Snell's Law — Physica

Equivalent forms

\frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1}

\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2}

A single product equality governs every bending of light at every interface in the universe.

Symbol	Name	SI unit	Dimension	Range
$n_1$	Refractive index (medium 1) Refractive index of the incident medium	dimensionless	1	1 – 2.5
$theta_1$	Angle of incidence Angle between incident ray and the normal	rad	1	0 – 1.5708
$n_2$	Refractive index (medium 2) Refractive index of the refracting medium	dimensionless	1	1 – 2.5
$theta_2$	Angle of refractionoutput Angle between refracted ray and the normal	rad	1	0 – 1.5708

Unit systems

Symbol	SI	CGS	Imperial
$theta_1$	rad	rad	deg
$theta_2$	rad	rad	deg

Where it holds

Valid for isotropic, homogeneous media. Breaks down for metamaterials with negative refractive index and at scales comparable to wavelength (diffraction regime).

Dimensional analysis

n_{1}\cdot \sin (\theta _{1}) \to [1]\cdot [1] = [1]

n_{2}\cdot \sin (\theta _{2}) \to [1]\cdot [1] = [1]

Both sides are dimensionless.

\checkmark

Discovery

Willebrord Snellius · 1621

Discovered empirically by Snellius; independently derived by Descartes in 1637 using corpuscular theory. Ibn Sahl described an equivalent law in 984 AD.

Try this

Why does a swimming pool look shallower than it really is?

A light ray passes from air (n=1.00) into water (n=1.33) at 45 degrees. Find the refraction angle using Snell's law.

Research status: stable

Real-world applications

Optical fiber design: total internal reflection keeps light trapped inside the core.
Corrective lenses: curved surfaces use refraction to focus light on the retina.
Mirage formation: continuous refraction in heated air layers bends light upward.
Diamond cutting: high refractive index $(n=2.42)$ creates total internal reflection sparkle.

Common misconceptions

Light 'decides' to bend — it follows the path of least time naturally, not by choice.
Refraction only applies to light — it applies to all waves crossing an interface (sound, seismic, etc.).
The angle is measured from the surface — it is always measured from the normal to the surface.

Experimental verification

Verified by Snellius (1621) using protractors and glass prisms. Modern verification uses laser beams at precise angles through calibrated optical media with angular resolution < 0.01 degrees.

Derivation

From Fermat's principle of least time: light takes the path that minimizes travel time between two points.

At an interface, minimizing

t = d_{1}/(c/n_{1}) + d_{2}/(c/n_{2})

with respect to the entry point yields

n_{1}*\sin (\theta 1) = n_{2}*\sin (\theta 2)

.

Alternatively, from Huygens' principle: wavelets in the second medium travel at

v_{2} = c/n_{2}

, causing the wavefront to tilt, producing the same relation.

Limiting cases

\theta _1 \to 0

⟶

\theta _2 \to 0

Normal incidence means no bending.

n_1 = n_2

⟶

\theta _2 = \theta _1

Same medium on both sides means no refraction.

\theta _1 \to \arcsin (n_2/n_1) (n_1

> n_2)⟶

\theta _2 \to 90^{\circ}

Critical angle: total internal reflection begins.

What if…

What if n2 < n1 (denser to rarer)?

Light bends away from the normal. Beyond the critical angle, total internal reflection occurs — no light crosses the boundary.

What if the medium has a negative refractive index?

In metamaterials with n < 0, the refracted ray bends to the same side of the normal — 'negative refraction' enables superlenses.

What if wavelength changes?

Refractive index is wavelength-dependent (dispersion). Different colors refract at different angles — this is how prisms split white light into a rainbow.

1

Light entering water from air

Given ·

n 1:: 1
n 2:: 1.33
theta 1:: 0.7854

Find · theta_2

Steps

Write Snell's law: $n_{1}*\sin (\theta 1) = n_{2}*\sin (\theta 2)$
Substitute: $1.00 * \sin (0.7854) = 1.33 * \sin (\theta 2)$
Compute $\sin (0.7854) = 0.7071$
$\sin (\theta 2) = 0.7071 / 1.33 = 0.5317$
$\theta 2 = \arcsin (0.5317) = 0.5605\,\mathrm{rad} \approx 32.1^{\circ}$

Result ·

\theta _2 = \arcsin (n_{1}*\sin (\theta 1)/n_{2}) = \arcsin (1.00 * \sin (45^{\circ}) / 1.33) = \arcsin (0.5317) = 32.1^{\circ} = 0.5605\,\mathrm{rad}

2

Critical angle for glass-to-air

Given ·

n 1:: 1.5
n 2:: 1

Find · theta_critical

Steps

At the critical angle, $\theta 2 = 90^{\circ}$ , so $\sin (\theta 2) = 1$
$n_{1}*\sin (\theta _c) = n_{2}*1$
$\sin (\theta _c) = n_{2}/n_{1} = 1.0/1.5 = 0.6667$
$\theta _c = \arcsin (0.6667) = 0.7297\,\mathrm{rad} \approx 41.8^{\circ}$

Result ·

\theta _c = \arcsin (n_{2}/n_{1}) = \arcsin (1.0/1.5) = \arcsin (0.6667) = 41.8^{\circ} = 0.7297\,\mathrm{rad}

Law of Reflection→Brewster's Angle→Wave Speed Equation→