Geometric Optics: Refraction and Reflection

AP Physics 2· AP Physics 2 CED — Geometric and Physical Optics· 14 min read

1. Law of Reflection and Index of Refraction★★☆☆☆⏱ 3 min

The most fundamental quantity in geometric optics is the index of refraction $n$ , which describes how much slower light travels in a medium compared to vacuum. All angles are measured relative to the normal (perpendicular to the boundary), not the boundary itself.

📘 Definition

Index of Refraction

$n$

Ratio of the speed of light in vacuum to the speed of light in the medium. $n \geq 1$ for all physical media, with $n_{vacuum} = 1$ and $n_{air} \approx 1$ for nearly all exam problems.

n = \frac{c}{v}

When light crosses a boundary, its frequency $f$ does not change (set by the source), so wavelength changes proportionally to speed: $λ_{n} = \frac{λ _{0}}{n}$ , where $λ_{0}$ is the wavelength in vacuum/air.

📘 Definition

Law of Reflection

$θ_{i} = θ_{r}$

For reflection at a smooth flat boundary, the angle of incidence (between incident ray and normal) equals the angle of reflection (between reflected ray and normal). All three (incident ray, reflected ray, normal) lie in the same plane.

📐 Worked Example

A laser beam hits a flat glass window at an angle of 28° measured from the surface of the glass. What is the angle between the incident ray and the reflected ray?

1
Convert the given surface-relative angle to a normal-relative angle of incidence:
2
$\theta_i = 90^\circ - 28^\circ = 62^\circ$
3
By the law of reflection, angle of reflection equals angle of incidence:
4
$\theta_r = \theta_i = 62^\circ$
5
The total angle between the two rays is the sum of the angles, since they sit on opposite sides of the normal:
6
$\theta_{\text{total}} = \theta_i + \theta_r = 62^\circ + 62^\circ = 124^\circ$

Exam tip:

Always double-check whether a problem gives the angle relative to the boundary or the normal. If it's relative to the boundary, subtract from 90° before doing any calculations.

2. Snell's Law of Refraction★★★☆☆⏱ 4 min

When light transmits across a boundary from medium 1 to medium 2, it bends (refracts) because its speed changes. The relationship between incident and refracted angles is given by Snell's Law.

📘 Definition

Snell's Law of Refraction

Relates incident and refracted angles to the refractive indices of the two media.

n_1 \sin\theta_1 = n_2 \sin\theta_2

A simple rule of thumb for bending direction: if $n_{2} > n_{1}$ (light moves into a slower medium), light bends toward the normal. If $n_{2} < n_{1}$ , light bends away from the normal.

📐 Worked Example

Light travels from air ( $n_{1} = 1.00$ ) into olive oil ( $n_{2} = 1.47$ ). The incident angle is 30° relative to the normal, and the light has a wavelength of 630 nm in air. Find the refracted angle and the wavelength of the light in olive oil.

1
Start with Snell's Law, rearrange to solve for $sin θ_{2}$ :
2
$\sin\theta_2 = \frac{n_1 \sin\theta_1}{n_2} = \frac{1.00 \cdot \sin30^\circ}{1.47} \approx 0.340$
3
Calculate the refracted angle:
4
$\theta_2 = \arcsin(0.340) \approx 19.9^\circ$
5
This matches our expectation: light bends toward the normal moving from lower $n$ (air) to higher $n$ (oil). Calculate wavelength in oil:
6
$\lambda_{\text{oil}} = \frac{\lambda_{\text{air}}}{n_2} = \frac{630 \text{ nm}}{1.47} \approx 429 \text{ nm}$

Exam tip:

When asked for direction of bending, always compare the indices first: higher $n$ = slower speed = bend toward the normal. Don't guess from diagrams, which are often not drawn to scale.

3. Total Internal Reflection and Critical Angle★★★☆☆⏱ 3 min

Total internal reflection (TIR) is a phenomenon where all incident light reflects back into the original medium, with no refraction into the second medium. TIR only occurs when light travels from a higher index medium to a lower index medium ( $n_{1} > n_{2}$ ). If the incident angle is large enough, Snell's Law would require $sin θ_{2} > 1$ , which is impossible, so no refracted ray exists.

📘 Definition

Critical Angle

$θ_{c}$

The minimum incident angle that causes total internal reflection. At the critical angle, the refracted angle is exactly $9 0^{\circ}$ , so $sin θ_{2} = 1$ .

\sin\theta_c = \frac{n_2}{n_1} \quad (n_1 > n_2)

TIR is the operating principle behind fiber optic communications, diamond sparkle, and reflecting prisms in binoculars.

📐 Worked Example

Water in a fish tank has $n_{water} = 1.33$ , air has $n_{air} = 1.00$ . A fish looks up toward the surface at an angle of 45° from the normal (light travels from water to air). Does the fish see light from above the water, or a reflection of the tank bottom?

1
Confirm TIR conditions: light travels from higher $n$ (water) to lower $n$ (air), so TIR is possible. Calculate critical angle:
2
$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.33} \approx 0.752 \implies \theta_c = \arcsin(0.752) \approx 48.8^\circ$
3
Compare incident angle to critical angle: $4 5^{\circ} < 48. 8^{\circ}$ , so TIR does not occur. Therefore, the fish sees light from above the water.

Exam tip:

TIR can never occur when light moves from lower $n$ to higher $n$ . Always check the direction of travel first before calculating critical angle.

4. AP-Style Additional Worked Examples★★★★☆⏱ 4 min

📐 Worked Example

A ray of light travels from corn syrup ( $n_{1} = 1.48$ ) into an unknown clear liquid. Incident angle is 40° relative to normal, refracted angle is 47° relative to normal. (a) Calculate the index of refraction of the unknown. (b) Does light bend toward or away from the normal? (c) Find the critical angle for this boundary.

1
(a) Rearrange Snell's Law to solve for $n_{2}$ :
2
$n_2 = \frac{n_1 \sin\theta_1}{\sin\theta_2} = \frac{1.48 \cdot \sin40^\circ}{\sin47^\circ} \approx 1.30$
3
(b) Light bends away from the normal. Light moves from higher $n$ (1.48) to lower $n$ (1.30), so it speeds up and bends away from the normal.
4
(c) Critical angle exists because $n_{1} > n_{2}$ :
5
$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.30}{1.48} \approx 0.878 \implies \theta_c \approx 61^\circ$

📐 Worked Example

A step-index fiber optic cable has a glass core ( $n_{1} = 1.52$ ) surrounded by polymer cladding ( $n_{2} = 1.49$ ). What is the maximum angle of incidence (from air into the core) that results in total internal reflection at the core-cladding boundary?

1
First find critical angle for TIR at the core-cladding boundary:
2
$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.49}{1.52} \approx 0.980 \implies \theta_c \approx 78.5^\circ$
3
By geometry, the angle of the ray inside the core relative to the input face normal is:
4
$\theta_2 = 90^\circ - 78.5^\circ = 11.5^\circ$
5
Apply Snell's Law at the input face (air to core):
6
$n_{\text{air}} \sin\theta_{\text{max}} = n_1 \sin\theta_2 \implies \sin\theta_{\text{max}} \approx 0.304 \implies \theta_{\text{max}} \approx 17.7^\circ$

5. Common Pitfalls

Wrong move:

Using the angle given relative to the boundary directly in Snell's law or the law of reflection

Why:

Problems often intentionally give angles relative to the surface to test convention knowledge

Correct move:

Always check the problem's angle reference; if given relative to boundary, subtract from 90° before any calculation

Wrong move:

Calculating critical angle for light moving from lower $n$ to higher $n$

Why:

Students memorize the formula but forget TIR only occurs when going from higher to lower index

Correct move:

Explicitly confirm $n_{1} > n_{2}$ before using the critical angle formula; if not, TIR is impossible

Wrong move:

Changing the frequency of light when calculating wavelength or speed in a new medium

Why:

Students confuse wavelength and frequency changes, incorrectly assuming frequency scales with $n$

Correct move:

Frequency is always determined by the source, it never changes across a boundary; only speed and wavelength change

Wrong move:

Claiming light bends away from the normal when moving from air to glass

Why:

Students mix up the relationship between $n$ , speed, and bending direction

Correct move:

Follow the rule: higher $n$ = slower speed = smaller angle = bend toward the normal; lower $n$ = faster speed = larger angle = bend away

Wrong move:

Writing the critical angle formula as $sin θ_{c} = n_{1} / n_{2}$ instead of $n_{2} / n_{1}$

Why:

Students mix up which index is which when memorizing

Correct move:

Always derive from Snell's law from scratch: start with $n_{1} sin θ_{c} = n_{2} sin 9 0^{\circ}$ , so $sin θ_{c} = n_{2} / n_{1}$

Wrong move:

Adding incident and refracted angles to get the total angle between them

Why:

Students confuse reflection geometry with refraction geometry

Correct move:

For reflection, add incident and reflected angles; for refraction, subtract the smaller angle from the larger to get the angle between rays

6. Quick Reference Cheatsheet

Category	Formula	Notes
Index of Refraction	$n = \frac{c}{v}$	$c = 3 \times 1 0^{8}$ m/s, $n \geq 1$ always, $n_{ai r} \approx 1$ for most problems
Wavelength in Medium	$λ_{n} = \frac{λ _{0}}{n}$	$λ_{0}$ = wavelength in vacuum/air; frequency $f$ is unchanged across boundaries
Law of Reflection	$θ_{i} = θ_{r}$	All angles measured relative to the normal (perpendicular to boundary)
Snell's Law of Refraction	$n_{1} sin θ_{1} = n_{2} sin θ_{2}$	$θ_{1}$ = incident angle in medium 1, $θ_{2}$ = refracted angle in medium 2
Bending Direction Rule	N/A	If $n_{2} > n_{1}$ : bend toward normal; if $n_{2} < n_{1}$ : bend away from normal
Critical Angle for TIR	$sin θ_{c} = \frac{n _{2}}{n _{1}}$	Only valid when $n_{1} > n_{2}$ (light travels from higher n to lower n)
TIR Occurrence Condition	$θ_{1} > θ_{c}$	No refraction occurs when TIR happens; all light reflects back into incident medium

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
TIR critical angle calculation
2022 · FRQ
Snell's law application problem
2021 · MCQ
Bending direction concept question

What's Next

This topic is the foundational prerequisite for all remaining content in Unit 6. Next, you will apply these reflection and refraction rules to curved mirrors and thin lenses, where you extend the ray model to find image positions, sizes, and magnifications. Without mastering angle conventions, Snell's law, and TIR here, you cannot correctly draw ray diagrams or solve image formation problems, which make up a large portion of the unit's exam score. After geometric optics, you move to physical optics, where you drop the ray approximation to study interference and diffraction, relying on the wavelength-index relationship you learned here. This topic connects electromagnetic wave behavior to real-world optical technologies like fiber optics, cameras, and telescopes.