K323 / 2027
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Lesson 3 of 5

Critical angle and optical fibres

  • I can explain critical angle and total internal reflection.Syllabus

    Syllabus K323, 12(f). Explain critical angle and total internal reflection.

  • I can explain optical fibres in telecommunications and medicine, including advantages.Syllabus

    Syllabus K323, 12(g). Explain optical fibres in telecommunications and medicine, including advantages.

Make a guess. You are not marked.

Light inside a glass block reaches the glass-air boundary at an angle of incidence of 60 degrees. The critical angle for this glass is 42 degrees. What happens?

  1. Most of it refracts out into the air, bending away from the normal at the boundary.
  2. All of it reflects back into the glass, because 60 degrees is above the critical angle.
  3. It refracts out along the boundary.
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All of it reflects back into the glass, because 60 degrees is above the critical angle.

The angle of incidence is larger than the critical angle, so no light can refract out. Total internal reflection sends the whole ray back into the glass.

Light travelling from a higher-index medium towards a lower-index one can be completely reflected at the boundary. The direction of travel and the incidence angle both matter.

In refraction, the transmitted ray bends away from the normal when it enters a lower-index medium. As the internal incidence angle increases, the refracted angle can reach 90°.

Distinguish the critical case from total internal reflection

The critical angle c is the incidence angle in the higher-index medium for which the refracted ray is at 90° to the normal: along the boundary.

For an incidence angle greater than c, total internal reflection occurs. At an ideal boundary all the incident light is reflected back into the higher-index medium; there is no transmitted propagating ray in the simple ray model.

These rays travel from n = 1.50 towards air

Angles are measured from the normal. Ray thickness does not give the fraction of light reflected or transmitted.

Below the critical angle

Below the critical angle at a material-to-air boundaryAt i = 30 degrees, a refracted ray enters air at 48.6 degrees to the normal. Some light is also reflected. The boundary is horizontal, air is above it and the index-1.50 material is below it. The incoming ray approaches from below left and the reflected ray leaves below right, at equal angles from the downward normal.AirNormalir48.6°Material: n = 1.50

i = 30°; reflected r = 30°.
Refraction into air: 48.6°.

At the critical angle

At the critical angle at a material-to-air boundaryAt i = 41.8 degrees, the ideal refracted ray runs along the boundary: its refraction angle is 90 degrees. The boundary is horizontal, air is above it and the index-1.50 material is below it. The incoming ray approaches from below left and the reflected ray leaves below right, at equal angles from the downward normal.AirNormalir90° to normalMaterial: n = 1.50

i = c = 41.8°. The refracted ray runs along the boundary.

Above the critical angle

Above the critical angle at a material-to-air boundaryAt i = 50 degrees, no refracted ray enters the air. Total internal reflection gives a reflection angle of 50 degrees. The boundary is horizontal, air is above it and the index-1.50 material is below it. The incoming ray approaches from below left and the reflected ray leaves below right, at equal angles from the downward normal.AirNormalirNo refractedray into airMaterial: n = 1.50

i = 50° > c. All the light is reflected at this ideal boundary, with r = 50°.

These three cases all start inside the n = 1.50 material and meet air. Below c a ray is transmitted, at c the refracted ray is along the boundary, and above c the light is totally internally reflected.

Critical-angle calculation

An n = 1.50 material meets air

At the critical case, the refraction relationship gives:

1.50 sin c = 1 x sin 90°, so sin c = 1/1.50 and c = 41.8°.

  • At internal i = 30°, the transmitted angle is 48.6°. Some reflection generally occurs as well.
  • At i = 41.8°, the ideal refracted ray is along the surface.
  • At i = 50°, total internal reflection occurs. The reflected angle is also 50°.

The special relationship sin c = 1/n here uses air on the other side. For another pair of media, retain both indices.

Explore

Push the angle past critical

air, n = 1.00glass, n = 1.50

Refraction out. Angle of incidence 30 degrees: the ray refracts out at 48.6 degrees, bent away from the normal. A weaker reflected ray also returns into the glass.

Try it step by step

Work it out in steps: sin c = 1 / n, then Critical angle c in degrees.

  1. Problem 1. Water has a refractive index of 1.33. Find the critical angle for a water-air boundary.
  2. Problem 2. Diamond has a refractive index of 2.42. Find the critical angle for a diamond-air boundary.
  3. Problem 3. A type of glass has a refractive index of 1.60. Find the critical angle for a glass-air boundary.
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Problem 1.

sin c = 1 / n: sin c = 1 / 1.33 = 0.752

Critical angle c in degrees: c = inverse sin of 0.752 = 48.8 degrees, which is 49 degrees to 2 sf

Problem 2.

sin c = 1 / n: sin c = 1 / 2.42 = 0.413

Critical angle c in degrees: c = inverse sin of 0.413 = 24.4 degrees, which is 24 degrees to 2 sf

Problem 3.

sin c = 1 / n: sin c = 1 / 1.60 = 0.625

Critical angle c in degrees: c = inverse sin of 0.625 = 38.7 degrees, which is 39 degrees to 2 sf

Both conditions are needed: light travels from higher to lower refractive index, and i is greater than c. A large angle for light entering glass from air does not satisfy the first condition. Equality is the critical case, not the above-critical case.

Observe the change with a semicircular block

Direct a narrow ray through the curved surface along a radius towards the centre of the flat face. It enters the curved surface normally, so it does not bend there. The flat face is the boundary being investigated.

Vary the ray's internal incidence at the flat face while keeping its path through the curved face radial. Measure i from the normal to the flat face at its centre. Observe the refracted ray approach the surface, identify the grazing case, then increase i to see total internal reflection.

A ray entering away from a radius can bend at the curved face, changing the angle that actually reaches the flat face. A broad ray also makes the boundary between the grazing and above-critical cases harder to locate. Use the actual ray path and a carefully centred protractor to estimate the critical angle.

Guide light along a fibre

An optical fibre has a higher-index core surrounded by lower-index cladding. A suitable ray inside the core meets their boundary above its critical angle and is guided by repeated total internal reflection.

A higher-index core can guide light

For this supplied model, ncore = 1.50 and ncladding = 1.40. The core/cladding critical angle is about 69.0°.

Two total internal reflections inside a straight optical-fibre coreA straight core of index 1.50 is bounded above and below by cladding of index 1.40. A ray inside the core travels right and reflects from the upper and then the lower core boundary. At each reflection it is 70 degrees to the local vertical normal, or 20 degrees to the horizontal fibre axis. This exceeds the 69.0 degree core-to-cladding critical angle. A dashed local normal and two equal angle arcs are drawn at the first reflection. The fibre continues beyond the cropped side edges; the final ray segment does not depict exit into air.NormalCladding: n = 1.4070°70°Core: n = 1.50At each bounce: 70° > 69.0°Fibre continues beyond the side edges

The critical angle belongs to this pair of materials. It is different from the 41.8° material-to-air example.

The supplied core and cladding indices are 1.50 and 1.40. Their critical angle is about 69.0°, so a core ray incident at 70° to the normal is totally internally reflected. This is a different boundary from the material-air example.

For this fibre, 1.50 sin c = 1.40 sin 90°, so c = sin-1(1.40/1.50) = 69.0°. The shown 70° incidence is above that value. Measure incidence to the local normal, not along the fibre wall.

Telecommunications
Light signals carry information along fibres. Low attenuation supports long links, and a high bandwidth allows a large information rate. The optical signal is not affected by electromagnetic interference in the same way as an electrical signal in a metal cable.
Medical fibrescope
Flexible fibres can deliver illumination through a small opening to a difficult-to-view region. In a traditional fibrescope, an ordered bundle carries an image back; keeping the fibres in corresponding positions preserves the image arrangement. Some modern endoscopes instead use a camera at the tip.

A fibre is not lossless merely because the ideal core boundary gives total internal reflection. Absorption, scattering and connections can reduce the transmitted signal. A sufficiently sharp bend can also change the incidence conditions and cause light to escape.

Watch out for this

A student says: "Total internal reflection can happen when light passes from air into glass at a large angle." What is wrong with this?

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Total internal reflection needs light travelling from a denser medium to a less dense one, such as glass to air, at an angle of incidence greater than the critical angle. Going from air into glass, light always refracts in.

Check your understanding

Light travels from a transparent material into air. Its critical angle at this boundary is 41.8 degrees. Which statement correctly compares internal incidence angles of 30, 41.8 and 50 degrees?

  1. Both 41.8 and 50 degrees give the same above-critical case.
  2. 30 degrees: transmitted; 41.8 degrees: critical case; 50 degrees: total internal reflection.
  3. 50 degrees gives a transmitted ray bent towards the normal because the angle is large.
Show the answer

30 degrees: transmitted; 41.8 degrees: critical case; 50 degrees: total internal reflection.

At the critical angle the refracted angle is 90 degrees to the normal. Total internal reflection requires an incidence angle greater than the critical angle, with light travelling from higher to lower refractive index.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

Define critical angle. [2]

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Exam-style question

The refractive index of a glass is 1.50. The speed of light in air is 3.0 x 108 m/s. Calculate the speed of light in the glass and the critical angle of the glass. [4]

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Exam-style question

A ray of light inside a glass block of refractive index 1.50 meets the glass-air boundary at an angle of incidence of 45°. Explain why it does not leave the glass. [3]

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