Reflection and Refraction

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Reflection and Refraction is part of the AP Physics 2 course on Study Mondo, specifically in the Optics section. You can explore the full course for more related topics and practice resources.

Q: Are there practice problems for Reflection and Refraction?

Yes, this page includes 20 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

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

Written and reviewed by the Study Mondo Education TeamLast updated January 6, 2026

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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: $\theta_1 = 30°$

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!

8Problem 8medium

❓ 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!

9Problem 9medium

❓ 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!

10Problem 10medium

❓ 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!

11Problem 11medium

❓ 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!

12Problem 12medium

❓ 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!

13Problem 13medium

❓ 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.

14Problem 14hard

❓ 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

15Problem 15hard

❓ 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

16Problem 16hard

❓ 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

17Problem 17hard

❓ 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!).

18Problem 18hard

❓ 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

19Problem 19hard

❓ 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

20Problem 20hard

❓ 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

Explain using:

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❓ Frequently Asked Questions

What is Reflection and Refraction?▾

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

How can I study Reflection and Refraction effectively?▾

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 20 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Is this Reflection and Refraction study guide free?▾

Yes — all study notes, flashcards, and practice problems for Reflection and Refraction on Study Mondo are free to access. No account is needed.

What course covers Reflection and Refraction?▾

Reflection and Refraction is part of the AP Physics 2 course on Study Mondo, specifically in the Optics section. You can explore the full course for more related topics and practice resources.

Are there practice problems for Reflection and Refraction?▾

Yes, this page includes 20 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Reflection and Refraction

Try the Interactive Version!

🔆 Reflection and Refraction

Nature of Light

Law of Reflection

Plane Mirrors

Index of Refraction

Snell's Law

Light entering denser medium (n₂ > n₁):

Light entering less dense medium (n₂ < n₁):

Total Internal Reflection

Dispersion

Curved Mirrors

Concave Mirror (converging):

Convex Mirror (diverging):

Mirror Equation

Sign Conventions (Mirrors)

Ray Diagrams (Concave Mirror)

Problem-Solving Strategy

Common Mistakes

📚 Practice Problems

1Problem 1easy

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2Problem 2easy

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3Problem 3easy

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4Problem 4easy

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5Problem 5easy

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6Problem 6easy

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7Problem 7medium

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8Problem 8medium

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9Problem 9medium

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10Problem 10medium

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11Problem 11medium

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12Problem 12medium

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13Problem 13medium

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14Problem 14hard

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15Problem 15hard

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16Problem 16hard

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17Problem 17hard

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18Problem 18hard

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19Problem 19hard

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20Problem 20hard

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