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PHYS 1120 — Exam 3 Optics Formula Reference

Chapters 23 · 24 · 26  |  "In Physics explain means: Words & Diagrams & Equations"

§ Ch 23 — Light in Media
Index of Refraction
n = c / v
c = 3.00×10⁸ m/s  ·  n ≥ 1 always
Higher n → slower light
Wavelength in Medium
λ = λ₀ / n
λ₀ = vacuum wavelength
Higher n → shorter λ
Frequency (Invariant)
f = c / λ₀ = v / λ
Frequency NEVER changes
when crossing any boundary
Snell's Law
n₁ sin θ₁ = n₂ sin θ₂
All angles from the NORMAL (not surface).
n₂ > n₁ → θ₂ < θ₁ (bends toward normal, slows down)
n₂ < n₁ → θ₂ > θ₁ (bends away from normal, speeds up)
Total Internal Reflection
sin θ_c = n₂ / n₁  (n₁ > n₂ required)
Only possible: denser → less dense (n₁ > n₂)
Impossible: air → water (n_air < n_water)
glass→air: θ_c = sin⁻¹(1/1.52) = 41.1°
glass→water: θ_c = sin⁻¹(1.33/1.52) = 61.0°
Index of Refraction — Common Values
Vacuum
1.000
Air
1.0003
Water
1.33
Benzene
1.50
Glass (typ.)
1.52
Glass (n=1.6)
1.60
Crown Glass
1.52
Diamond
2.42
Reading Diagram: Which medium is faster/denser?
θ₂ < θ₁→ bends toward normal → n₂ > n₁, v₂ < v₁, λ₂ < λ₁
θ₂ > θ₁→ bends away from normal → n₂ < n₁, v₂ > v₁, λ₂ > λ₁
θ₂ = θ₁→ no bending → n₁ = n₂, v₁ = v₂, λ₁ = λ₂
§ Ch 24 — Mirrors & Lenses
Mirror / Lens Equation
1/xₒ + 1/xᵢ = 1/f
Same equation for both!
Mirrors: f = R/2
Focal Length (Mirror)
f = R / 2
R = radius of curvature
Concave: f > 0  ·  Convex: f < 0
Magnification
m = hᵢ/hₒ = −xᵢ/xₒ
m < 0 → inverted image
m > 0 → upright image
|m| > 1 → magnified
Sign Conventions
QuantityPositive (+)Negative (−)
Object distance xₒalways positive—
Image distance xᵢ (mirror)real image (in front of mirror)virtual image (behind mirror)
Image distance xᵢ (lens)real image (behind lens, far side)virtual image (same side as object)
Focal length f (mirror)concave mirrorconvex mirror
Focal length f (lens)converging (convex) lensdiverging (concave) lens
Image height hᵢuprightinverted
Mirror Image Cases (Concave, f = +5 cm as example)
Object positionImage xᵢReal/VirtualOrientationSize
xₒ → ∞xᵢ = fRealInvertedPoint
xₒ > 2ff < xᵢ < 2fRealInvertedReduced
xₒ = 2fxᵢ = 2fRealInvertedSame size
f < xₒ < 2fxᵢ > 2fRealInvertedMagnified
xₒ = fxᵢ = ∞——No image
xₒ < fxᵢ < 0VirtualUprightMagnified
Convex Mirror
f < 0 → image is always virtual, upright, reduced. Wide field of view.
Diverging Lens
f < 0 → image is always virtual, upright, reduced, same side as object.
Converging Lens (xₒ < f)
f > 0, object inside f → virtual, upright, magnified. This is a magnifying glass.
3-Ray Rules
Concave Mirror — 3 Rays
Ray 1: Parallel to axis → reflects through F
Ray 2: Through F → reflects parallel to axis
Ray 3: Through C (= 2F) → reflects straight back
Image = where the 3 reflected rays cross
Converging Lens — 3 Rays
Ray 1: Parallel to axis → refracts through far F₂
Ray 2: Through near F₁ → refracts parallel to axis
Ray 3: Through center of lens → straight through (no bend)
Diverging lens: rays diverge — extend backwards (dashed) to find virtual image
§ Ch 26 — Wave Optics: Interference & Diffraction
Two-Source Constructive
δ = mλ    m = 0, ±1, ±2, …
δ = path difference = |r₂ − r₁|
Waves arrive in phase
Two-Source Destructive
δ = (m + ½)λ    m = 0, ±1, ±2, …
= λ/2, 3λ/2, 5λ/2, …
Waves arrive out of phase
Fringe Spacing (Double Slit)
Δy = λL / d
d = slit separation
L = screen distance
Bright: y = mλL/d
Single Slit — Dark Fringes
sin θ = mλ/a    m = ±1, ±2, …
a = slit width. Center IS bright.
1st dark: y₁ = λL/a
Central max width = 2λL/a
Double Slit — Bright Fringes
d sin θ = mλ    m = 0, ±1, ±2, …
y_bright = mλL/d
Fringe spacing Δy = λL/d (uniform)
Slit Width from Exam
a = λL / y₁
y₁ = (total central max width) / 2
Δy = 4.5 mm → y₁ = 2.25 mm
Thin Film Interference — Phase Shift Rules
Phase Shift Rule
Reflection at LOW→HIGH n boundary: 180° phase shift (= ½λ equivalent)
Reflection at HIGH→LOW n boundary: No phase shift

Count total phase shifts:
• 0 or 2 shifts → constructive: 2t = mλ/n  (m = 1, 2, 3, …)
• 1 shift → constructive: 2t = (m + ½)λ/n  (m = 0, 1, 2, …)
Benzene on Water (Exam Q1)
Air(1.00) → Benzene(1.50): n increases → shift ✓
Benzene(1.50) → Water(1.33): n decreases → no shift
Net: 1 shift → constructive: 2t = (m+½)λ₀/n
Minimum (m=0): t = λ₀ / (4n) = 575/(4×1.50) = 95.8 nm
Effect of Medium on Diffraction/Interference
When apparatus is submerged in medium with index n:   λ_medium = λ₀/n
All fringe spacings scale as:   y_new = y_air / n
→ Fringes move toward center (pattern compresses by factor n)
→ Central maximum gets narrower
→ This is always true for both single-slit and double-slit setups
Exam 3 Quick Answers
QQuestionKey EquationAnswer
1Benzene (n=1.50) on water (n=1.33), λ=575nm — min thickness?t = λ/(4n)t ≈ 95.8 nm
2Concave R=10cm, obj 16mm at 10cm — image?1/xᵢ = 1/f − 1/xₒxᵢ=10cm, hᵢ=−16mm, real/inverted/same size
3Thin lens: obj 8cm left, image 3cm left — type & height?1/f = 1/xₒ + 1/xᵢf=−4.8cm, diverging, hᵢ=2.44mm
4Single slit λ=600nm, L=3m, 2y₁=4.5mm — slit width?a = λL/y₁a = 0.8 mm
5Two-source constructive: x=y, x+y=2λ, x−y=λ, x−y=5λδ = mλa, c, d correct (not b)
6aDestructive path difference =δ = (m+½)λλ/2, 3λ/2, 5λ/2, … (odd multiples of λ/2)
6bConstructive path difference =δ = mλ0, λ, 2λ, 3λ, …
7TIR from air to water possible?sin θ_c = n₂/n₁NO — n_air < n_water, sin θ_c > 1
8Single slit in water — dark spots?y₁ = λL/a, λ=λ₀/n(c) Move toward center spot
9λ=256nm vacuum → glass n=1.6 — f and λ in glass?f=c/λ₀, λ=λ₀/nf=1.17×10¹⁵ Hz, λ_glass=160nm
10Wavelength in medium 1 vs 2 (θ₁ > θ₂ in diagram)?λ=λ₀/n, n∝1/sinθλ₁ > λ₂ (medium 1 wavelength larger)
11Speed in medium 2 vs 1 (same diagram)?v=c/nv₂ < v₁ (medium 2 is slower)
Common Exam Traps
Trap 1 — f ≠ R
R = 10 cm → f = 5 cm, not 10 cm. Always divide by 2!
Trap 2 — Frequency Changes
Frequency is constant in all media. Only λ and v change.
Trap 3 — TIR direction
TIR requires going denser → less dense. Air→water: impossible.
Trap 4 — Single slit bright/dark
a sinθ = mλ gives DARK fringes (m≠0). Center is bright!
Trap 5 — x+y ≠ path diff
Constructive needs |x−y| = mλ. Sum x+y is NOT path difference.
Trap 6 — 2y₁ vs y₁
Exam says distance between dark fringes = 4.5mm. So y₁ = 2.25mm.