New YorkPhysicsSyllabus dot point

How do waves bounce off surfaces and bend when they pass between media?

State the law of reflection, define the absolute index of refraction $n = c/v$ , and apply Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ to refraction, including the bending of light between media.

A Regents Physics answer on reflection and refraction: the law of reflection, the absolute index of refraction, and Snell's law for the bending of light between media, using the Reference-Table equations, with worked examples.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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Quick answer

In reflection, a wave bounces off a surface so that the angle of incidence equals the angle of reflection, both measured from the normal. In refraction, a wave bends as it passes into a medium where its speed changes. The absolute index of refraction of a medium is $n = \dfrac{c}{v}$ , the ratio of the speed of light in a vacuum to its speed in the medium; $n$ has no units and is at least 1. Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$ , relates the angles and indices on the two sides of a boundary. Light bends toward the normal entering a denser (higher- $n$ ) medium and away from the normal entering a less dense one. The frequency stays the same; the speed and wavelength change. The equations $\theta_i = \theta_r$ , $n = \dfrac{c}{v}$ and Snell's law are printed on the Reference Tables.

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What this topic is asking
Reflection
The index of refraction
Snell's law and the bending of light
Reference Tables note
Try this

What this topic is asking

This dot point covers two ways waves change direction at a boundary: reflection (bouncing off a surface) and refraction (bending when passing into a new medium). The Physical Setting/Physics course asks you to state the law of reflection, to define the absolute index of refraction $n = \dfrac{c}{v}$ , and to apply Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ to find how light bends between media. All angles are measured from the normal, the line perpendicular to the surface.

Reflection

The law applies to all waves, including light and sound. A smooth (specular) surface reflects parallel rays in parallel, forming a clear image (a mirror); a rough surface scatters them (diffuse reflection). The crucial convention is that angles are always measured from the normal, not from the surface, an easy mark to lose if reversed.

The index of refraction

A larger index means light travels more slowly in that medium and bends more sharply at its surface. The Reference Tables include a table of indices for common materials (water about $1.33$ , crown glass about $1.52$ , diamond about $2.42$ ). A higher index ("optically denser") medium slows the light more.

Snell's law and the bending of light

Refraction is why a straw looks bent in water and why a pool appears shallower than it is. The direction of bending follows from Snell's law and the change in speed: slower means bending toward the normal. Since frequency is fixed by the source, and $v = f\lambda$ , a slower speed means a shorter wavelength in the denser medium.

Refraction from water into air

Light travels from water ( $n_1 = 1.33$ ) into air ( $n_2 = 1.00$ ), striking the surface at $20.$ degrees from the normal. Find the angle of refraction in air.

step 1 Write Snell's law

$n_1\sin\theta_1 = n_2\sin\theta_2$ .

step 2 Solve for the refraction angle

$\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} = \dfrac{(1.33)\sin 20^\circ}{1.00} = (1.33)(0.342) = 0.455$ .

step 3 Find the angle

$\theta_2 = \sin^{-1}(0.455) = 27^\circ$ .

step 4 Check the direction of bending

The light goes from water (denser) into air (less dense), so it bends away from the normal: the refraction angle ( $27^\circ$ ) is larger than the incidence angle ( $20^\circ$ ), as expected.

Reference Tables note

The Reference Tables print the law of reflection $\theta_i = \theta_r$ , the index of refraction $n = \dfrac{c}{v}$ and Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ in the Waves section, along with a table of absolute indices of refraction for common materials and the speed of light constant. Note that the thin-lens and mirror equation is not on these tables: geometric optics on the Regents is handled by ray diagrams and the relationships above, not by a lens formula. You supply the convention that angles are measured from the normal and the reasoning for which way light bends.

Try this

Q1. State the law of reflection. [1 point]

Cue. The angle of incidence equals the angle of reflection, both measured from the normal.

Q2. Light travels at $1.5 \times 10^8$ m/s in a medium. Calculate the index of refraction ( $c = 3.00 \times 10^8$ m/s). [2 points]

Cue. $n = \dfrac{c}{v} = \dfrac{3.00 \times 10^8}{1.5 \times 10^8} = 2.0$ .

Exam-style practice questions

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

Regents (style)2 marksPart B-2 (constructed response). Light travels at

2.0 \times 10^8

m/s in a transparent material. Using

c = 3.00 \times 10^8

m/s, calculate the absolute index of refraction of the material. Show the equation, substitution and answer.

Show worked answer →

A 2-point constructed-response calculation using the Reference-Table equation $n = \dfrac{c}{v}$ .

Equation: $n = \dfrac{c}{v}$ .
Substitution: $n = \dfrac{3.00 \times 10^8}{2.0 \times 10^8}$ .
Answer: $n = 1.5$ (the index of refraction has no units).

Markers reward the equation from the tables, correct substitution, and a dimensionless answer. The index of refraction is always greater than or equal to 1, since light travels fastest in a vacuum.

Regents (style)3 marksPart C (extended response). Light travels from air (

n_1 = 1.00

) into glass (

n_2 = 1.50

), striking the surface at an angle of incidence of

30.

degrees. (a) State the law of reflection. (b) Use Snell's law to calculate the angle of refraction in the glass. Show all work.

Show worked answer →

A 3-point Part C item on reflection and refraction.

(a) Law of reflection (1 point): the angle of incidence equals the angle of reflection (both measured from the normal).
(b) Angle of refraction (2 points): Snell's law gives $n_1\sin\theta_1 = n_2\sin\theta_2$ , so $\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} = \dfrac{(1.00)\sin 30^\circ}{1.50} = \dfrac{0.500}{1.50} = 0.333$ , giving $\theta_2 = 19^\circ$ .

Markers reward the law of reflection and applying Snell's law to find the refraction angle. The ray bends toward the normal entering the denser medium, so the refraction angle is smaller than the incidence angle.

Related dot points

Sources & how we know this

Reference Tables for Physical Setting/Physics — NYSED (2006)
Physical Setting/Physics Core Curriculum — NYSED (2010)