Refraction and total internal reflection

Refraction. When a ray of light passes from one transparent medium into another, its speed changes; if it strikes the boundary at an angle to the normal, this change of speed bends its path. The refractive index of medium 2 relative to medium 1 compares the angle of incidence $i$ in medium 1 with the angle of refraction $r$ in medium 2 (both measured from the normal to the surface):

where: $i$ is the angle of incidence, $r$ is the angle of refraction, and $v_1,v_2$ are the speeds of light in medium 1 and medium 2. This is Snell's law. The absolute refractive index of a single medium is $n=c/v$, the ratio of the speed of light in vacuum to the speed in that medium — a denser medium (higher $n$) always slows light down more.

Lateral displacement through a slab. A ray crossing a parallel-sided rectangular glass slab bends once entering and bends back by an equal amount leaving, so it emerges parallel to the original ray but shifted sideways. For a slab of thickness $t$ and refractive index $n$, with $i$ the angle of incidence and $r$ the angle of refraction inside the slab:

where: $d$ is the perpendicular lateral shift between the incident ray's line and the emergent ray's line, $t$ is the slab thickness, and $i-r$ is the net bending the ray undergoes on entry (it undoes exactly that bend on exit, which is why the emergent ray is parallel to the incident one).

Total internal reflection. TIR happens only when light travels from a denser medium toward a rarer one (glass to air, water to air) and strikes the boundary steeply enough. As the angle of incidence in the denser medium increases, the refracted ray bends further away from the normal and grows dimmer while the reflected ray grows brighter. At one particular angle, the critical angle $\theta_c$, the refracted ray grazes along the boundary itself, at $r=90°$; past that angle no refracted ray exists at all, and essentially all the light is reflected back into the denser medium. Two conditions must both hold for TIR: light must be going from denser to rarer, and the angle of incidence must exceed $\theta_c$. The figure below shows the three regimes at a single boundary: a shallow ray that partly refracts and partly reflects, the critical ray whose refracted beam grazes the surface, and a steep ray beyond $\theta_c$ that is totally reflected with no refracted beam at all.

Applications. A 45°–45°–90° right-angled prism turns light through 90° or 180° by TIR at its hypotenuse face, since 45° exceeds the critical angle (about 42° for ordinary glass); such totally reflecting prisms replace mirrors in periscopes and binoculars because TIR reflects essentially 100% of the light with no silvering to tarnish. Optical fibres guide light along a thin core of higher refractive index surrounded by cladding of lower refractive index; a ray entering within the fibre's acceptance cone strikes the core–cladding boundary beyond the critical angle every time, undergoes repeated TIR, and travels the full length of the fibre with very little loss.

A ray of light is incident on a rectangular glass slab of refractive index 1.5 and thickness 6 cm at an angle of incidence of 60°.

Step 1. Find the angle of refraction inside the slab using Snell's law, $n=\sin i/\sin r$, rearranged as $\sin r=\sin i/n$.

Step 2. Substitute into the lateral displacement formula with $t=6$ cm, $i=60°$, $r=35.3°$.

Step 3. Check whether this same glass could support TIR. Its critical angle is $\theta_c=\sin^{-1}(1/1.5)\approx41.8°$. Since this ray meets the far (glass-to-air) face of the slab travelling almost straight through (angle of incidence there equals $r\approx35.3°$, well below $\theta_c$), it refracts out normally rather than reflecting — TIR is not triggered inside this slab at this geometry.

Answer. The ray refracts to about 35.3° inside the slab and emerges parallel to the incident ray but laterally shifted by about 3.1 cm.

Competitive stretch

The critical angle for a certain glass–air interface is 42°. A ray inside the glass strikes the surface at 50°. Since $n=1/\sin42°\approx1.495$ and the incidence angle (50°) exceeds $\theta_c$ (42°) while travelling from the denser glass toward air, the ray cannot refract out at all — it undergoes total internal reflection, leaving the surface at 50° to the normal on the reflected side, obeying the ordinary law of reflection.

• Running TIR in the wrong direction. TIR only ever happens going from a denser medium into a rarer one; a ray travelling from air into glass never totally internally reflects, whatever the angle.
• Treating the critical angle as "the angle of maximum bending." $\theta_c$ is specifically the angle of incidence at which the refracted ray grazes the boundary at exactly 90°, not some vague largest-bend angle.
• Forgetting the emergent ray from a slab is offset, not superimposed. It is parallel to the incident ray but shifted sideways by $d$; it does not return to the same line.
• Mixing up the direction of $n_{21}$. $n_{21}=\sin i/\sin r$ is defined for light travelling from medium 1 into medium 2; reversing the direction gives $n_{12}=1/n_{21}$.

Ask the tutor: want to see how the lateral-displacement formula is derived step by step from the slab geometry, or work through finding the maximum acceptance angle for light entering an optical fibre?