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Experiment: Determining the refractive index of a glass block

To determine the refractive index of a glass block by measuring the angle of incidence and the angle of refraction.

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Aim

To determine the refractive index of a glass block by measuring the angle of incidence and the angle of refraction.

Variables

  • Manipulated: Angle of incidence, i
  • Responding: Angle of refraction, r
  • Constant: The same glass block (same medium)

Apparatus & materials

  • Rectangular glass block
  • Ray box with a single slit (or optical pins)
  • White paper
  • Protractor
  • Ruler
  • Sharp pencil
  • Drawing board

Procedure

  1. Place the glass block on the white paper and draw its outline; mark a point on one long side and draw the normal there.
  2. Direct a single ray at the point so the angle of incidence is i = 30°.
  3. Mark the incident ray and the emergent ray with dots or pins, then remove the block.
  4. Join the marks and draw the refracted ray inside the outline; measure the angle of refraction r with the protractor.
  5. Repeat for i = 40°, 50°, 60° and 70°.
  6. Record i and r, and calculate sin i and sin r for each angle.

Tabulating results

Record the angle of incidence i and the angle of refraction r in degrees, and the derived values sin i and sin r. Keep the same number of decimal places for the sine values.

The graph

Plot sin i (y-axis) against sin r (x-axis). A straight line through the origin shows sin i is directly proportional to sin r; the gradient is the refractive index.

Analysis

By Snell's law, n = sin i / sin r. The gradient of the sin i against sin r graph gives the refractive index n of the glass.

Precautions

  • Use a sharp pencil so the rays and outline are thin and accurate.
  • Mark the ray positions with dots far apart (or pins upright) to draw the ray accurately.
  • Measure all angles from the normal, not from the surface.

Light and Optics · Graph skills

Sample results and what they show

These are example readings for a single light ray entering a rectangular glass block, with the angle of incidence i and the angle of refraction r measured from the normal.

  • i = 30°, r = 19.5°, sin i = 0.500, sin r = 0.334
  • i = 40°, r = 25.4°, sin i = 0.643, sin r = 0.429
  • i = 50°, r = 30.7°, sin i = 0.766, sin r = 0.511
  • i = 60°, r = 35.3°, sin i = 0.866, sin r = 0.578
  • i = 70°, r = 38.8°, sin i = 0.940, sin r = 0.627

The refracted ray always bends towards the normal, so r is smaller than i every time, showing the light slows as it enters the denser glass. Working out sin i ÷ sin r for each row gives almost exactly 1.50 throughout.

That constant ratio is the message of the experiment: it is Snell's law in action, and the number 1.50 is the refractive index of the glass. The individual angles change a lot, but their sine ratio stays fixed.

Reading the graph and finding the answer

Plot sin i on the y-axis against sin r on the x-axis. The points lie on a straight line passing through the origin, which shows sin i is directly proportional to sin r.

Take a large triangle on the best-fit line using two well-separated points, for example (0.334, 0.500) and (0.627, 0.940):

  • gradient = (0.940 − 0.500) ÷ (0.627 − 0.334) = 0.440 ÷ 0.293 = 1.50

The gradient has no unit because both sin i and sin r are pure ratios, and this gradient is the refractive index n of the glass. So n = 1.50.

Because the line passes through the origin, you should read the coordinates off the drawn line, not from a single data point. The SPM Physics 4531 papers give no formula sheet, so recall n = sin i / sin r yourself.

Choose axis scales that spread the plotted points across more than half of the grid, and draw the best-fit line so the points are balanced evenly on both sides of it. If one point is clearly off the line, treat it as an anomalous reading, ring it, and leave it out when drawing the line and taking the gradient.

Marks examiners look for

Precautions that keep the angles accurate:

  • Use a sharp pencil so the ray lines and outline are thin and precise.
  • Mark the incident and emergent rays with two well-spaced dots each before removing the block.
  • Measure every angle from the normal, not from the glass surface.
  • Keep the ray narrow so its centre is easy to trace.

For the Paper-3 science process skills, examiners award marks for: identifying the manipulated variable (i), responding variable (r) and constant (the same glass block); tabulating i, r, sin i and sin r with consistent decimal places; plotting at least five points with labelled axes and a best-fit straight line through the origin; and finding the gradient with a large triangle. State the relationship, that sin i is proportional to sin r, and quote the refractive index as your conclusion.

Repeat each reading and take the average to reduce random error, and quote the final answer to a sensible number of significant figures with its correct unit. In the discussion, name one source of error together with a matching improvement, and note that any point off the line was ignored so the conclusion rests on consistent data.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)

Written by the spmphysics.com.my editorial team.· Updated 5 Sept 2026

Frequently asked questions

Do I need a lab to practise?
No, the Paper 3 graph and analysis skills can be practised from home with example data.
Why plot sin i against sin r instead of i against r?
The angles i and r are not directly proportional, so an i–r graph would be a curve. Snell's law says n = sin i / sin r, so it is the sines that are proportional. Plotting sin i against sin r therefore gives a straight line through the origin whose gradient is the refractive index.
Why does the graph pass through the origin?
When the light hits the surface along the normal, i = 0° and r = 0°, so sin i and sin r are both zero. That point (0, 0) lies on the line, which is why a correct sin i–sin r graph passes through the origin.
Why measure angles from the normal and not from the glass surface?
The refractive index is defined using angles measured from the normal, the line drawn perpendicular to the surface. Measuring from the surface would give the complementary angle and a wrong sine, so every incidence and refraction angle must be taken from the normal.

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