Total internal reflection happens when light in a denser material meets a boundary with a less dense material at an angle of incidence greater than the critical angle. Then none of the light is refracted out, and all of it reflects back inside.
It builds on refraction and the refractive index, so learn that first. It is the idea behind optical fibres, reflecting prisms and bicycle reflectors, and it belongs to light and imaging.
What are the two conditions, and why are both needed?
Condition 1: the light is travelling from a denser material toward a less dense one, for example glass to air or water to air. Light going from air into glass bends toward the normal and can never be totally reflected at that surface, because there is no angle at which it fails to enter.
Condition 2: the angle of incidence is greater than the critical angle, c. Below c, part of the light refracts out (while a little also reflects).
At exactly c, the refracted ray runs along the boundary at 90° to the normal. Above c, no refracted ray can exist, so all of the light reflects.
How do I find the critical angle?
At the critical angle, the angle of refraction in the less dense material (air) is 90°. Using n = sin i ÷ sin r with the ray going from air into the material, the critical angle in the material satisfies sin c = 1 ÷ n, where n is the refractive index of the dense material.
For a denser material with a larger n, the critical angle is smaller, so light is trapped more easily.
How do I explain total internal reflection step by step?
- Name the direction: state that the light is travelling in the denser material toward a less dense one.
- Give the critical angle for that material, or find it from sin c = 1 ÷ n.
- Compare the angle of incidence with c: if it is greater, say so.
- State the result: all the light is reflected, and the angle of reflection equals the angle of incidence.
- Draw it: incident ray, normal, reflected ray with arrows, and i labelled.
Worked example
Glass has a refractive index of 1.5. (Invented example data.) A ray inside the glass meets a glass-to-air boundary at 45°.
Explain what happens. Then find the path of a second ray that meets the same boundary at 30°.
Step 1, critical angle: sin c = 1 ÷ 1.5 = 0.6667. c = sin⁻¹(0.6667) = 41.8°.
Step 2, compare, first ray: the ray is in glass heading for air, which is the dense-to-less-dense direction. 45° is greater than 41.8°, so both conditions hold.
Step 3, conclude: the ray undergoes total internal reflection. It reflects back into the glass at an angle of reflection of 45°.
Step 4, second ray: 30° is less than 41.8°, so the ray refracts out. Using sin r = n × sin i = 1.5 × sin 30° = 1.5 × 0.5 = 0.75, r = sin⁻¹(0.75) = 48.6°.
Step 5, check: in the second case the ray bends away from the normal (48.6° is greater than 30°), as it must when leaving glass. The critical angle is below 45° but above 30°, so the first ray is trapped and the second escapes.
Where is total internal reflection used?
In an optical fibre, light enters one end and meets the wall at angles greater than the critical angle. It reflects along the fibre again and again, with very little escaping at the sides. Fibres carry telephone, internet and cable signals, and are used in endoscopes.
In a simple reflecting prism, a right-angled glass prism with two 45° angles reflects light through 90° or 180° without a mirror. The light hits an internal face at 45°, which is greater than c = 41.8° for n = 1.5. That is how a periscope turns light through the tube.
The mistake to watch for
Students often say total internal reflection occurs “whenever the angle is large”, or draw it for light going into glass.
Mistaken answer: “Light in air hits the glass at 60°, which is greater than the critical angle, so it is totally internally reflected.”
The light is travelling from the less dense material to the denser one, so condition 1 fails.
The correction is to state both conditions in order. Ask: which material is the ray in, and is it heading for a less dense one?
Only then compare the angle with c. Also avoid saying the ray “reflects at the critical angle”. At c the refracted ray grazes the surface; the trapping starts above c.
Check yourself
Try these, then open each answer.
1. Find the critical angle for a plastic with refractive index 1.6.
Show answer
sin c = 1 ÷ 1.6 = 0.625. c = sin⁻¹(0.625) = 38.7°, so c ≈ 39°.
2. Can light travelling in air undergo total internal reflection at the surface of a glass block? Explain in one sentence.
Show answer
No. Total internal reflection needs light to travel from a denser to a less dense material, and going from air to glass is the other way round, so the ray always refracts into the glass.
3. A ray in glass (c = 41.8°) meets the glass-to-air boundary at 38°. Does total internal reflection occur?
Show answer
No. The direction is right, but 38° is less than the critical angle 41.8°, so most of the light refracts out into the air (with only a weak reflected ray).
Where this leads next
Now move to images made by lenses, starting with drawing a simple lens image. If you want to see how refraction and reflection combine in questions, the light and imaging practice set has mixed examples.
Some students can recite both conditions but mix them up when a diagram shows light crossing two boundaries. Our teachers work through exactly that kind of diagram in online one-to-one Physics tuition.