Refraction and Snell's law

Refractive index

Light travels at c = 3.00 × 10⁸$c = 3.00 \times 10^8$c = 3.00 × 10^8 m s⁻¹ in a vacuum and more slowly in everything else. The refractive index n of a material compares the two speeds.

where c_s$c_s$c_s is the speed of light in the material. Water has n ≈ 1.33 and typical glass n ≈ 1.5, and a larger n means a slower, optically denser material. Air slows light so little that its refractive index is taken as 1.

When light crosses into a new medium its frequency cannot change, because the wavefronts arrive at the boundary at a fixed rate and must leave at the same rate. With c = fλ$c = f\lambda$c = fλ, a smaller speed therefore means a proportionally smaller wavelength.

Snell's law

A ray crossing a boundary at an angle bends, because one side of each wavefront slows down before the other. All angles are measured between the ray and the normal, the line at right angles to the surface. The bend obeys Snell's law.

The direction of the bend follows from the equation. Going into a higher n, sin θ$\sin\theta$sin θ must shrink, so the ray bends towards the normal; going into a lower n it bends away from the normal. A ray along the normal itself passes straight through, slowed but unbent.

Measuring the refractive index of a solid

Snell's law is also a recipe for measuring n, and a rectangular glass block, a protractor and four pins are the whole apparatus. Air has n = 1.00 to three figures, so for a ray going from air into a solid the law reduces to a ratio of two sines,

with θ₁$\theta_1$θ_1 the angle in air and θ₂$\theta_2$θ_2 the angle in the solid, both measured from the normal.

The difficulty is that you cannot see the ray inside the glass, so the method reconstructs it. Lay the block on paper and draw round it, then mark a point on one long face and rule the normal there. Send a narrow ray in at that point, either from a ray box or by sighting along two optical pins pushed into the paper, and mark the incoming ray. Walk round to the far side, look through the block, and push in two more pins so that all four appear in a line through the glass. Now lift the block off. Joining the entry point to the exit point draws the path the light took inside, the one segment you were never able to watch, and the two angles can be measured with a protractor.

Describe the drawing

On the left, a ray traced through a rectangular glass block: it bends towards the normal on entering, crosses the block, and leaves parallel to its original direction but shifted sideways, with two pins marking each straight part and the angles of incidence and refraction marked at the top face. On the right, a graph of the sine of the angle in air against the sine of the angle in the glass: the points lie on a straight line through the origin whose gradient is the refractive index.

One reading would give you an answer, and a poor one. Instead repeat the tracing for perhaps six angles of incidence spread between 20° and 70°, and plot sin θ₁$\theta_1$θ_1 against sin θ₂$\theta_2$θ_2. Snell's law says the points should lie on a straight line through the origin, so the graph tests the law and measures the constant in the same stroke: the gradient is n. Averaging six points through a line of best fit beats any single protractor reading, and an intercept away from the origin indicates that the block outline moved between runs.

Three habits carry most of the precision. Use a sharp pencil and the thinnest ray the box will give, because a fat ray has no single edge to measure from. Space the pins as far apart as the paper allows, since the same one-millimetre slip in a pin position turns into a smaller angular error over a longer baseline. And keep off the very small angles, where sin θ$\sin\theta$sin θ is small and a half-degree protractor error is a large fraction of the reading.

The tracing checks itself. A block with parallel faces bends the ray towards the normal going in and away from it by the same angle coming out, so the emergent ray must be parallel to the incident ray and merely displaced sideways. Lay a ruler along both: if they are not parallel, something in the tracing is wrong and no amount of arithmetic afterwards will fix it.

There is a second route to the same number, and it is worth knowing because it needs no protractor work in air. Find the angle at which light inside the solid stops escaping, the critical angle of the next section, and n follows from sin θ_c = 1/n$\sin\theta_c = 1/n$sin θ_c = 1/n. It suits a semicircular block, where the ray can be aimed at the curved face along a radius and enters without bending at all.