K326 / K327 / 2027
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Lesson 2 of 4

Refraction and refractive index

  • I can use the normal and incidence and refraction angles at a boundary.Syllabus

    Syllabus K326 / K327, 11(c). Use the normal and incidence and refraction angles at a boundary.

  • I can apply the constant sine ratio to new situations and calculations.Syllabus

    Syllabus K326 / K327, 11(d). Apply the constant sine ratio to new situations and calculations.

  • I can define refractive index as the ratio of vacuum light speed to speed in the medium.Syllabus

    Syllabus K326 / K327, 11(e). Define refractive index as the ratio of vacuum light speed to speed in the medium.

Make a guess. You are not marked.

A ray of light travels from air into a glass block along the normal, so that the angle of incidence is 0 degrees. What happens to the ray as it enters the glass?

  1. It slows down but does not change direction.
  2. It bends towards the normal, because glass is optically denser than air.
  3. It reflects straight back.
Show the answer

It slows down but does not change direction.

Refraction bends a ray only when it meets the boundary at an angle. Along the normal, sin i = 0, so sin r = 0 and the ray carries straight on, although it still slows down in the glass.

When light crosses a boundary between media, its speed can change. At oblique incidence, the transmitted ray then changes direction.

Draw the normal at the particular point where the ray crosses the surface. The angle of incidence i is between the arriving ray and this normal; the angle of refraction r is between the transmitted ray and the normal in the new medium.

Connect refractive index to speed

Refractive index   n = c/vc is the speed of light in vacuum; v is its speed in the named medium. The ratio has no unit.

For the same light, a higher refractive index means a lower speed. It does not mean the material must have a greater mass density. When light enters a higher-index medium obliquely, it bends towards the normal. Entering a lower-index medium bends it away from the normal.

Speed and index

Light travels at 2.0 x 108 m/s in a material

Using the supplied vacuum speed 3.0 x 108 m/s:

n = (3.0 x 108)/(2.0 x 108) = 1.50.

Keep both speeds in the same units. The index is 1.50, not 1.50 m/s.

Use the sines of the angles

For a fixed pair of media and the same light, sin i / sin r is constant, with the direction of travel specified. If the first medium has index n1 and the second has index n2, the constant is n2/n1.

sin i / sin r = n2/n1Equivalently, n1 sin i = n2 sin r. For air into a material, taking nair as 1 gives n = sin i / sin r.

Entering the material

From air into n = 1.50 at i = 45°

  1. Choose the direction: air is the first medium; the material is the second.
  2. Rearrange: sin r = sin 45°/1.50 = 0.4714.
  3. Use inverse sine in degree mode: r = 28.1°.

The smaller angle agrees with bending towards the normal. Dividing 45° directly by 1.50 would not apply the sine relationship.

Calculated air-to-material predictions for n = 1.50
Incidence isin r = sin i / 1.50Refraction r
30°0.333319.5°
45°0.471428.1°
60°0.577435.3°

Follow both faces of a block

Apply refraction at each face

Air is modelled with n = 1.00 and the block with n = 1.50. The dashed lines are local normals, not rays.

Oblique entry: the emergent ray is parallel but displaced

A ray refracts twice through a parallel-sided blockA ray enters the upper face from air at 45 degrees to the downward normal, refracting to 28.1 degrees inside an index-1.50 block. It travels to the lower parallel face, where its internal incidence is 28.1 degrees. On re-entering air it refracts to 45 degrees. The two external rays are parallel but laterally displaced. Each angle is measured from the normal at its own point of incidence.irirAirBlockn = 1.50Air

At entry: i = 45°, r = 28.1°.
At exit: i = 28.1°, r = 45°.

Normal incidence: speed changes, direction does not

A ray along the normal passes straight through the blockA vertical ray enters and leaves the horizontal parallel faces of the same block at zero incidence angle. It travels along each surface normal and remains straight. Its speed is lower in the block and increases again in air. The superimposed normal and ray do not imply that the speed is unchanged.AirBlockn = 1.50AirRay followsthe normal
At oblique entry into the block the ray bends towards the local normal; at exit into air it bends away. Parallel faces give a parallel but sideways-displaced emergent ray. The separate normal-incidence path is undeviated.

For a parallel-sided block with air on both sides, the emergent ray is parallel to the incident ray. Its sideways displacement remains: being parallel does not mean following the same line. Each face has its own normal at the crossing point.

At normal incidence, the ray travels along the normal. Its speed changes but its direction does not. Substituting i = r = 0 into the sine ratio would give 0/0, so this case cannot determine the index from that ratio.

Leaving the material

From n = 1.50 into air at internal i = 30°

Now n1 = 1.50 and n2 = 1:

sin r = 1.50 sin 30° = 0.750; r = 48.6°.

The transmitted ray bends away from the normal. The material's refractive index remains 1.50; reversing the ray reverses the direction-dependent sine ratio, not the definition n = c/v.

Try it step by step

Work it out in steps: sin i, then sin r = sin i / n, then Angle of refraction r in degrees.

  1. Problem 1. A ray enters water (n = 1.33) from air at an angle of incidence of 30 degrees. Find the angle of refraction.
  2. Problem 2. A ray enters glass (n = 1.50) from air at an angle of incidence of 50 degrees. Find the angle of refraction.
  3. Problem 3. A ray enters water (n = 1.33) from air at an angle of incidence of 40 degrees. Find the angle of refraction.
Show the answer

Problem 1.

sin i: sin 30 = 0.500

sin r = sin i / n: sin r = 0.500 / 1.33 = 0.376

Angle of refraction r in degrees: r = inverse sin of 0.376 = 22.1 degrees, which is 22 degrees to 2 sf

Problem 2.

sin i: sin 50 = 0.766

sin r = sin i / n: sin r = 0.766 / 1.50 = 0.511

Angle of refraction r in degrees: r = inverse sin of 0.511 = 30.7 degrees, which is 31 degrees to 2 sf

Problem 3.

sin i: sin 40 = 0.643

sin r = sin i / n: sin r = 0.643 / 1.33 = 0.483

Angle of refraction r in degrees: r = inverse sin of 0.483 = 28.9 degrees, which is 29 degrees to 2 sf

Measure refraction through a glass block

  1. Trace the block on paper and keep it on that outline. Direct a narrow ray obliquely at one face.
  2. Mark entry and exit points, plus well-separated points along the incident and emergent rays.
  3. Remove the block. Join entry to exit to draw the internal ray, and draw each external ray from its marks.
  4. Construct a normal at the entry point and use a protractor centred there to measure i and r from the normal.
  5. Repeat for several nonzero incident angles. Calculate sin i / sin r and compare the values.

A pin-tracing method can also recover the path: align separated upright pins with their viewed images through the block so they lie along the same sight line, without relative parallax. Keep the block fixed while tracing that path.

Measuring from the surface, shifting the block or using marks too close together changes the inferred angles. A thin ray, a careful outline and a longer marked baseline address different sources of uncertainty. Choose a suitable protractor scale and read it at the correct centre.

For an air-to-material investigation, a graph with sin i vertically and sin r horizontally should be approximately a straight line through the origin, with gradient n. A small scatter in measured ratios is expected; do not present the calculated table above as experimental observations.

Label the first and second media before substituting. The air-to-material shortcut cannot be used unchanged for a ray leaving the material.

Check your understanding

Light enters a material of refractive index 1.50 from air at 45 degrees to the normal. Taking the refractive index of air as 1, which calculation gives the refracted angle?

  1. 45 x 1.50 = 67.5 degrees.
  2. Inverse sine of (sin 45/1.50), giving 28.1 degrees.
  3. 45/1.50 = 30 degrees, dividing the angle by the refractive index.
Show the answer

Inverse sine of (sin 45/1.50), giving 28.1 degrees.

For air into this material, 1.50 = sin i/sin r. Rearranging gives sin r = sin 45/1.50 = 0.4714, so r = 28.1 degrees. This smaller angle agrees with bending towards the normal.

Practise an exam question

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

Exam-style question

A student shines light from air into a plastic block at several angles. She plots sin i (vertical axis) against sin r (horizontal axis). The best-fit line is straight, passes through the origin and through (0.40, 0.60). Find the refractive index of the plastic and explain how the graph shows this. [3]

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