Total internal reflection and the critical angle

Key ideas

When light travels from a denser material toward a less dense one (e.g. glass → air), it bends away from the normal. Increase the angle of incidence enough and the refracted ray bends until it runs along the surface — at the critical angle $\theta_c$, the angle of refraction is $90°$.

Beyond the critical angle, no light escapes — it is all reflected back inside. This is total internal reflection (TIR).

The two conditions for TIR

TIR happens only when both are true:

• Light is going from a more dense to a less dense medium ($n_1 > n_2$).
• The angle of incidence is greater than the critical angle ($\theta_1 > \theta_c$).

Finding the critical angle

At the critical angle, $\theta_2 = 90°$, so Snell's law becomes:

$n_1 \sin\theta_c = n_2 \sin 90°$

Since $\sin 90° = 1$:

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

• $n_1$ — refractive index of the denser medium (where the light starts)
• $n_2$ — refractive index of the less dense medium

:::tip TIR only works "denser to less dense". If a question sends light from air into glass, the critical angle idea does not apply — light always refracts into the denser medium. Check the direction before reaching for $\sin\theta_c = n_2/n_1$. :::

Practice question

Water has a refractive index of $1.33$. Find the critical angle for a water–air boundary.

Worked solution: $\sin\theta_c = 1.00/1.33 = 0.752$, so $\theta_c = 48.8°$.