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Why does light bend when it enters a new medium, and when does it reflect entirely?

Topic 13.3 Refraction: apply Snell's law and the index of refraction, and find the critical angle for total internal reflection.

A focused answer to AP Physics 2 Topic 13.3, covering the index of refraction, Snell's law for the bending of light at a boundary, the link between index and speed, total internal reflection and the critical angle, and the direction of bending, with full worked examples.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The index of refraction $n = \dfrac{c}{v}$ measures how much a medium slows light: a higher $n$ means slower light and stronger bending. Snell's law, $n_1 \sin\theta_1 = n_2 \sin\theta_2$ , gives how a ray bends crossing a boundary (angles from the normal). Light entering a denser medium (higher $n$ ) slows and bends toward the normal; entering a less dense medium it speeds up and bends away from the normal. The frequency stays constant across a boundary while the speed and wavelength change. When light goes from a denser to a less dense medium, increasing the angle eventually reaches the critical angle $\theta_c$ , where $\sin\theta_c = \dfrac{n_2}{n_1}$ ; beyond it the light undergoes total internal reflection, reflecting entirely back. This is how optical fibers trap light. Refraction is the bending that makes lenses work.

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What this topic is asking
The index of refraction and the speed of light
Snell's law and the direction of bending
Total internal reflection
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What this topic is asking

The College Board (Topic 13.3) wants you to apply Snell's law and the index of refraction to the bending of light at a boundary, relate the index to the speed of light in a medium, and find the critical angle for total internal reflection.

The index of refraction and the speed of light

The index of refraction encodes how a medium slows light: $v = c/n$ , so glass ( $n \approx 1.5$ ) slows light to two thirds of its vacuum speed. This slowing is the cause of bending. When light crosses a boundary, its frequency stays the same (set by the source), but its speed and wavelength change, and the change of speed at an angle is what redirects the ray.

Snell's law and the direction of bending

Snell's law is the quantitative rule of refraction, and the bending direction follows a simple guide: "slow down, bend toward the normal; speed up, bend away." Light passing from air into glass bends toward the normal (it slows); emerging from glass into air it bends away (it speeds up). A ray hitting head-on (along the normal) does not bend at all. As always in optics, the angles are measured from the normal.

Total internal reflection

When light travels from a denser to a less dense medium, it bends away from the normal, and as the angle of incidence grows, the refracted ray bends closer and closer to the surface. At the critical angle $\theta_c$ , given by $\sin\theta_c = \dfrac{n_2}{n_1}$ (with $n_1 > n_2$ ), the refracted ray would skim along the boundary. Beyond the critical angle, no refraction is possible, and the light reflects entirely back into the denser medium: total internal reflection. This traps light inside the medium, which is exactly how optical fibers carry signals over long distances and how diamonds sparkle ( $n \approx 2.4$ , a small critical angle). The strategic role of this topic is that refraction is the mechanism behind lenses: a lens bends light by refraction at its curved surfaces, so Snell's law here underlies the thin-lens image formation of Topic 13.4. Mastering the bending direction and the critical-angle condition is what makes lens behavior intuitive rather than memorized.

Critical angle for a glass-air boundary

Light travels inside glass ( $n = 1.50$ ) toward a boundary with air ( $n = 1.00$ ). Find the critical angle, and state what happens at $50$ degrees of incidence.

step 1 Critical-angle condition

$\sin\theta_c = \dfrac{n_2}{n_1} = \dfrac{1.00}{1.50} = 0.667$ .

step 2 Solve for the critical angle

$\theta_c = \sin^{-1}(0.667) = 41.8$ degrees.

step 3 Compare with 50 degrees

$50$ degrees is greater than the critical angle of $41.8$ degrees, so the light undergoes total internal reflection: it reflects entirely back into the glass, with no ray escaping into the air.

step 4 Interpret

This is the principle of optical fibers: light striking the wall above the critical angle is trapped, bouncing along the fiber with no loss to refraction.

Try this

Q1. Light passes from air into water (higher index). State whether it bends toward or away from the normal. [1 point]

Cue. Toward the normal (it slows entering the denser medium).

Q2. Calculate the speed of light in a medium of refractive index $2.0$ . [1 point]

Cue. $v = c/n = (3.0 \times 10^8)/2.0 = 1.5 \times 10^8$ m/s.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). Light travels from air (

n = 1.00

) into glass (

n = 1.50

), striking the surface at

40

degrees from the normal. (a) Use Snell's law to find the angle of refraction in the glass. (b) State and justify whether the light bends toward or away from the normal. (c) Calculate the speed of light in the glass (speed in vacuum

c = 3.0 \times 10^8

m/s).

Show worked answer →

A 7-point FRQ on refraction.

(a) Angle of refraction (3 points): Snell's law $n_1 \sin\theta_1 = n_2 \sin\theta_2$ , so $\sin\theta_2 = \dfrac{n_1 \sin\theta_1}{n_2} = \dfrac{(1.00)\sin 40^\circ}{1.50} = \dfrac{0.643}{1.50} = 0.428$ , giving $\theta_2 = 25.4$ degrees.
(b) Direction (2 points): the light enters a denser medium (higher $n$ ), so it slows and bends toward the normal ( $\theta_2 < \theta_1$ ).
(c) Speed (2 points): $v = \dfrac{c}{n} = \dfrac{3.0 \times 10^8}{1.50} = 2.0 \times 10^8$ m/s.

Markers reward Snell's law for the angle, bending toward the normal into a denser medium, and the speed from $v = c/n$ .

AP 2023 (style)1 marksSection I (multiple choice). Total internal reflection can occur when light travels (A) from a less dense to a more dense medium at any angle (B) from a more dense to a less dense medium above the critical angle (C) along the normal (D) only in a vacuum. Justify your reasoning.

Show worked answer →

A 1-point MCQ on total internal reflection. The answer is (B).

Total internal reflection happens only when light goes from a higher-index (denser) medium to a lower-index one, at an angle of incidence greater than the critical angle, where Snell's law has no refracted solution. The trap is (A): going into a denser medium, light always refracts and never totally reflects.

Related dot points

Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)