Reflection refraction and total internal reflection

Light is an electromagnetic wave: in vacuum it travels at $c \approx 3\cdot 10^{8}$ m/s and more slowly in a transparent medium, which is why the medium is described by the refractive index $n = c/v$.

At the boundary between two media part of the light is reflected and part is transmitted into the second medium with a changed direction — this is refraction; all angles are measured from the normal, the perpendicular to the surface at the point of incidence.

Law of reflection: the angle of reflection equals the angle of incidence, $\theta_r = \theta_i$, and the incident ray, the reflected ray and the normal lie in one plane.

Snell's law relates the angles to the refractive indices: $n_1\sin\theta_1 = n_2\sin\theta_2.$

When light travels from a denser medium into a less dense one ($n_1 > n_2$), the refracted ray bends further away from the normal than the incident ray, and at the critical angle $\sin\theta_c = \frac{n_2}{n_1}$ it grazes along the boundary.

For $\theta_1 > \theta_c$ there is no refracted ray: the light returns entirely into the first medium — this is total internal reflection, which requires $n_1 > n_2$.

A ray of light strikes a plane mirror at an angle $\theta_i = 30^\circ$ to the normal. Find the angle of reflection $\theta_r$ and the angle between the incident ray and the mirror surface.

Light passes from air ($n_1 = 1{,}00$) into glass ($n_2 = 1{,}50$) and strikes the boundary at $\theta_1 = 40^\circ$ to the normal. Find the angle of refraction $\theta_2$ and decide whether the ray bent towards or away from the normal.

An optical fibre consists of a core ($n_1 = 1{,}50$) and a cladding ($n_2 = 1{,}45$). (a) Find the critical angle at the core–cladding boundary. (b) Explain why a ray striking this boundary at $80^\circ$ stays inside the core.

Self-check in a simulation (safe, no laser).

1. Open a geometrical-optics simulation (for example PhET's "Bending Light") in the refraction mode.
2. Put air on top ($n = 1{,}00$) and glass below ($n = 1{,}50$). Vary the angle of incidence from $10^\circ$ to $70^\circ$ and record the pairs $(\theta_1, \theta_2)$ in a table.
3. For each pair compute $n_1\sin\theta_1$ and $n_2\sin\theta_2$ and check that Snell's law holds within the reading error of the on-screen angles.
4. Swap the media: glass on top, air below. Increase $\theta_1$ until the refracted ray disappears and compare the measured critical angle with $\arcsin(n_2/n_1)$.
5. Household observation: put a pencil in a glass of water — it looks broken at the water–air boundary. That is refraction; carry out total-internal-reflection measurements only in the simulation and never point a laser pointer at eyes or at mirror-like surfaces.

To show the geometry of reflection and refraction at a boundary between two media and the threshold at which the refracted ray disappears and only total internal reflection remains.

Diagram (cross-section) and two graphs.

On the left: a horizontal boundary between two media, a vertical dashed normal, an incident ray at angle $\theta_1$ to the normal, a reflected ray at $\theta_r = \theta_1$ and a refracted ray at $\theta_2 < \theta_1$ (transition into the denser medium); the angles are marked with arcs at the normal and $n_1$, $n_2$ are labelled.

On the right: two graphs of $\theta_2$ versus $\theta_1$ — for $n_1 < n_2$ the curve rises smoothly and stays below $\theta_1$; for $n_1 > n_2$ the curve stops at $\theta_c$ (a dashed vertical labelled "$\sin\theta_c = n_2/n_1$"), and beyond it only the reflected branch is drawn, marked "total internal reflection".

An inset below shows the core and cladding of an optical fibre with a zigzag ray reflecting at the boundary at an angle greater than $\theta_c$.

On a summer afternoon the asphalt of a road heats up to about $60\,^\circ\mathrm{C}$, while the air a metre above it stays near $30\,^\circ\mathrm{C}$. Sunlight arrives from the sky at only a few degrees to the road surface. To a driver $15\ \mathrm{m}$ away the road looks flooded with a bright "puddle", but it vanishes as soon as the car reaches it.

The hot air near the asphalt is less dense, so its refractive index is smaller: between air at $60\,^\circ\mathrm{C}$ and air at $30\,^\circ\mathrm{C}$ the difference is only $\Delta n \approx 2\times10^{-5}$. At each thin layer refraction obeys $n_1\sin\theta_1 = n_2\sin\theta_2$, and total internal reflection needs an angle larger than the critical angle $\theta_c = \arcsin(n_2/n_1)$, which for those two layers is $89.6^\circ$. Only rays within about $0.4^\circ$ of grazing the road are turned back; because the layers form a continuous gradient the ray curves upwards gradually, and the eye traces it straight and sees a bright "puddle".

The optical fibre that brings internet to a house has a glass core with $n_1 = 1.48$ surrounded by cladding with $n_2 = 1.46$. Inside the core light travels at $v = c/n_1 = 3.00\times10^8/1.48 = 2.03\times10^8\ \mathrm{m\,s^{-1}}$. A laser at $1550\ \mathrm{nm}$ sends pulses lasting $10^{-10}\ \mathrm{s}$ along a $50\ \mathrm{km}$ span between two amplifiers, and about $0.2\ \mathrm{dB}$ is lost per kilometre.

The critical angle at the core–cladding boundary is $\theta_c = \arcsin(1.46/1.48) = 80.6^\circ$: any ray that hits the wall at a larger angle to the normal is totally internally reflected and cannot leave the core. The numerical aperture $\mathrm{NA} = \sqrt{n_1^2-n_2^2} = 0.24$ fixes the acceptance half-angle $\arcsin(0.24) = 14^\circ$ — rays launched at a larger angle leak into the cladding.

A pool is filled with water of refractive index $n = 1.33$ to a depth of $2.0\ \mathrm{m}$. Standing at the edge and looking almost straight down, you see the floor at an apparent depth of only $2.0/1.33 = 1.5\ \mathrm{m}$. A swimmer below the surface wearing goggles looks up and sees the whole sky squeezed into a bright cone of half-angle $48.8^\circ$, while everything outside that cone is a mirror image of the pool itself.

The apparent depth follows from Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ applied to near-vertical rays coming from the floor. The edge of the cone is the critical angle for the water–air boundary: $\theta_c = \arcsin(1.00/1.33) = 48.8^\circ$. At angles greater than $48.8^\circ$ from the vertical, light from the air cannot enter the water at all, and the water surface becomes a perfect mirror of the pool bottom by total internal reflection.