Total internal reflection prisms

A right-angled prism cut at exactly $45°$ , $45°$ and $90°$ can turn a beam of light through a clean $90°$ or a full $180°$ using nothing but glass — no silvering, no metal backing, no glue-on mirror. The trick is total internal reflection (TIR): inside the glass, light striking the slanted hypotenuse face at $45°$ finds that angle far steeper than the critical angle of ordinary glass, so it cannot escape and bounces back with essentially no loss. This one geometric fact — a $45°$ prism angle beating a roughly $42°$ critical angle — is why periscopes, binoculars and optical fibres all favour prisms over mirrors: brighter images, no ghosting, and a compact fold in the light path. This article grounds that fact in the critical-angle formula, walks the ray through both prism arrangements, and works a full numeric example.

The core insight

A $45°$ - $45°$ - $90°$ glass prism totally internally reflects light striking its hypotenuse because $45°$ exceeds glass's critical angle of about $42°$ , giving brighter, ghost-free $90°$ or $180°$ deviations without any mirror coating.

The core idea

Total internal reflection (TIR) happens only when both conditions below are met together — dropping either one means the light simply refracts out instead of reflecting.

Light must travel from an optically denser medium toward a rarer one — inside the glass prism, heading toward the glass-air surface.
The angle of incidence at that surface must be greater than the critical angle $C$ , the special angle at which the refracted ray would graze along the surface at exactly $90°$ .

The critical angle itself comes straight out of Snell's law, applied at the one angle where the refracted ray bends all the way to $90°$ :

\sin C = \frac{1}{n}

where: $C$ is the critical angle for the denser medium, and $n$ is its refractive index relative to the rarer medium (air, taken as $n = 1$ ). For ordinary crown glass, $n \approx 1.50$ , which gives $C \approx 41.8°$ — comfortably below $45°$ . That single number is what makes the totally reflecting prism work. Cut as an isosceles right triangle ( $45°$ , $45°$ , $90°$ ), the prism sends light onto its hypotenuse at exactly $45°$ of incidence — a geometric consequence of the triangle's shape, not a choice the light makes. Since $45° > C$ , the ray cannot refract out; it reflects totally, as if the hypotenuse were a perfect mirror. Two standard arrangements exploit this:

90° deviation prism: light enters normally (straight in, no bending) through one short face, travels to the hypotenuse, strikes it at $45°$ , undergoes TIR, and emerges normally through the other short face — turned through exactly $90°$ .
180° deviation prism: light enters normally through the hypotenuse itself, strikes each of the two short faces in turn at $45°$ , reflects twice by TIR, and emerges parallel to but reversed from its original direction — a $180°$ turn.

Against an ordinary silvered mirror, the prism wins on two counts: the image is brighter, since TIR reflects essentially all the light with no absorption by a metal backing, and it is free of ghost images, since a back-silvered mirror also gives a faint partial reflection off its front glass surface, while a TIR prism has only the one genuine reflecting interface.

Figure

Ray path through a 45-45-90 prism showing normal entry, a 45 degree strike on the hypotenuse producing total internal reflection, and normal exit turned through 90 degrees

Applications (name and one-line use each, as prescribed by the syllabus): periscope — two such prisms let an observer see over an obstacle, as in a submarine; binoculars — pairs of prisms fold the light path so the instrument stays compact while keeping a long optical path; optical fibres — repeated TIR along a thin glass core carries light signals over long distances with very little loss, used in telecommunications and endoscopy.

Derivation AWhy 45 degrees guarantees total internal reflection in a glass prism

The goal is to show, starting only from Snell's law at the critical angle, why an ordinary $45°$ - $45°$ - $90°$ glass prism reliably totally-internally-reflects a ray striking its hypotenuse, and to compute the numeric margin by which $45°$ clears the critical angle for glass of refractive index $n = 1.50$ — the same margin that makes prisms a dependable, non-negotiable design choice rather than a lucky coincidence.

Step 1. Start from Snell's law at the glass-air boundary, written for a ray inside the glass (denser, index $n$ ) meeting air (rarer, index $1$ ) at angle of incidence $i$ : $n \sin i = 1 \times \sin r$ , where $r$ is the angle of refraction in air, always larger than $i$ since light bends away from the normal on entering the rarer medium.

Step 2. As $i$ increases, $r$ increases faster and reaches $90°$ (the refracted ray grazing along the surface) before $i$ does. Define the critical angle $C$ as the particular value of $i$ at which $r$ reaches exactly $90°$ . Substituting $r = 90°$ gives $n \sin C = \sin 90° = 1$ , so $\sin C = 1/n$ .

Step 3. Insert the refractive index of ordinary crown glass, $n = 1.50$ : $\sin C = 1/1.50 = 0.6667$ , so $C = \sin^{-1}(0.6667) = 41.8°$ .

Step 4. Compare with the prism's fixed geometric angle of incidence at the hypotenuse, $45°$ . Since $45° > 41.8° = C$ , and the ray is travelling from glass (denser) to air (rarer), both TIR conditions are satisfied — not narrowly, but with a margin of about $3.2°$ to spare, which is why the prism performs reliably even allowing for small manufacturing tolerances in the glass or the cut angle. Result: $\sin C = 1/n = 0.6667 \Rightarrow C = 41.8°$ ; since $45° > C$ , TIR at the hypotenuse is guaranteed.

Worked example

A ray of light enters a $45°$ - $45°$ - $90°$ glass prism (refractive index $n = 1.50$ ) perpendicular to one of its shorter faces, exactly as in a periscope. Find the critical angle for the glass, confirm that TIR occurs at the hypotenuse, and state the total deviation of the ray on emerging from the prism.

Step 1. Find the critical angle from $\sin C = 1/n = 1/1.50 = 0.6667$ , giving $C = 41.8°$ .

Step 2. Trace the ray. It enters perpendicular to the short face, so $i = 0°$ there and it passes straight through without bending. It travels on and strikes the hypotenuse at $45°$ to the normal — fixed by the prism's shape, not by the entry angle.

Step 3. Compare $45°$ with $C = 41.8°$ . Since $45° > 41.8°$ , the angle-of-incidence condition for TIR is satisfied, and the ray is travelling from the denser glass toward the rarer air, satisfying the direction condition too — so TIR occurs at the hypotenuse.

Step 4. After reflecting, the ray travels to the second short face and strikes it perpendicularly ( $i = 0°$ again), so it emerges without further bending. The overall deviation between the entering ray and the emerging ray is $90°$ , and essentially all the light is preserved since the reflection is total.

Competitive stretch

If the same prism were instead used in its $180°$ arrangement (light entering normally through the hypotenuse), the ray would strike each short face at $45°$ in turn. Since $45° > 41.8° = C$ at both faces, TIR occurs twice, and the ray emerges parallel to but reversed from its original direction — a total deviation of $180°$ , with the same near-total brightness preserved through two reflections instead of one.

Common traps

Thinking TIR prisms are silvered like a mirror — there is no reflective coating at all; the reflection is entirely due to TIR at the plain glass-air surface.
Applying only one TIR condition — both "denser to rarer" and "angle greater than $C$ " must hold; stating just one is an incomplete answer.
Confusing which faces carry light in the two arrangements — in the $90°$ prism, light enters and exits through the two short faces; in the $180°$ prism, it enters and exits through the hypotenuse.
Assuming any prism angle works — $45°$ is chosen specifically because it exceeds the roughly $42°$ critical angle of glass; a shallower angle could fail to produce TIR.

Ask the tutor: Want to sketch the ray diagrams for a periscope and a pair of binocular prisms side by side and check the deviations yourself?

Why is 45 degrees special — would any prism angle greater than the critical angle work just as well

The $45°$ - $45°$ - $90°$ shape is chosen because it is easy to cut precisely and gives clean $90°$ or $180°$ deviations with light entering and leaving perpendicular to a face, so there is no extra bending to account for. Other angles above the critical angle would also produce TIR, but the light would enter or leave at some slanted angle, refracting at those faces and complicating the ray path and the final image orientation. The $45°$ prism keeps the geometry simple while still comfortably clearing glass's roughly $42°$ critical angle.

Does a totally reflecting prism ever lose any light at all

In the idealised picture used at this level, no — TIR reflects effectively $100\%$ of the incident light, unlike a silvered mirror, which loses some light to absorption in the metal coating and to partial reflection at its front glass surface. In practice, real prisms have tiny losses from surface imperfections, dust or slight absorption within the glass itself, but these are negligible compared to a mirror's losses, which is exactly why prisms are preferred in precision optical instruments like binoculars and periscopes.

What happens if the glass used has a critical angle greater than 45 degrees

Then the prism would fail. If $C \geq 45°$ , a ray striking the hypotenuse at $45°$ would no longer satisfy $i > C$ , so it would partially or wholly refract out through the hypotenuse instead of reflecting — the device would leak light and the periscope or binocular image would dim or vanish. This is exactly why ordinary crown glass, with $C \approx 41.8°$ , safely below $45°$ , is the standard choice for these prisms; a glass with a much higher refractive index would push $C$ even lower and give an even larger safety margin.

Total internal reflection prisms

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads