Wavelength $\lambda$ and frequency $f$: $c = \lambda f$

Visible spectrum: 400 nm (violet) to 700 nm (red)

Law of Reflection

When light reflects from smooth surface:

$\theta_i = \theta_r$

Angle of incidence = Angle of reflection

Angles measured from normal (perpendicular to surface).

Specular reflection: Smooth surface (mirror) - parallel rays stay parallel Diffuse reflection: Rough surface - parallel rays scatter

Plane Mirrors

Image characteristics:

• Virtual: Behind mirror (light doesn't actually go there)
• Upright: Same orientation as object
• Same size: $h_i = h_o$
• Same distance: $d_i = d_o$ (but behind mirror)

Lateral inversion: Left and right reversed (but not up/down!)

Index of Refraction

Speed of light in a material: $v = \frac{c}{n}$

where n is index of refraction (n ≥ 1, dimensionless)

Common values:

• Vacuum: n = 1 (exactly)
• Air: n ≈ 1.0003 ≈ 1
• Water: n = 1.33
• Glass: n ≈ 1.5
• Diamond: n = 2.42

Higher n → slower light → more bending

Snell's Law

When light crosses boundary between media:

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

where:

• $n_1$, $\theta_1$ = index and angle in medium 1
• $n_2$, $\theta_2$ = index and angle in medium 2

Angles from normal to surface!

Light entering denser medium (n₂ > n₁):

• Bends toward normal (θ₂ < θ₁)
• Speed decreases, wavelength decreases
• Frequency stays same!

Light entering less dense medium (n₂ < n₁):

• Bends away from normal (θ₂ > θ₁)
• Speed increases, wavelength increases

💡 Mnemonic: Fast → Slow: toward normal. Slow → Fast: away from normal.

Total Internal Reflection

When light goes from denser to less dense (n₁ > n₂):

If θ₁ > θ_c (critical angle), light reflects completely back!

Critical angle: $\sin\theta_c = \frac{n_2}{n_1}$

(Only exists when n₁ > n₂)

Applications:

• Fiber optics (light trapped inside fiber)
• Diamonds sparkle (n = 2.42 → small θ_c → lots of TIR)
• Prisms in binoculars

For water-air (n₁ = 1.33, n₂ = 1): $\theta_c = \sin^{-1}(1/1.33) = 48.6°$

Dispersion

Different wavelengths have different n (for same material)!

White light → prism → spectrum (ROYGBIV)

• Violet: higher n → bends more
• Red: lower n → bends less

Rainbow: Water droplets act as prisms

1. Refraction entering drop (dispersion)
2. Reflection inside drop
3. Refraction exiting drop (more dispersion)

Curved Mirrors

Concave Mirror (converging):

• Parallel rays converge at focal point
• Focal length: $f = R/2$ (R = radius of curvature)
• Can form real or virtual images

Convex Mirror (diverging):

• Parallel rays appear to diverge from focal point behind mirror
• $f$ is negative
• Always forms virtual, upright, reduced images

Mirror Equation

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

where:

• $f$ = focal length
• $d_o$ = object distance (positive)
• $d_i$ = image distance (positive if real, negative if virtual)

Magnification: $m = -\frac{d_i}{d_o} = \frac{h_i}{h_o}$

• $m > 0$: upright image
• $m < 0$: inverted image
• $|m| > 1$: enlarged
• $|m| < 1$: reduced

Sign Conventions (Mirrors)

Ray Diagrams (Concave Mirror)

Draw any 2 of these 3 rays:

1. Parallel ray → reflects through F
2. Focal ray (through F) → reflects parallel
3. Center ray (through C) → reflects back on itself

Where rays intersect = image location!

Problem-Solving Strategy

Snell's Law:

1. Draw diagram with normal
2. Identify n₁, θ₁, n₂
3. Apply: $n_1 \sin\theta_1 = n_2 \sin\theta_2$
4. Check for TIR if going to less dense medium

Mirrors:

1. Identify f (positive for concave, negative for convex)
2. Use mirror equation: $1/f = 1/d_o + 1/d_i$
3. Find magnification: $m = -d_i/d_o$
4. Interpret signs

Common Mistakes

❌ Measuring angles from surface instead of normal ❌ Forgetting to check for total internal reflection ❌ Wrong sign for f (convex mirrors have negative f!) ❌ Confusing d_i sign (positive = real/front, negative = virtual/behind) ❌ Thinking frequency changes during refraction (only v and λ change!) ❌ Using degrees instead of checking calculator mode

Solution:

Apply Snell's Law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$ $(1.00)\sin(30°) = (1.33)\sin\theta_2$ $0.50 = 1.33 \sin\theta_2$ $\sin\theta_2 = \frac{0.50}{1.33} = 0.376$ $\theta_2 = \sin^{-1}(0.376) = 22.1°$

Check: Entering denser medium (n₂ > n₁), so bends toward normal. ✓ θ₂ = 22.1° < θ₁ = 30° ✓

Answer: θ₂ = 22.1° (bends toward normal)

Solution:

Apply Snell's Law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$ $(1.00)\sin(30°) = (1.33)\sin\theta_2$ $0.50 = 1.33 \sin\theta_2$ $\sin\theta_2 = \frac{0.50}{1.33} = 0.376$ $\theta_2 = \sin^{-1}(0.376) = 22.1°$

Check: Entering denser medium (n₂ > n₁), so bends toward normal. ✓ θ₂ = 22.1° < θ₁ = 30° ✓

Answer: θ₂ = 22.1° (bends toward normal)

Solution:

Apply Snell's Law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$ $(1.00)\sin(30°) = (1.33)\sin\theta_2$ $0.50 = 1.33 \sin\theta_2$ $\sin\theta_2 = \frac{0.50}{1.33} = 0.376$ $\theta_2 = \sin^{-1}(0.376) = 22.1°$

Check: Entering denser medium (n₂ > n₁), so bends toward normal. ✓ θ₂ = 22.1° < θ₁ = 30° ✓

Answer: θ₂ = 22.1° (bends toward normal)

Solution:

Apply Snell's Law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$ $(1.00)\sin(30°) = (1.33)\sin\theta_2$ $0.50 = 1.33 \sin\theta_2$ $\sin\theta_2 = \frac{0.50}{1.33} = 0.376$ $\theta_2 = \sin^{-1}(0.376) = 22.1°$

Check: Entering denser medium (n₂ > n₁), so bends toward normal. ✓ θ₂ = 22.1° < θ₁ = 30° ✓

Answer: θ₂ = 22.1° (bends toward normal)

Solution:

Apply Snell's Law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$ $(1.00)\sin(30°) = (1.33)\sin\theta_2$ $0.50 = 1.33 \sin\theta_2$ $\sin\theta_2 = \frac{0.50}{1.33} = 0.376$ $\theta_2 = \sin^{-1}(0.376) = 22.1°$

Check: Entering denser medium (n₂ > n₁), so bends toward normal. ✓ θ₂ = 22.1° < θ₁ = 30° ✓

Answer: θ₂ = 22.1° (bends toward normal)

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$ $\theta_c = \sin^{-1}(0.667) = 41.8°$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$ $\theta_c = \sin^{-1}(0.667) = 41.8°$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$ $\theta_c = \sin^{-1}(0.667) = 41.8°$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$ $\theta_c = \sin^{-1}(0.667) = 41.8°$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$ $\theta_c = \sin^{-1}(0.667) = 41.8°$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$ $\theta_c = \sin^{-1}(0.667) = 41.8°$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{20} = \frac{1}{60} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{20} - \frac{1}{60} = \frac{3-1}{60} = \frac{2}{60} = \frac{1}{30}$ $d_i = 30 \text{ cm}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: Reduced (half size)
• Located 30 cm in front of mirror

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{20} = \frac{1}{60} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{20} - \frac{1}{60} = \frac{3-1}{60} = \frac{2}{60} = \frac{1}{30}$ $d_i = 30 \text{ cm}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: Reduced (half size)
• Located 30 cm in front of mirror

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{20} = \frac{1}{60} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{20} - \frac{1}{60} = \frac{3-1}{60} = \frac{2}{60} = \frac{1}{30}$ $d_i = 30 \text{ cm}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: Reduced (half size)
• Located 30 cm in front of mirror

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{20} = \frac{1}{60} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{20} - \frac{1}{60} = \frac{3-1}{60} = \frac{2}{60} = \frac{1}{30}$ $d_i = 30 \text{ cm}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: Reduced (half size)
• Located 30 cm in front of mirror

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{20} = \frac{1}{60} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{20} - \frac{1}{60} = \frac{3-1}{60} = \frac{2}{60} = \frac{1}{30}$ $d_i = 30 \text{ cm}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: Reduced (half size)
• Located 30 cm in front of mirror

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{20} = \frac{1}{60} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{20} - \frac{1}{60} = \frac{3-1}{60} = \frac{2}{60} = \frac{1}{30}$ $d_i = 30 \text{ cm}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: Reduced (half size)
• Located 30 cm in front of mirror

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size