Recall and use the equation n = 1 / sin c

3.2.2 Refraction of Light

Learning Objective

Recall and use the critical‑angle equation

\[

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

\]

where n is the refractive index of the denser medium and c is the critical angle for total internal reflection (TIR).

Key Definitions

  • Normal – an imaginary line drawn perpendicular to the surface at the point of incidence.
  • Angle of incidence (i) – the angle between the incident ray and the normal.
  • Angle of refraction (r) – the angle between the refracted ray and the normal.
  • Critical angle (c) – the smallest angle of incidence in the denser medium for which the refracted ray emerges at \(90^{\circ}\) (i.e. just grazes the interface).
  • Total internal reflection (TIR) – the phenomenon that occurs when the angle of incidence exceeds the critical angle, causing the light to be reflected back into the denser medium.

Conceptual Overview

  • Refraction – change in direction of a light ray when it passes from one transparent medium to another with a different optical density.
  • Boundaries only – the Cambridge syllabus requires discussion of the two discrete interfaces (e.g. air ↔ glass). Gradual‑index media are not examined.

Refractive Index

The refractive index is a dimensionless quantity that compares the speed of light in a medium with that in vacuum:

\[

n=\frac{c_{0}}{v}

\]

where \(c_{0}=3.00\times10^{8}\,\text{m s}^{-1}\) (speed of light in vacuum) and \(v\) is the speed in the medium.

Using Snell’s law it can also be written in terms of angles:

\[

n=\frac{\sin i}{\sin r}

\]

Snell’s Law

For a ray passing from medium 1 (\(n{1}\)) to medium 2 (\(n{2}\)):

\[

n{1}\sin i=n{2}\sin r

\]

Critical Angle and Total Internal Reflection

When light travels from a denser medium (\(n{1}\)) to a rarer medium (\(n{2}\)), set \(r=90^{\circ}\) (\(\sin r=1\)) in Snell’s law:

\[

\sin c=\frac{n{2}}{n{1}}\qquad (n{1}>n{2})

\]

If the second medium is air (\(n_{2}\approx1\)), the expression simplifies to the target form:

\[

n_{1}= \frac{1}{\sin c}

\]

Derivation of \(\displaystyle n=\frac{1}{\sin c}\)

  1. Start with Snell’s law: \(n{1}\sin i=n{2}\sin r\).
  2. For the critical condition the refracted ray grazes the interface, so \(r=90^{\circ}\) and \(\sin r=1\).
  3. Replace the angle of incidence by the critical angle \(c\): \(n{1}\sin c=n{2}\).
  4. When the second medium is air (\(n{2}\approx1\)), rearrange to obtain \(n{1}=1/\sin c\).

Typical Refractive Indices (20 °C, 1 atm)

MediumRefractive Index, \(n\)
Air (standard conditions)1.00
Water1.33
Glass (crown)1.52
Glass (flint)1.62
Diamond2.42

Experiments

1. Refraction Through a Rectangular Glass Slab

Objective: Verify Snell’s law and determine the slab’s refractive index.

Apparatus: Ray box or laser pointer, rectangular glass slab, protractor, white paper, ruler, pins.

Procedure:

  1. Place the slab on the paper and trace its outline.
  2. Draw the normal at one face of the slab.
  3. Set the incident ray to a known angle of incidence \(i\) and mark the incident, refracted, and emergent rays.
  4. Measure the angle of refraction \(r\) inside the slab.
  5. Repeat for several values of \(i\) (e.g. 20°, 30°, 40°, 50°).
  6. Calculate \(\sin i / \sin r\) for each trial; the ratio should be constant and equal to the slab’s \(n\).

Observation: The emergent ray is parallel to the incident ray but laterally displaced; the calculated ratio gives \(n\approx1.5\) for typical crown glass.

2. Refraction Through a Triangular Prism (different shape)

Objective: Show that when the two interfaces are not parallel the emergent ray is not parallel to the incident ray.

Apparatus: Ray box, equilateral glass prism, protractor, paper, pins.

Procedure:

  1. Draw the normal at the first face and set an incident angle \(i\).
  2. Measure the first refraction angle \(r_{1}\) inside the prism.
  3. At the second face draw the normal, measure the angle of incidence inside the prism \(i{2}\) and the emergent angle \(r{2}\).
  4. Repeat for several values of \(i\) and compare the measured emergent angles with those predicted by Snell’s law.

Observation: The emergent ray is deviated from the original direction, illustrating that the shape of the transparent object influences the overall path of the light.

3. Classroom Demonstration of Total Internal Reflection

Objective: Observe TIR directly and determine the critical angle of water.

Apparatus: Shallow rectangular tank, water, laser pointer, protractor, white screen.

Procedure:

  1. Fill the tank with water and place the laser so that the beam strikes the water‑air interface from within the water.
  2. Gradually increase the angle of incidence and note the angle at which the beam ceases to emerge into air and instead reflects back into the water.
  3. Record this angle as the experimental critical angle \(c_{\text{exp}}\).
  4. Calculate the refractive index of water using \(n{\text{water}}=1/\sin c{\text{exp}}\) and compare with the accepted value (1.33).

Observation: Light is totally internally reflected for angles greater than the measured critical angle (≈ 48°), confirming the equation \(n=1/\sin c\).

Everyday Applications of Total Internal Reflection

  • Optical fibres – repeated TIR keeps light confined to the core, enabling long‑distance data transmission.
  • Diamond sparkle – a high refractive index (2.42) gives a very small critical angle, producing many internal reflections.
  • Fish eyes – the cornea‑lens interface reflects light internally, improving underwater vision.
  • Periscopes and binocular prisms – use TIR to change the direction of light without loss.

Worked Example

Question: Light travelling in glass (\(n=1.52\)) strikes the glass‑air interface. Calculate the critical angle \(c\) for total internal reflection.

  1. Use \(\displaystyle \sin c=\frac{n{\text{air}}}{n{\text{glass}}}=\frac{1.00}{1.52}\).
  2. \(\sin c = 0.658\).
  3. \(c = \sin^{-1}(0.658) \approx 41.1^{\circ}\).

Any incident angle greater than \(41.1^{\circ}\) will produce TIR.

Practice Questions

  1. Given a critical angle of \(48^{\circ}\) for an unknown liquid, determine its refractive index using \(\displaystyle n=\frac{1}{\sin c}\).
  2. A ray of light passes from water into glass. If the angle of incidence in water is \(30^{\circ}\), calculate the angle of refraction in glass (use \(n{\text{water}}=1.33\), \(n{\text{glass}}=1.52\)).
  3. Explain why a diamond sparkles more than glass, referring to the values of \(n\) and the resulting critical angles.
  4. In the water‑tank experiment, the measured critical angle is \(47^{\circ}\). Calculate the experimental refractive index of water and comment on the difference from the accepted value.

Common Mistakes to Avoid

  • Confusing the critical angle with the angle that gives the greatest refraction; the critical angle marks the onset of total internal reflection.
  • Applying \(\displaystyle n=\frac{1}{\sin c}\) when the second medium is not air. Use the full form \(\sin c=n{2}/n{1}\) in those cases.
  • Forgetting to set the calculator to degree mode (or to convert degrees to radians when required).
  • Assuming the refractive index varies continuously inside a medium; the syllabus only requires discussion of the two discrete boundaries.

Suggested diagram: (a) Ray diagram at a glass‑air interface showing the normal, incident ray, refracted ray, and critical angle \(c\); (b) Sketch of the rectangular‑slab experiment with measured angles; (c) Sketch of the water‑tank TIR demonstration.