Reflection and Refraction

Law of reflection, Snell's law, total internal reflection, dispersion, mirrors

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🔆 Reflection and Refraction

Nature of Light

Light is an electromagnetic wave:

Electric and magnetic fields oscillate perpendicular to propagation
Travels at $c = 3.0 \times 10^8$ m/s in vacuum
Also exhibits particle properties (photons) - wave-particle duality!

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

Refraction entering drop (dispersion)
Reflection inside drop
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)

| Quantity | Positive | Negative | |----------|----------|----------| | $f$ | Concave | Convex | | $d_o$ | Real object | (rare) | | $d_i$ | Real image (front) | Virtual image (behind) | | $m$ | Upright | Inverted |

Ray Diagrams (Concave Mirror)

Draw any 2 of these 3 rays:

Parallel ray → reflects through F
Focal ray (through F) → reflects parallel
Center ray (through C) → reflects back on itself

Where rays intersect = image location!

Problem-Solving Strategy

Snell's Law:

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

Mirrors:

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

📚 Practice Problems

1Problem 1easy

❓ Question:

Light travels from air (n = 1.00) into water (n = 1.33) at an angle of 30° from the normal. What is the angle of refraction?

💡 Show Solution

Given:

Medium 1 (air): $n_1 = 1.00$
Medium 2 (water): $n_2 = 1.33$
Incident angle: $\theta_1 = 30°$

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)

2Problem 2easy

❓ Question:

Light travels from air (n = 1.00) into water (n = 1.33) at an angle of 30° from the normal. What is the angle of refraction?

💡 Show Solution

Given:

Medium 1 (air): $n_1 = 1.00$
Medium 2 (water): $n_2 = 1.33$
Incident angle: $\theta_1 = 30°$

Solution:

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

Answer: θ₂ = 22.1° (bends toward normal)

3Problem 3medium

❓ Question:

What is the critical angle for light going from glass (n = 1.50) to air (n = 1.00)? What happens at 45° incidence?

💡 Show Solution

Given:

Medium 1 (glass): $n_1 = 1.50$
Medium 2 (air): $n_2 = 1.00$

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!

4Problem 4medium

❓ Question:

What is the critical angle for light going from glass (n = 1.50) to air (n = 1.00)? What happens at 45° incidence?

💡 Show Solution

Given:

Medium 1 (glass): $n_1 = 1.50$
Medium 2 (air): $n_2 = 1.00$

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!

5Problem 5medium

❓ Question:

Light travels from air (n = 1.00) into water (n = 1.33) at an incident angle of 40°. (a) What is the angle of refraction? (b) What is the speed of light in water? (c) Does the wavelength increase or decrease? Use c = 3.0 × 10⁸ m/s.

💡 Show Solution

Solution:

Given: n₁ = 1.00, n₂ = 1.33, θ₁ = 40°, c = 3.0 × 10⁸ m/s

(a) Angle of refraction (Snell's Law): n₁ sin θ₁ = n₂ sin θ₂ (1.00) sin 40° = (1.33) sin θ₂ 0.643 = 1.33 sin θ₂ sin θ₂ = 0.483 θ₂ = 28.9° or 29°

Light bends toward the normal entering denser medium.

(b) Speed in water: v = c/n = (3.0 × 10⁸)/1.33 = 2.26 × 10⁸ m/s

Wavelength decreases to 75% of its value in air.

6Problem 6hard

❓ Question:

A concave mirror has focal length 20 cm. An object is placed 60 cm from the mirror. Find (a) image distance, (b) magnification, (c) describe the image.

💡 Show Solution

Given:

Focal length: $f = 20$ cm (positive for concave)
Object distance: $d_o = 60$ cm

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

7Problem 7hard

❓ Question:

Light in a glass fiber (n = 1.50) strikes the glass-air boundary at 50°. (a) Will total internal reflection occur? (b) What is the critical angle for this fiber? Use n_air = 1.00.

💡 Show Solution

Solution:

Given: n_glass = 1.50, n_air = 1.00, θ = 50°

(a) Will TIR occur? First find critical angle: n₁ sin θ_c = n₂ sin 90° (1.50) sin θ_c = (1.00)(1) sin θ_c = 1/1.50 = 0.667 θ_c = 41.8°

Since θ = 50° > θ_c = 41.8°, YES, total internal reflection occurs.

(b) Critical angle: θ_c = sin⁻¹(n₂/n₁) = sin⁻¹(1.00/1.50) = 41.8° or 42°

For angles greater than 42°, all light is reflected (this is how fiber optics work!).

8Problem 8hard

❓ Question:

A concave mirror has focal length 20 cm. An object is placed 60 cm from the mirror. Find (a) image distance, (b) magnification, (c) describe the image.

💡 Show Solution

Given:

Focal length: $f = 20$ cm (positive for concave)
Object distance: $d_o = 60$ cm

Part (a): Image distance

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

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