Light and Sound » Practical: Refractive Index of Glass

The aim of the practical

This practical is named in the specification, so it can be examined. You must be able to describe the method, handle the results and judge the errors, even if you have never done it.

The aim is to find the refractive index of glass by experiment. Remember, n = sin i ÷ sin r, where i is the angle in the air and r is the angle in the glass. A good experiment uses a range of angles and a graph, because that gives a value you can trust.

Apparatus: a ray box with a single slit, a power supply, a rectangular glass block, plain white paper, a sharp pencil, a ruler, a protractor and a calculator.

Key terms:

• Range: the spread of values you choose for the independent variable, from the smallest to the largest. Here it is 10° to 70°.
• Gradient triangle: a large right-angled triangle drawn under a straight line on a graph. You use it to read off the rise and the run.

The method

The ray is traced as in Practical: Investigating Refraction: draw round the block, mark each ray with two crosses, then remove the block and join the marks.

1. Place the block on plain paper and draw round it. Take the block away for a moment. Near the middle of one long side, draw a normal as a dashed line at 90° to the side. The point where the normal meets the side is called the foot of the normal.
2. With the protractor, draw faint pencil lines to the foot of the normal at 10°, 20°, 30°, 40°, 50°, 60° and 70° from the normal. These are your angles of incidence, i.
3. Put the block back on its outline. Dim the room and shine the ray along the 10° line, so it enters the glass at the foot of the normal.
4. Mark two crosses on the ray that comes out of the far side of the block.
5. Repeat for each of the other lines. Label each pair of crosses so you know which ray they belong to.
6. Switch off the ray box and remove the block. For each ray, join the two crosses and carry the line back to the outline. This shows where the ray left the glass.
7. Join each of those points to the foot of the normal. These lines are the rays inside the glass.
8. Measure the angle of refraction, r, for each ray. It is the angle between the ray inside the glass and the normal.

The results table

Give the table four columns: i, r, sin i and sin r. Sines have no unit. Check your calculator is in degrees mode, then write each sine to three decimal places.

Angle of incidence i in °	Angle of refraction r in °	sin i	sin r
10	7	0.174	0.122
20	13	0.342	0.225
30	19	0.500	0.326
40	25	0.643	0.423
50	34	0.766	0.559
60	35	0.866	0.574
70	39	0.940	0.629

Look down the r column. It jumps to 34° at 50° and barely moves at 60°. The 50° reading does not fit the pattern, so it is an anomalous result. The student traces that ray again and gets r = 31°, so sin r = 0.515. The new reading replaces the old one.

Finding n from a graph

Plot sin i on the y-axis against sin r on the x-axis. Do not plot the angles themselves, because that graph is a curve.

1. Choose scales that use more than half of the grid. Here both axes could run from 0 to 1.0.
2. Plot each point as a small neat cross.
3. Draw a straight line of best fit with a ruler. It should pass through the origin, because a ray along the normal has i = 0° and r = 0°.
4. Draw a large gradient triangle and work out rise ÷ run.

The gradient is the change in sin i divided by the change in sin r. For a straight line through the origin that is the same as sin i ÷ sin r, so the gradient is the refractive index.

Take your two points from the line, not from the table, and choose them far apart.

The other way: averaging n

You can also work out n for every row and take the mean. Using the sines in the table, with the repeated 50° reading:

i in °	10	20	30	40	50	60	70
n = sin i ÷ sin r	1.43	1.52	1.53	1.52	1.49	1.51	1.49

Both ways give about 1.5. The graph is usually better: it shows an anomalous point at a glance, and the line of best fit lets you leave it out. A mean includes every value you put into it.

Why several angles are better than one

Sources of error

Source of error	Why it matters here	Improvement
The protractor reads only to the nearest 1°	At 10° the angle of refraction is only about 7°, so 1° is a large fraction of it. That is why the 10° row gave 1.43	Rely most on the larger angles, and use the graph so one weak point has little effect
The ray has width	The crosses may not sit on the true path, so the line inside the glass is drawn at the wrong angle	Use a narrow slit and mark the middle of the ray, with the crosses far apart
The ray misses the foot of the normal	Angles are then measured from the wrong point	Check where the ray enters the glass before marking each ray

Safety

The ray box gets hot, so do not touch it near the lamp. The room is dim, so keep bags off the floor. A dropped glass block can chip and leave sharp edges, so do not use a chipped block.

Exam-style question

A student investigates the refractive index of glass. She shines a ray of light from a ray box into a rectangular glass block and measures the angle of refraction, r, for six angles of incidence, i.

(a) State the measuring instrument she uses for the angles, and state the line from which both angles are measured. [2 marks]

(b) For one ray, i = 65° and r = 37°. Calculate the refractive index of the glass from this pair of readings. Give your answer to 3 significant figures. [3 marks]

(c) The student plots a graph of sin i (y-axis) against sin r (x-axis). Her line of best fit passes through the points (0.10, 0.16) and (0.60, 0.92). Use the graph to determine the refractive index of the glass. [3 marks]

(d) Explain why the value from the graph is more reliable than the value from part (b). [2 marks]

(e) One of her points lies well away from the line of best fit. State what she should do about this result. [1 mark]