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E. Physics: Light and Geometric Optics · Light and Geometric Optics

Partial Reflection and Total Internal Reflection

E3.4 E2.1 E1.2 A1.4About 70 minutes
Learning goals
  • I can explain partial reflection and refraction and give examples.
  • I can state the conditions for total internal reflection.
  • I can calculate a critical angle using sin θc = n₂/n₁.
  • I can explain how optical fibres, prisms and diamonds use total internal reflection.

Words to know

partial reflection and refraction
When light meets a boundary, some is reflected and the rest is refracted.
critical angle (θc)
The angle of incidence in the slower medium that gives an angle of refraction of 90°.
total internal reflection (TIR)
The complete reflection of light back into a slower medium when θi is greater than the critical angle.
optical fibre
A thin strand of glass or plastic that carries light by total internal reflection.
cladding
The outer layer of an optical fibre, with a lower index of refraction than the core.
endoscope
A medical instrument that uses optical fibres to see inside the body.
anti-reflective coating
A thin layer on a lens that reduces partial reflection and glare.

A light that cannot escape

Shine a flashlight up through the side of an aquarium towards the water's surface. At a shallow angle, the light does not come out of the water at all: the underside of the surface shines like a perfect mirror. This is total internal reflection, and it is the reason you can stream a video over a thin glass fibre.

Partial reflection and refraction

Whenever light reaches a boundary between two transparent media, some light is reflected and the rest is refracted. This is called partial reflection and refraction.

  • Look through a window during the day and you see outside clearly. At night, with the lights on inside, the window shows your reflection: the same small amount of light is reflected as in the daytime, but now there is little light coming in from outside to hide it.
  • Each surface of an ordinary glass lens reflects a few percent of the light. This causes glare on eyeglasses and faint "ghost" images in cameras. Anti-reflective coatings on lenses reduce partial reflection, so more light passes through.
  • The larger the angle of incidence, the more light is reflected and the less is refracted.

From slow to fast: the critical angle

When light passes from a slower medium into a faster one (larger n to smaller n, such as water to air), it bends away from the normal, so θR > θi. As you increase θi, θR grows faster, until θR reaches 90°: the refracted ray skims along the surface.

The angle of incidence that gives θR = 90° is called the critical angle (θc). Using Snell's law with θ₂ = 90° (sin 90° = 1):

sin θc = n₂ / n₁ (only when n₁ > n₂)

If the angle of incidence is larger than θc, no light is refracted. All of it is reflected back into the first medium. This is total internal reflection (TIR).

The two conditions for total internal reflection:

  1. Light travels from a medium with a higher index of refraction into one with a lower index (from slower to faster).
  2. The angle of incidence is greater than the critical angle.
41.7°30°θi < θc: refracts (weak reflection)airwaterrefracted ray at 90°48.8°θi = θc = 48.8°airwater62°θi > θc: total internal reflectionairwater
Light in water meeting the surface. Left: below the critical angle, most light refracts out (bending away from the normal) and a little reflects. Middle: at θc = 48.8°, the refracted ray travels along the surface. Right: above θc, all the light reflects back into the water.
Worked example

Example 1: Critical angle for water

Find the critical angle for light travelling from water (n = 1.33) into air.

  1. sin θc = n₂ / n₁ = 1.00 / 1.33 = 0.752.
  2. θc = sin⁻¹(0.752) = 48.8°.

θc = 48.8°. Any ray in water hitting the surface at more than 48.8° is totally internally reflected.

BoundaryCritical angle
water → air48.8°
acrylic (1.49) → air42.2°
crown glass (1.52) → air41.1°
diamond (2.42) → air24.4°
crown glass → water61.0°
Critical angles. The bigger the difference in n, the smaller the critical angle.
Worked example

Example 2: Will the light escape?

A ray inside a glass block (n = 1.52) hits the glass–air surface at 45.0°. Does it leave the block? What if the block is under water (n = 1.33)?

  1. Glass to air: θc = sin⁻¹(1.00/1.52) = 41.1°. Since 45.0° > 41.1°, there is total internal reflection: no light leaves.
  2. Glass to water: θc = sin⁻¹(1.33/1.52) = sin⁻¹(0.875) = 61.0°. Now 45.0° < 61.0°, so the light refracts out (with partial reflection).
  3. Check with Snell's law: sin θ₂ = 1.52 sin 45.0° ÷ 1.33 = 1.075 ÷ 1.33 = 0.808, θ₂ = 53.9°: a real angle, so the light escapes.

In air the ray is totally internally reflected; under water it escapes at 53.9°. Total internal reflection depends on both media.

Technologies that use total internal reflection

Optical fibres. An optical fibre is a thin strand of very pure glass or plastic. A core with a higher index of refraction is surrounded by a cladding with a lower index. Light entering the end hits the core's wall at more than the critical angle, so it reflects again and again, travelling along the fibre even around gentle bends, with very little loss. Pulses of laser light in fibres carry most of the world's internet and phone traffic, including the cables under the oceans. Doctors use bundles of fibres in endoscopes to see inside the body without major surgery.

Prisms in optical instruments. A glass prism with angles of 45°, 45° and 90° can turn light through 90° or 180° by total internal reflection. Because θi = 45° is greater than glass's critical angle (41.1°), the reflection is total, giving a brighter image than a silvered mirror and never tarnishing. Binoculars, periscopes and some cameras use these prisms.

Diamonds. Diamond has a very small critical angle (24.4°), so light entering a cut diamond reflects many times inside before escaping through the top. Together with strong dispersion, this gives diamonds their sparkle.

Bicycle and road reflectors. Many plastic reflectors are made of tiny corner-shaped prisms that send light back towards its source using total internal reflection.

cladding, n = 1.46θi = 80°, more than θc = 73.8°Light enters the end and reflects off the walls by total internal reflection, even around gentle bends.15°core, n = 1.52
An optical fibre (drawn much wider than real; real fibres are about as thick as a hair). Inside the core the light travels at 10° to the axis, so it meets the wall at θi = 80°. That is more than the core-to-cladding critical angle (73.8°), so all of it reflects and stays in the core.
turns light 90°45°turns light 180°45°45°
Right-angle prisms. Left: light enters a short face straight on, meets the long face at 45° (more than θc = 41.1°) and is totally reflected, turning 90°, as in a periscope. Right: two total internal reflections turn light 180°, as in binoculars.
Just for fun

Why did the light ray refuse to leave the diamond?

It found the situation too critical.

Hands-on activity

Trap light in water

You need: a clear glass or straight-sided plastic container, water, a few drops of milk, a flashlight, and (only with a grown-up) a low-power laser pointer and a clear plastic bottle with a small hole near the bottom.

Part A: The silver ceiling. Fill the glass with water and hold it a little above eye level. Look up through the side wall at the underside of the water's surface. At a steep enough angle it looks like shiny silver. You are seeing total internal reflection.

Part B: The bending beam (grown-up needed). Over a sink, fill the bottle with water and let a smooth stream pour out of the hole. In a dark room, shine the laser through the bottle from the opposite side so the beam enters the stream at the hole. The light follows the curving stream, reflecting along its inside, and makes a bright spot where the stream lands.

Part C: Calculate. Calculate the critical angle for water, and explain why the light in the stream cannot escape.

What to look for: the underside of the surface looks mirror-like in part A; the stream glows and the spot of light moves with it in part B. The critical angle is 48.8°.

Safety (grown-up note): Never shine a laser at anyone's eyes, at pets, or out a window, and don't look into the beam or its reflections. Keep the laser and any electrical devices dry and away from the sink.

Your turn

1

State the two conditions needed for total internal reflection.

2

Define the critical angle.

3

Calculate the critical angle for acrylic (n = 1.49) to air.

4

Can total internal reflection happen when light goes from air into water? Explain.

5

Why do you see your reflection in a window at night but not in the day?

6

Give two technologies that use total internal reflection.

More practice: warm-up

1

What happens to light that hits a boundary at more than the critical angle (going from slower to faster)?

2

What is the critical angle for water to air?

3

What is the core of an optical fibre surrounded by?

4

What do anti-reflective coatings on eyeglasses reduce?

5

Which has a smaller critical angle in air: glass or diamond?

More practice: on your own

1

Find the critical angle for ice (n = 1.31) to air.

2

A material has a critical angle of 40.0° in air. Find its index of refraction.

3

A ray in glass (n = 1.52) hits the surface at 50.0°. Use Snell's law to show that it cannot refract into air.

4

Explain why a 45°–45°–90° glass prism works as a mirror in air but not under water.

5

Why are prisms better than silvered mirrors in high-quality binoculars?

Challenge

1

Light inside an optical fibre core (n = 1.52) travels at 25.0° to the fibre's axis. (a) What is the angle of incidence on the wall? (b) If the cladding has n = 1.46, is the light totally reflected?

2

Explain why diamonds sparkle more than glass replicas of the same cut.

3

A fish looks up at the water's surface. Explain why it sees the whole world above the water squeezed into a circle, with the rest of the surface looking like a mirror.

Check your understanding

Quiz

1.

Total internal reflection can occur when light travels from:

2.

The critical angle for water (n = 1.33) to air is:

3.

At the critical angle, the angle of refraction is:

4.

Optical fibres keep light inside the core by:

5.

Why do diamonds sparkle so much?

6.

A ray in glass (θc = 41.1°) hits the surface at 35°. What happens?

For parents and teachers

Big idea: Expectation E3.4 asks students to explain the conditions for partial reflection/refraction and for total internal reflection, with labelled ray diagrams. The key formula is sin θc = n₂/n₁ (only when light goes from higher to lower n). Expectation E1.2 is supported by technologies such as optical fibres, prisms and reflectors.

Common difficulties:

  • Direction. TIR happens only from slow to fast (e.g., water to air), never from air into water.
  • Calculator errors. If sin θ comes out greater than 1, that is not a mistake in arithmetic: it means total internal reflection.
  • Angles in fibres. The angle of incidence on the wall is measured from the normal to the wall, not from the axis.

How to help: Try the "silver ceiling" with a glass of water; it needs no equipment. The laser water-stream demonstration is memorable but needs close supervision.