Geometric Optics: Refraction and Reflection — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Index of refraction definition, law of reflection at plane boundaries, Snell’s law of refraction, total internal reflection, critical angle calculation, and boundary behavior for light moving between transparent media.

You should already know: 1. Wave speed, wavelength, and frequency relationships for electromagnetic waves. 2. Basic ray modeling of straight-line light propagation. 3. Geometric angle measurement conventions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Geometric Optics: Refraction and Reflection?

Geometric optics models light as propagating in straight-line rays, ignoring wave effects like diffraction that are covered later in physical optics. This subtopic describes how light behaves when it encounters a flat boundary between two transparent media with different optical properties. Per the AP Physics 2 Course and Exam Description (CED), Unit 6 (Geometric and Physical Optics) makes up 16–24% of the total exam score, and this refraction and reflection subtopic accounts for roughly one-third of that unit weight. It appears regularly in both multiple-choice questions (MCQ) and as a foundational component of free-response questions (FRQ) covering lenses, mirrors, or optical technologies.

Standard notation conventions are used here: all angles are measured relative to the normal, the line perpendicular to the boundary at the point of incidence, not relative to the boundary itself. The index of refraction $n$ is defined relative to vacuum ($n_{\text{vacuum}}=1$, $n_{\text{air}}\approx 1$ for nearly all exam problems). Reflection describes light bouncing back into the original medium, while refraction describes light bending as it transmits into the new medium. Mastery of the rules here is required for all subsequent geometric optics content.

2. Law of Reflection and Index of Refraction

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. The definition is: $$n=\frac{c}{v}$$ where $c=3.0\times 10^8\text{ m/s}$ is the speed of light in vacuum, and $v$ is the speed of light in the medium. Since $v$ can never exceed $c$, $n$ is always greater than or equal to 1. When light crosses a boundary, its frequency $f$ does not change (it is set by the source), so its wavelength changes proportionally to speed: ${\lambda }_n=\frac{v}{f}=\frac{c/n}{f}=\frac{{\lambda }_0}{n}$, where ${\lambda }_0$ is the wavelength in vacuum.

For reflection at a flat, smooth boundary, the law of reflection states that the angle of incidence ${\theta }_i$ (angle between incident ray and normal) equals the angle of reflection ${\theta }_r$ (angle between reflected ray and normal), or ${\theta }_i={\theta }_r$. The incident ray, reflected ray, and normal all lie in the same plane, called the plane of incidence. This rule comes from the requirement that the parallel component of the light’s wave vector is continuous across the boundary.

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. All angles for geometric optics must be measured relative to the normal, not the surface. The angle from the surface is 28°, so the angle of incidence is ${\theta }_i=90^{\circ }-28^{\circ }=62^{\circ }$.
2. By the law of reflection, the angle of reflection equals the angle of incidence: ${\theta }_r={\theta }_i=62^{\circ }$, measured from the normal on the opposite side of the normal from the incident ray.
3. The total angle between the two rays is the sum of the two angles, since they sit on opposite sides of the normal: ${\theta }_{\text{total}}={\theta }_i+{\theta }_r=62^{\circ }+62^{\circ }=124^{\circ }$.
4. Check for consistency: a result greater than 90° makes sense here, since we started with a shallow angle relative to the surface.

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.

3. Snell's Law of Refraction

When light transmits across a boundary from medium 1 (index $n_1$) to medium 2 (index $n_2$), it bends (refracts) because its speed changes. The relationship between the incident angle ${\theta }_1$ and refracted angle ${\theta }_2$ (both measured relative to the normal) is given by Snell’s Law: $$n_1\sin {\theta }_1=n_2\sin {\theta }_2$$ Snell’s Law also comes from conservation of the parallel component of the wave vector across the boundary, so it is a direct consequence of wave behavior. There is a simple rule of thumb for bending direction: if $n_2>n_1$ (light moves into a slower medium), then $\sin {\theta }_2=\frac{n_1}{n_2}\sin {\theta }_1<\sin {\theta }_1$, so ${\theta }_2<{\theta }_1$: 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. The light has a wavelength of 630 nm in air. Find the refracted angle and the wavelength of the light in olive oil.

1. Write Snell’s Law for the boundary: $n_1\sin {\theta }_1=n_2\sin {\theta }_2$.
2. Rearrange to solve for $\sin {\theta }_2$: $\sin {\theta }_2=\frac{n_1\sin {\theta }_1}{n_2}=\frac{1.00\cdot \sin 30^{\circ }}{1.47}=\frac{0.5}{1.47}\approx 0.340$.
3. Take the inverse sine to get ${\theta }_2=\arcsin (0.340)\approx 19.9^{\circ }$. This is less than 30°, which matches our expectation: light bends toward the normal when moving from lower $n$ (air) to higher $n$ (oil).
4. Wavelength in oil is ${\lambda }_{\text{oil}}=\frac{{\lambda }_{\text{air}}}{n_2}=\frac{630\text{ nm}}{1.47}\approx 429\text{ nm}$, which makes sense: wavelength decreases when light enters a slower medium.

Exam tip: When asked for direction of bending, always compare the indices first: slower medium (higher $n$) means bending toward the normal, faster (lower $n$) means away. Don’t guess from the diagram alone, as diagrams are often not drawn to scale.

4. Total Internal Reflection and Critical Angle

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 in one specific case: when light is traveling 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 $\sin {\theta }_2>1$, which is impossible, so no refracted ray exists.

The minimum incident angle that causes TIR is called the critical angle ${\theta }_c$. At ${\theta }_c$, the refracted angle is exactly $90^{\circ }$, so $\sin {\theta }_2=1$. Substituting into Snell’s Law gives: $$\sin {\theta }_c=\frac{n_2}{n_1}(n_1>n_2)$$ If the incident angle ${\theta }_1>{\theta }_c$, TIR occurs. TIR is the operating principle behind fiber optic communications, diamond sparkle, and reflecting prisms in binoculars and periscopes.

Worked Example

The water in a fish tank has an index of refraction $n_{\text{water}}=1.33$, and air is $n_{\text{air}}=1.00$. A fish looks up toward the surface of the water at an angle of 45° from the normal (the fish’s line of sight is the light ray traveling from water to air). Does the fish see light from above the water, or does it see a reflection of the bottom of the tank?

1. First confirm TIR conditions: light is traveling from water (higher $n=1.33$) to air (lower $n=1.00$), so TIR is possible if the angle is large enough.
2. Calculate the critical angle: $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.00}{1.33}\approx 0.752$, so ${\theta }_c=\arcsin (0.752)\approx 48.8^{\circ }$.
3. Compare the incident angle to the critical angle: the incident angle is 45°, which is less than 48.8°.
4. TIR does not occur here, so light refracts from air into the water: the fish sees light from above the water, not a reflection.

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.

5. Common Pitfalls (and how to avoid them)

• 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, and students forget to convert to normal-relative angles. 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 air (n=1) into water (n=1.33). Why: Students memorize the critical angle formula but forget the condition that TIR only occurs when going from higher to lower n. Correct move: Before writing the critical angle formula, explicitly confirm $n_1>n_2$ (incident medium has higher index than refracted medium); if not, TIR is impossible and no critical angle exists.
• Wrong move: Changing the frequency of light when calculating wavelength or speed in a new medium. Why: Students confuse wavelength and frequency changes; many incorrectly assume frequency scales with n. Correct move: Remember that frequency is determined by the source, so 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 from the normal.
• Wrong move: Writing the critical angle formula as $\sin {\theta }_c=n_1/n_2$ instead of $n_2/n_1$. Why: Students mix up which index is which when rearranging Snell's law. Correct move: Always derive critical angle from Snell's law from scratch instead of memorizing the inverted ratio: start with $n_1\sin {\theta }_c=n_2\sin 90^{\circ }$, so $\sin {\theta }_c=n_2/n_1$.
• Wrong move: Adding incident and refracted angles to get the total angle between them. Why: Students confuse reflection (both angles on same side of boundary, opposite sides of normal) with refraction (both angles on opposite sides of boundary), so they incorrectly add angles. Correct move: For reflected angle, add incident and reflected; for refracted angle, subtract the smaller from the larger to get the angle between incident and refracted rays.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A student shines a laser through a rectangular block of transparent plastic sitting in air. The laser enters the left side of the block, then exits the right side back into air. Which of the following correctly compares the exiting laser beam to the incident beam before it entered the block? A) The exiting beam is parallel to the incident beam, but shifted laterally B) The exiting beam is bent toward the normal relative to the incident beam C) The frequency of the exiting beam is lower than the incident beam D) The speed of the exiting beam is slower than the incident beam

Worked Solution: We apply Snell's law to both boundaries. When entering the plastic: $n_{\text{air}}\sin {\theta }_1=n_{\text{plastic}}\sin {\theta }_2$. When exiting back into air: $n_{\text{plastic}}\sin {\theta }_2=n_{\text{air}}\sin {\theta }_3$. Equating the two expressions gives $\sin {\theta }_1=\sin {\theta }_3$, so ${\theta }_1={\theta }_3$: the exit angle equals the incident angle, so the beams are parallel, just shifted because of refraction inside the block. Frequency never changes across boundaries, and speed in exiting air is the same as incident air. Only option A is correct.

Question 2 (Free Response)

A ray of light travels from corn syrup (n = 1.48) into an unknown clear liquid. The incident angle in corn syrup is 40° relative to the normal, and the refracted angle in the unknown liquid is 47° relative to the normal. (a) Calculate the index of refraction of the unknown liquid. (b) Does the light bend toward or away from the normal when crossing the boundary? Justify your answer. (c) What is the critical angle for the boundary between corn syrup and the unknown liquid, for light traveling from corn syrup into the unknown? If no critical angle exists, explain why.

Worked Solution: (a) Apply Snell's law: $n_1\sin {\theta }_1=n_2\sin {\theta }_2$. Rearrange for $n_2$: $$n_2=\frac{n_1\sin {\theta }_1}{\sin {\theta }_2}=\frac{1.48\cdot \sin 40^{\circ }}{\sin 47^{\circ }}\approx \frac{1.48\cdot 0.6428}{0.7314}\approx 1.30$$ (b) The light bends away from the normal. The incident angle (40°) is smaller than the refracted angle (47°), which occurs when moving from a higher n medium (corn syrup, 1.48) to a lower n medium (unknown, 1.30), so light speeds up and bends away from the normal. (c) Critical angle exists because light travels from higher n (1.48) to lower n (1.30). Calculate: $$\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.30}{1.48}\approx 0.878⟹{\theta }_c\approx 61^{\circ }$$

Question 3 (Application / Real-World Style)

A step-index fiber optic cable has a glass core with $n_1=1.52$, surrounded by a polymer cladding with $n_2=1.49$. What is the maximum angle of incidence (measured from air into the core at the end of the cable) that results in total internal reflection at the core-cladding boundary? Use $n_{\text{air}}=1.00$.

Worked Solution: First, find the critical angle for TIR at the core-cladding boundary: $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.49}{1.52}\approx 0.980⟹{\theta }_c\approx 78.5^{\circ }$. By geometry, the angle of the ray inside the core relative to the normal of the input face is ${\theta }_2=90^{\circ }-78.5^{\circ }=11.5^{\circ }$. Apply Snell's law at the input face: $n_{\text{air}}\sin {\theta }_{\text{max}}=n_1\sin {\theta }_2⟹\sin {\theta }_{\text{max}}=1.52\cdot \sin (11.5^{\circ })\approx 1.52\cdot 0.20\approx 0.304⟹{\theta }_{\text{max}}\approx 17.7^{\circ }$. This maximum angle (called the acceptance angle) means that any light entering the fiber at an angle larger than ~18° will leak out of the core and not be transmitted to the other end.

7. Quick Reference Cheatsheet

8. 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 will extend the ray model to find image positions, sizes, and magnifications. Without mastering angle conventions, Snell's law, and TIR here, you will not be able to correctly draw ray diagrams or solve image formation problems for lenses and mirrors, which make up a large portion of the unit's exam score. After geometric optics, you will move to physical optics, where you will drop the ray approximation and study interference and diffraction of light, which relies on understanding how wavelength changes with index of refraction that you learned here. The bigger picture: this topic connects electromagnetic wave behavior to real-world optical technologies like fiber optics, cameras, and telescopes.

• Image formation by plane and curved mirrors
• Thin lens equation and image formation by lenses
• Physical Optics: Interference and Diffraction