Critical angle

Refraction sets the stage. When light travels from a denser medium into a rarer one — glass into air, or water into air — it bends away from the normal, so the angle of refraction $r$ is always larger than the angle of incidence $i$. As $i$ grows, $r$ grows faster.

The critical angle is where $r$ hits $90°$. At one particular value of $i$, called the critical angle $C$, the refracted ray no longer leaves the surface at an angle — it grazes exactly along the boundary itself:

Snell's law pins down the formula. For light going from the denser medium (refractive index $\mu$, relative to the rarer medium) into the rarer medium, Snell's law at $i = C$ reads $\mu \sin C = 1 \times \sin 90°$, so

where: $C$ is the critical angle, and $\mu$ is the refractive index of the denser medium with respect to the rarer one (for a glass-air surface, $\mu$ is simply the refractive index of glass, since air's index is taken as $1$).

Beyond $C$, refraction breaks down entirely. If $i$ is pushed past $C$, there is no possible angle of refraction — sine cannot exceed $1$ — so light cannot leave the surface at all. Every bit of it is reflected back into the denser medium instead: this is total internal reflection, picked up in the next topic. Because $\sin C = 1/\mu$, a larger $\mu$ forces a smaller $C$: denser media trap light more easily. For glass ($\mu \approx 1.5$), $C \approx 41.8°$; for water ($\mu \approx 1.33$), $C \approx 48.8°$; for diamond ($\mu \approx 2.42$), $C \approx 24.4°$ — diamond's tiny critical angle means light entering it is very easily trapped and bounced around inside many times before it finally escapes, which is a major reason diamonds sparkle so brilliantly.

A ray of light travels from glass ($\mu = 1.5$) toward a glass-air boundary. Find the critical angle for this boundary, and state what happens to a ray inside the glass striking the boundary at $50°$ to the normal.

Step 1. Apply $\sin C = 1/\mu$ with $\mu = 1.5$: $\sin C = 1/1.5 = 0.667$.

Step 2. Take the inverse sine: $C = \sin^{-1}(0.667) \approx 41.8°$.

Step 3. Compare the given angle of incidence, $50°$, with $C = 41.8°$. Since $50° > C$, the angle of incidence exceeds the critical angle.

Step 4. Apply the physical rule: whenever $i > C$ at a denser-to-rarer boundary, refraction is impossible and the light undergoes total internal reflection instead. So the critical angle is $41.8°$, and the ray striking at $50°$ is totally internally reflected back into the glass rather than escaping into the air.

Competitive stretch

A transparent liquid has a critical angle of $45°$ with respect to air. Find its refractive index and the speed of light inside it (take $c = 3\times10^8\ \text{m/s}$).

Step 1. Use $\mu = 1/\sin C$ with $C = 45°$: $\mu = 1/\sin 45° = 1/0.707 \approx 1.41$.

Step 2. Use $\mu = c/v$ to find the speed: $v = c/\mu = (3\times10^8)/1.41 \approx 2.13\times10^8\ \text{m/s}$. The liquid's refractive index is about $1.41$, and light travels through it at about $2.13\times10^8\ \text{m/s}$ — consistent with this liquid sitting between water ($\mu \approx 1.33$) and glass ($\mu \approx 1.5$) in density.

• Confusing the critical angle with a general angle of incidence — $C$ is one specific value, the one at which $r$ becomes exactly $90°$, not a range or an approximation.
• Writing $\sin C = \mu$ instead of $\sin C = 1/\mu$ — a very common inversion error that gives an impossible answer whenever $\mu > 1$.
• Forgetting that there is no critical angle going the other way — light entering a denser medium from a rarer one always has $r < i$, so $r$ never reaches $90°$ and this relation simply does not apply.
• Assuming total internal reflection happens at any denser-to-rarer boundary regardless of angle — it only happens once $i$ exceeds $C$; at smaller angles, the light still refracts out as usual, just partly reflected too.

Ask the tutor: Want to see how the critical angle you just calculated leads straight into total internal reflection diagrams for prisms, periscopes and optical fibres?