Reflection and refraction.

Step 1 of 5

An angle from the normal

When a ray of light meets a surface, its direction is measured from the normal: an imaginary line square to the surface where the ray strikes. The angle between the arriving ray and the normal is the angle of incidence, and the angle between the ray that went into the material and the normal is the angle of refraction. Why not from the surface itself? Because a surface can be curved, like a lens or a water drop, while the normal has one clear direction at every point.

Some of the light does not go into the glass but reflects off it: that is the dim ray. It leaves on the other side of the normal, at the same angle the ray came in at. This is the law of reflection: the angle of reflection equals the angle of incidence, for any smooth surface, from glass to a mirror.

In the Scene the ray strikes a glass block, and the angle control is hidden. Answer, then choose On under Show normals and angles to check.

Step 2 of 5

Towards the normal

The ray now meets the glass head on, along the normal, and goes straight through without bending. What happens when it arrives at a slant?

Like any wave that passes from one medium into another, light changes speed at the boundary. In glass it is slower than in air. The refractive index $n$⁠ of a material says how many times slower: $n = \dfrac{c}{v}$⁠, where $c$⁠ is the speed of light in a vacuum (and very nearly in air) and $v$⁠ its speed in the material. Glass has $n \approx 1.5$⁠, so light crosses it at about 200,000 km/s instead of about 300,000 km/s.

Predict, then watch the Scene tilt the ray to $40^\circ$⁠.

Step 3 of 5

Snell’s law

The ray turns because one side of its wavefront enters the glass and slows before the other. The exact rule is Snell’s law:

$n_1 \sin\theta_1 = n_2 \sin\theta_2$$n_1 \sin\theta_1 =$$n_2 \sin\theta_2$⁠

$n_1$⁠ and $\theta_1$⁠ belong to the material the light comes from (in air $n_1 = 1$⁠), and $n_2$⁠ and $\theta_2$⁠ to the one it enters. When $n_2$⁠ is larger than $n_1$⁠, the sine of the angle of refraction is smaller, and so is the angle: the ray bends towards the normal. To find $\theta_2$⁠, work out $\sin\theta_2$⁠ first, then the angle with $\sin^{-1}$⁠.

The angle, the glass and the direction are now fixed. Work out the angle of refraction, then check with Show normals and angles.

Step 4 of 5

Out of the glass

Now the light leaves the glass. It enters the curved side of the half-cylinder along a radius, so it does not bend there, and meets the flat side from within. Snell’s law works this way round too, with $n_1 = 1.5$⁠ and $n_2 = 1$⁠:

$\sin\theta_2 = 1.5 \sin\theta_1$$\sin\theta_2 =$$1.5 \sin\theta_1$⁠

The sine of the angle of refraction is larger than that of the angle of incidence, so the ray bends away from the normal: at $30^\circ$⁠ it leaves at about $48.5^\circ$⁠. But what happens when $1.5 \sin\theta_1$⁠ would come out larger than $1$⁠? Predict, then watch the Scene raise the angle to $50^\circ$⁠.

Step 5 of 5

The critical angle

The angle at which the refracted ray leaves exactly along the surface, $\theta_2 = 90^\circ$⁠, is the critical angle $\theta_c$⁠. From Snell’s law, $n \sin\theta_c = 1$⁠, so:

$\sin\theta_c = \frac{1}{n}$$\sin\theta_c =$$\frac{1}{n}$⁠

Past the critical angle Snell’s law has no solution, and no light gets out: all of it reflects back inside, at an angle of reflection equal to the angle of incidence. This is total internal reflection. Optical fibres use it to carry light for kilometres, and diamond’s small critical angle, about $24^\circ$⁠, traps light inside it and makes it sparkle.

The disc is now made of water. Work out its critical angle, then raise the Angle of incidence slowly and find the angle at which the light stops getting out.