Snell's Law — Refraction Formula & Critical Angle

Snell's law is the refraction formula for a light ray crossing from one medium into another. If you know the two refractive indices and one angle, it tells you the other angle.

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

Here $n_1$ and $n_2$ are the refractive indices, and $\theta_1$ and $\theta_2$ are measured from the normal, not from the surface. If light enters a higher-index medium, it bends toward the normal. If it enters a lower-index medium, it bends away from the normal.

The same idea also explains the critical angle. If light starts in the higher-index medium, there is a largest incident angle that still produces refraction. Above that angle, total internal reflection occurs instead.

Snell's law formula and what the angles mean

Snell's law is used in ray optics when a light ray reaches a boundary and enters the second medium. In plain language, the ray changes direction because light travels at different speeds in different media.

The order matters. $n_1$ and $\theta_1$ belong to the starting medium, while $n_2$ and $\theta_2$ belong to the second medium. Swapping them changes the physical situation.

How to tell which way the ray bends

You can often predict the direction before doing any algebra.

• If $n_2 > n_1$, then $\theta_2 < \theta_1$, so the ray bends toward the normal.
• If $n_2 < n_1$, then $\theta_2 > \theta_1$, so the ray bends away from the normal, provided refraction still occurs.

This quick check catches many setup mistakes before you trust a numerical answer.

Critical angle: when refraction stops

The critical angle exists only when light goes from a higher-index medium to a lower-index medium. At that angle, the refracted ray would travel along the boundary, so $\theta_2 = 90^\circ$.

Substituting that into Snell's law gives

$$n_1 \sin \theta_c = n_2$$

so

$$\sin \theta_c = \frac{n_2}{n_1}$$

This formula works only if $n_1 > n_2$. If $n_1 \le n_2$, there is no critical angle for that direction of travel, so total internal reflection cannot happen in that case.

Worked example: water to air

Suppose light travels from water into air with

$$n_1 = 1.33, \qquad n_2 = 1.00, \qquad \theta_1 = 40^\circ$$

First use Snell's law:

$$1.33 \sin 40^\circ = 1.00 \sin \theta_2$$

Using $\sin 40^\circ \approx 0.643$,

$$\sin \theta_2 \approx 1.33 \times 0.643 \approx 0.855$$ $$\theta_2 \approx \sin^{-1}(0.855) \approx 58.8^\circ$$

So the refracted angle is about $58.8^\circ$. That is larger than the incident angle, which makes sense because the light is moving from higher index to lower index and bends away from the normal.

Now find the critical angle for the same pair of media:

$$\sin \theta_c = \frac{1.00}{1.33} \approx 0.752$$ $$\theta_c \approx \sin^{-1}(0.752) \approx 48.8^\circ$$

Since $40^\circ < 48.8^\circ$, the ray refracts into the air. If the incident angle were larger than about $48.8^\circ$, this water-to-air setup would produce total internal reflection instead.

Common mistakes in Snell's law problems

• Measuring angles from the surface instead of from the normal.
• Reversing $n_1$ and $n_2$ after the diagram is already set.
• Assuming light always bends toward the normal.
• Using the critical-angle formula when light is entering the higher-index medium instead of leaving it.
• Accepting a sine value greater than $1$ as a normal refraction result instead of recognizing that total internal reflection should occur.

Where Snell's law is used

Snell's law appears in basic optics problems involving water surfaces, glass blocks, prisms, lenses, and optical fibers. It also explains why a straw looks bent in water and why fiber optics can trap light.

For most introductory physics questions, this law is the first tool to use whenever a ray crosses a boundary between two media.

Try A Similar Problem

Let light start in glass with $n_1 = 1.50$ and enter air with $n_2 = 1.00$. First find the critical angle, then decide whether an incident angle of $35^\circ$ gives refraction or total internal reflection.