 8 · Total Internal Reflection

Total Internal Reflection (TIR)

When light travels from a denser medium to a less dense medium (n₁ > n₂), it bends away from the normal. As the incident angle increases, the refracted angle approaches 90°. At the critical angle θ_c, the refracted ray skims the surface. Beyond θ_c, no refraction occurs at all — all light is reflected back into the denser medium. This is Total Internal Reflection.

At critical angle: n₁ sinθ_c = n₂ sin90° = n₂ × 1 Critical angle formula: sinθ_c = n₂ / n₁ (requires n₁ > n₂) Example: glass (n=1.52) → air (n=1.00): sinθ_c = 1.00/1.52 = 0.658 → θ_c = sin⁻¹(0.658) = 41.1° TIR occurs whenever θ_incident > θ_c = 41.1°
Air (n=1.00) Glass (n=1.52) θ=30° refracted θ_c≈41° grazes surface (90°) θ>θ_c TOTAL REFLECTION θ < θ_c partial refraction θ = θ_c refracted at 90° θ > θ_c TOTAL internal reflection
Three rays from glass (denser) to air (less dense). Left: θ < 41.1°, some light escapes. Center: θ = 41.1° (critical angle), refracted ray grazes the surface. Right: θ > 41.1°, 100% reflection — TIR.
What makes TIR possible — conceptually

Going from denser to less dense means n₁ > n₂, so the refracted angle is always larger than the incident angle. Keep increasing the incident angle and eventually the refracted angle would have to exceed 90° — which is impossible, so refraction ceases entirely and the light reflects completely. The critical angle is the incident angle that would produce a 90° refracted angle.

Slide 9 · TIR: The Air-to-Water Case

Why Air-to-Water TIR is Impossible (Exam Q7 Type)

This is one of the most commonly tested TIR concepts: "Can total internal reflection occur when light travels from air into water?" The answer is NO — and you need to know exactly why.

The proof in three steps

  • Step 1 — Check the condition: TIR requires n₁ > n₂ (traveling from denser to less dense). Here n_air = 1.00 and n_water = 1.33. Since 1.00 < 1.33, we are going from LESS dense to MORE dense. The condition is NOT met.
  • Step 2 — Attempt the formula: If we try sinθ_c = n₂/n₁, we get sinθ_c = n_water/n_air = 1.33/1.00 = 1.33. This is greater than 1. The sine function can never exceed 1 by definition, so no real angle satisfies this — there is no critical angle.
  • Step 3 — Physical interpretation: Going air → water, the ray bends TOWARD the normal (n increases). No matter how large θ₁ gets (even 89.9°), the refracted ray always exists, just barely entering the water at a large angle toward the normal. TIR simply cannot happen.
TIR condition: n₁ > n₂ (going from denser to less dense) For air → water: n₁ = n_air = 1.00, n₂ = n_water = 1.33 Check: 1.00 > 1.33? NO ✗ Attempted critical angle: sinθ_c = n₂/n₁ = 1.33/1.00 = 1.33 > 1 ← IMPOSSIBLE (sinθ ≤ 1 always) Therefore: NO critical angle exists, NO TIR is possible. ✓
Air (n=1.00) — less dense Water (n=1.33) — MORE dense θ₁=50° θ₂=35° always refracts θ₁=85° θ₂=49° still refracts! NO TIR possible n_air < n_water going less dense → denser ray always bends toward normal refraction always occurs
Air → water: no matter how large the incident angle is (even nearly 90°), the refracted ray always exists. TIR is physically impossible going from less dense to more dense media.
The exact exam answer

"No. Total internal reflection requires the incident medium to be denser than the transmitted medium (n₁ > n₂). Since n_air = 1.00 < n_water = 1.33, this condition is not satisfied. Attempting to apply the critical angle formula gives sinθ_c = 1.33/1.00 = 1.33 > 1, which has no solution. Therefore TIR cannot occur."

Slide 10 · TIR Applications

Total Internal Reflection: Real-World Applications

TIR is not just an exam curiosity — it underlies some of the most important technologies in modern life. Understanding these applications also reinforces the physics, since each example reveals the same principle at work.

Fiber Optics (the most important application)

A fiber optic cable consists of a glass core (n ≈ 1.50) surrounded by a glass cladding (n ≈ 1.46). Since n_core > n_cladding, any light ray that strikes the core-cladding boundary at an angle greater than θ_c undergoes TIR and bounces along the core without escaping.

θ_c (fiber optic) = sin⁻¹(n_cladding / n_core) = sin⁻¹(1.46 / 1.50) = sin⁻¹(0.973) = 76.7° Light bounces along the cable as long as θ_incident > 76.7° from the normal (i.e., the ray makes an angle of < 13.3° with the fiber axis)
Cladding (n ≈ 1.46) Core (n ≈ 1.50) Cladding (n ≈ 1.46) Light bounces by TIR — never escapes the core laser in signal out
Fiber optic cable: light undergoes TIR at every core-cladding boundary (red dots). The critical angle is 76.7°, so light must enter nearly along the fiber axis to stay trapped inside.

Other TIR Applications

  • Diamond (n=2.42): θ_c = sin⁻¹(1/2.42) = 24.4°. Very small! Most light that enters a diamond totally reflects off the back facets, bouncing around before exiting through the top — creating the characteristic sparkle. Diamonds are cut to maximize TIR.
  • Medical endoscopes: Flexible fiber bundles guide images from inside the body using TIR. Each fiber carries one pixel of the image.
  • Retroreflectors / prisms: A 45°-45°-90° glass prism reflects light through 90° or 180° via TIR (θ_c for glass-air ≈ 41°, and the ray hits at 45° > 41°).
Diamond sparkle — why cutting matters

With θ_c = 24.4°, any ray striking a diamond facet at more than 24.4° from the normal will totally internally reflect. A well-cut diamond ensures that most light entering the top undergoes multiple TIR bounces before exiting back through the top — this is "brilliance". A poorly cut diamond lets light escape through the sides and bottom, appearing dull.

Slide 11 · Dispersion

Dispersion: Why Prisms Make Rainbows

We have been treating the index of refraction n as a single number for a given material. In reality, n depends slightly on the wavelength (color) of the light. Violet light has a slightly larger n than red light in glass. This wavelength-dependence of n is called dispersion.

Dispersion: n depends on wavelength λ (or equivalently, on color) Typical glass values: n_violet ≈ 1.538 (shorter λ, higher f → bends MORE) n_green ≈ 1.523 n_red ≈ 1.514 (longer λ, lower f → bends LESS) Snell's Law: n₁ sinθ₁ = n(λ) × sinθ₂(λ) Different λ → different θ₂ → different output angles → separated colors

When white light hits a prism, each wavelength (color) is refracted by a slightly different amount. The result is the visible spectrum — red bends least, violet bends most.

white light red (λ≈700nm) orange yellow green blue violet (λ≈400nm) n_violet > n_red violet bends more at exit Prism: glass n≈1.50 (average)
White light disperses in a prism because n is slightly larger for violet than for red. Violet (high f, short λ) bends most; red (low f, long λ) bends least. This is the physics of rainbows.
Rainbows and dispersion in nature

A rainbow is dispersion by millions of water droplets. Sunlight enters each drop, reflects off the back inner surface, and exits — with different colors exiting at different angles (red at ~42°, violet at ~40° from the anti-solar point). Dispersion is also why white light under a CD or soap bubble creates color patterns (thin-film interference, combined with the wavelength-dependence of the phase shift).

Slide 12 · Prism TIR Problem

Prism TIR: Exam-Style Calculation

A classic exam problem: a glass prism with faces labeled. Light enters one face along the normal (no refraction on entry), then hits a second face at angle δ. Will TIR occur? How does the answer change if the prism is immersed in water?

Setup: glass prism (n = 1.52) in air

Critical angle for glass→air: sinθ_c = n_air/n_glass = 1.00/1.52 = 0.658 θ_c = sin⁻¹(0.658) = 41.1° TIR occurs on face AC if δ > 41.1° For a 45°-45°-90° prism: δ = 45° Since 45° > 41.1° → TIR occurs ✓

Modified setup: same prism immersed in water (n = 1.33)

Critical angle for glass→water: sinθ_c = n_water/n_glass = 1.33/1.52 = 0.875 θ_c = sin⁻¹(0.875) = 61.0° For a 45°-45°-90° prism: δ = 45° Since 45° < 61.0° → TIR does NOT occur ✗ Conclusion: immersing the prism in water raises θ_c and can prevent TIR.
Face AB Face BC Face AC enters along normal δ=45° TIR! IN AIR — TIR occurs 45° > θ_c = 41.1° immersed in water (n=1.33) δ=45° escapes! (no TIR) IN WATER — no TIR 45° < θ_c = 61.0°
Same 45-45-90 prism. In air (left): 45° > 41.1° critical angle → TIR, light exits through bottom face. In water (right): 45° < 61.0° critical angle → no TIR, light escapes through the hypotenuse face.
Exam strategy for prism TIR

Step 1: Find θ_c = sin⁻¹(n_outside/n_glass). Step 2: Find the angle δ at which the ray strikes the second face from geometry (often requires knowing the prism angles). Step 3: Compare δ vs. θ_c. If δ > θ_c → TIR; if δ < θ_c → no TIR. Watch out for problems where n_outside changes — a higher surrounding medium n raises θ_c and makes TIR harder.

Slide 13 · Polarization

Polarization of Light

Light is a transverse electromagnetic wave: the electric field E oscillates perpendicular to the direction of propagation. In unpolarized light, E oscillates in all directions within that perpendicular plane simultaneously. In polarized light, E oscillates in one specific direction only.

Methods of polarization

  • Polaroid filters: Transmit only the component of E aligned with the filter axis. Intensity after polarizer: I = I₀/2 (for unpolarized input).
  • Malus's Law: If polarized light of intensity I₀ passes through a polarizer at angle θ to the polarization direction: I = I₀ cos²θ.
  • Brewster's Law (reflection polarization): At Brewster's angle, reflected light is completely polarized parallel to the surface. tanθ_B = n₂/n₁. At this angle, the reflected and refracted rays are perpendicular to each other (θ_reflected + θ_refracted = 90°).
Brewster's angle: tanθ_B = n₂ / n₁ Example: air (n₁=1.00) → glass (n₂=1.52): tanθ_B = 1.52/1.00 = 1.52 θ_B = tan⁻¹(1.52) = 56.7° At θ_incident = 56.7°, reflected light is 100% polarized (horizontal) Refracted light is partially polarized (vertical component) Malus's Law: I = I₀ cos²θ
unpolarized light θ_B reflected (polarized) refracted (partial polarize) 90° between reflected & refracted n₁ = 1.00 (air) n₂ = 1.52 (glass)
At Brewster's angle (56.7° for air-glass), the reflected ray is 100% polarized. The reflected and refracted rays are perpendicular to each other. This is why polarized sunglasses cut glare from roads and water.
Why polarized sunglasses work

Light reflecting from horizontal surfaces (roads, water) is predominantly horizontally polarized via Brewster's Law. Polarized sunglasses have a vertically oriented polaroid axis, which blocks this horizontally polarized glare while transmitting most other light. The effect is dramatic — on water, you can suddenly see fish below the surface when you rotate polarized glasses to block the surface reflection.

Slide 14 · Worked Exam Problem 1

Exam Problem Type 1: Wavelength and Frequency in Glass

This type of problem is extremely common and straightforward once you know the three-step method. Do NOT skip steps — even if you know the answer, showing the method earns partial credit.

"Light with wavelength 256 nm in vacuum enters a glass with n = 1.60. Find: (a) the frequency in vacuum, (b) the frequency in glass, (c) the wavelength in glass, (d) the speed in glass."

Full Solution

Given: λ₀ = 256 nm = 256×10⁻⁹ m, n_glass = 1.60, c = 3.00×10⁸ m/s (a) Frequency in vacuum: f = c/λ₀ = (3.00×10⁸) / (256×10⁻⁹) = 1.172 × 10¹⁵ Hz (b) Frequency in glass: f_glass = f_vacuum = 1.172 × 10¹⁵ Hz (Frequency is invariant across any boundary — same source, same oscillations) (c) Wavelength in glass: λ_glass = λ₀ / n = 256 nm / 1.60 = 160.0 nm Alternative path: v_glass = c/n = 3.00×10⁸/1.60 = 1.875×10⁸ m/s λ_glass = v/f = 1.875×10⁸ / 1.172×10¹⁵ = 160.0 nm ✓ (d) Speed in glass: v_glass = c/n = (3.00×10⁸) / 1.60 = 1.875 × 10⁸ m/s
VACUUM n = 1.00 λ₀ = 256 nm f = 1.172×10¹⁵ Hz v = 3.00×10⁸ m/s n=1.60 GLASS n = 1.60 λ = 160 nm f = 1.172×10¹⁵ Hz ← SAME v = 1.875×10⁸ m/s Key results: λ shrinks by factor n f is unchanged v shrinks by factor n λ = 256/1.6 = 160 nm v = c/n = 1.875×10⁸
Summary of all wave properties crossing vacuum → glass (n=1.60). Wavelength and speed decrease by factor n. Frequency is invariant. The downward arrows show decreasing quantities.
Why does frequency stay constant? (deep understanding)

Consider the boundary as a fixed point in space. The number of wave crests arriving per second at the boundary from medium 1 must equal the number leaving into medium 2 per second. Otherwise charge would pile up at the boundary — which doesn't happen for EM waves. This boundary condition forces f₁ = f₂. Since the speed changed, the only way to maintain v = fλ is for λ to change proportionally.

Slide 15 · Worked Exam Problem 2

Exam Problem Type 2: Reading Diagrams for n, v, and λ

This is the highest-frequency exam skill in Chapter 23. Practice until you can do these in under 30 seconds. No calculations needed — just compare angles.

The three cases

Case A: equal angles θ₁ θ₂=θ₁ n₁ = n₂ v₁=v₂, λ₁=λ₂, same medium Case B: θ₂ < θ₁ (toward) θ₁ (large) θ₂ small n₂ > n₁ (medium 2 denser) v₂ < v₁, λ₂ < λ₁ Case C: θ₂ > θ₁ (away) θ₁ θ₂ big n₂ < n₁ (less dense) v₂ > v₁, λ₂ > λ₁
Three diagram cases. Case A: unchanged angle → same n. Case B: bends toward normal → n₂ > n₁. Case C: bends away from normal → n₂ < n₁. From n comparison, speed and wavelength comparisons follow immediately.

Practice drill — answer all three questions per diagram

  • For Case B: (1) n₂ > n₁ (denser). (2) v₂ < v₁ (slower). (3) λ₂ < λ₁ (shorter wavelength).
  • For Case C: (1) n₂ < n₁ (less dense). (2) v₂ > v₁ (faster). (3) λ₂ > λ₁ (longer wavelength).
  • For Case A: all three quantities are equal in both media.
How to guess from diagram without exact angles

On a multiple-choice exam, you don't need to measure — just look at the bend direction relative to the dashed normal. Toward normal = denser. Away from normal = less dense. Then apply: denser ↔ slower ↔ shorter wavelength ↔ larger n. These always move together.

Slide 16 · What-If Scenarios

What-If Scenarios: Testing Your Conceptual Depth

Exam problems often change one parameter to see if you understand the underlying physics. Work through each scenario carefully — each one targets a different potential misconception.

Scenario 1: What if λ₀ = 500 nm instead of 256 nm?

For glass n=1.60: f = c/λ₀ = 3×10⁸/500×10⁻⁹ = 6.00×10¹⁴ Hz (different frequency from the 256 nm case, but still invariant across the boundary). λ_glass = 500/1.60 = 312.5 nm. The ratio λ_glass/λ₀ = 1/n = 1/1.60 is the same regardless of the starting wavelength. Only the absolute values change.

Scenario 2: What if n = 1.0 (vacuum)?

If n = 1.0: v = c/1 = c, λ = λ₀/1 = λ₀, f = f₀ (unchanged) Light in vacuum: all three properties match the "free" values. This is the reference state — n=1 by definition.

Scenario 3: What if n = 2.0?

If n = 2.0: v = c/2 = 1.5×10⁸ m/s (half of c) λ = λ₀/2 (wavelength halved) f = f₀ (still unchanged — source determines this) Example: λ₀=256 nm → λ_medium = 128 nm, v = 1.5×10⁸ m/s

Scenario 4: Light goes from glass (n=1.52) to water (n=1.33)

Now n₁ > n₂ (glass is denser than water). Snell's Law still applies: 1.52 sinθ₁ = 1.33 sinθ₂. Since n₁ > n₂, sinθ₂ > sinθ₁, so θ₂ > θ₁ — ray bends away from normal. TIR is now possible!

θ_c (glass → water) = sin⁻¹(n_water/n_glass) = sin⁻¹(1.33/1.52) = sin⁻¹(0.875) = 61.0° TIR occurs if θ_incident > 61.0°

Scenario 5: What if the frequency of the incoming light is doubled?

Frequency doubles → wavelength halves (in vacuum: λ₀ = c/f). In glass (n=1.6): the new wavelength is still λ₀/n, just half as large. The speed in glass is still c/n — speed depends only on n, not on f. The index n for most materials barely changes with frequency (dispersion is small), so to a good approximation v_glass is the same regardless of the light's color.

What Changes vs. What Stays the Same in a Medium Property Changes? Formula Frequency f NO (constant) f = c/λ₀ (source) Speed v YES (decreases) v = c/n Wavelength λ YES (decreases) λ = λ₀/n Index n NO (material property) n = c/v (fixed for medium)
Master table: frequency and n are constant for a given source-medium pair. Speed and wavelength both decrease by factor n when entering a denser medium.
Key conceptual check

If you double the index of refraction (hypothetically n → 2n), what happens? Speed halves (v → c/2n), wavelength halves (λ → λ₀/2n), frequency is unchanged. All of this makes sense: light moves more slowly through a denser medium because it interacts more strongly with the material's electrons, but the driving oscillation frequency is still set by the original source.

Slide 17 · Common Mistakes & Exam Traps

Common Mistakes and Exam Traps: What to Watch For

These are the errors that cost students points on Chapter 23 exams. Read each one, internalize it, and promise yourself you won't make it. Each trap is paired with the correct statement and a quick diagnostic.

Trap 1 — "Frequency changes when entering a medium"

WRONG: "The frequency of light decreases as it enters glass." CORRECT: Frequency is invariant across any boundary. It is set by the source and never changes. WHY: The boundary cannot create or destroy oscillations (energy conservation). TEST: If you find yourself writing f_glass ≠ f_vacuum, stop and reconsider.

Trap 2 — "Angles are measured from the surface"

WRONG: θ = angle between ray and the interface surface CORRECT: θ = angle between ray and the NORMAL (perpendicular to surface) WHY: Snell's Law is derived using the normal as reference. TEST: If your "angle" for a ray traveling nearly straight through is large, you're measuring from the surface. A ray close to normal should have θ ≈ 0°.

Trap 3 — "TIR can happen going from air to water"

WRONG: "A ray in air hitting water can undergo TIR at a large enough angle." CORRECT: TIR requires n_incident > n_transmitted (denser to less dense). Air (1.00) → water (1.33): 1.00 < 1.33 → IMPOSSIBLE. WHY: sinθ_c = n_water/n_air = 1.33/1.00 = 1.33 > 1, no real solution exists. TEST: Always check n_incident > n_transmitted before attempting TIR.

Trap 4 — "Wavelength in a medium is λ₀ × n (multiply)"

WRONG: λ_medium = λ₀ × n (multiply by n) CORRECT: λ_medium = λ₀ / n (DIVIDE by n) WHY: Entering a denser medium (n > 1) means SLOWER speed, SHORTER wavelength. Multiplying by n > 1 would INCREASE the wavelength — impossible in denser medium. MEMORY AID: "n in the denominator — you're going into the den (denser medium)"
TRAP 1 Frequency changes in new medium TRUTH: Frequency is ALWAYS constant ✗ → ✓ TRAP 2 Measure θ from the surface TRUTH: Measure θ from the NORMAL ✗ → ✓ TRAP 3 TIR possible air → water TRUTH: TIR needs n₁>n₂ air(1.0)<water(1.33) IMPOSSIBLE ✗ → ✓ TRAP 4 λ_medium = λ₀ × n (multiply) TRUTH: λ_medium = λ₀ / n DIVIDE by n ✗ → ✓
The four most dangerous exam traps in Chapter 23, with corrections. Each trap exploits an intuitive-but-wrong shortcut. Memorize the corrections before your exam.
One more subtle trap: TIR direction confusion

Students sometimes confuse "total reflection" (which can happen at any interface at any angle, but partially) with "total INTERNAL reflection" (which requires specific n conditions). Any surface partially reflects light (Fresnel reflection). TIR is specifically the phenomenon where the transmitted wave completely disappears and 100% of the intensity reflects — and this only happens at θ > θ_c when n₁ > n₂.

Slide 18 · Chapter 23 Summary

Chapter 23 Complete Cheat Sheet

Everything you need for Exam 3, Chapter 23 — condensed to one slide. Study this until you can reproduce it from memory.

All Formulas

1. Index of Refraction: n = c / v (dimensionless, n ≥ 1) 2. Speed in Medium: v = c / n (always slows down) 3. Wavelength in Medium: λ = λ₀ / n (DIVIDE — it shrinks) 4. Frequency: f_medium = f_vacuum (NEVER changes) 5. Snell's Law: n₁ sinθ₁ = n₂ sinθ₂ (angles from NORMAL) 6. Critical Angle: sinθ_c = n₂ / n₁ (requires n₁ > n₂) 7. TIR condition: θ_incident > θ_c AND n₁ > n₂ 8. Brewster's Angle: tanθ_B = n₂ / n₁ (reflected ray polarized) 9. Malus's Law: I = I₀ cos²θ (polarized light through polarizer) 10. Ratio Chain: sinθ₁/sinθ₂ = v₁/v₂ = n₂/n₁ = λ₁/λ₂

Common n Values (memorize these)

Material n θ_c (to air) Key fact
Vacuum / Air 1.000 Reference medium
Water 1.33 48.8° TIR impossible air→water
Glass (typical) 1.52 41.1° Standard exam value
Diamond 2.42 24.4° Tiny θ_c → sparkle from TIR
n = c/v index Snell's Law n₁sinθ₁ = n₂sinθ₂ TIR sinθ_c = n₂/n₁ (n₁>n₂) Wavelength λ = λ₀ / n Frequency f = constant (invariant) v = c/n (always < c) Chapter 23 Concept Map All roads lead back to n — the index of refraction governs everything
Concept map for Chapter 23. The index of refraction n is the central quantity: it determines Snell's Law bending, the critical angle for TIR, the wavelength compression λ₀/n, and the speed reduction c/n. Frequency stands apart — it is invariant, set only by the source.
Final exam mantra — repeat until automatic

n = c/v (larger n = slower light).   λ = λ₀/n (DIVIDE — wavelength shrinks).   f never changes (source determines frequency).   Snell: n₁sinθ₁ = n₂sinθ₂ (angles from normal).   TIR: sinθ_c = n₂/n₁, needs n₁ > n₂ (denser to less-dense only).   Toward normal = denser; away from normal = less dense.

"In Physics: explain = Words + Diagrams + Equations"