Before you read this, I suggest you read post 22.4.

In post 19.21, we saw that a ray of light passing from air into glass (or any other dense, transparent medium) is bent towards the normal to the interface between them, as shown in the picture above. In this picture A is a point in air, B is the point where the ray meets the interface and C is a point in the glass. The angle that AB makes with the normal is denoted by θ_i, the angle of incidence; the angle that BC makes with the normal is denoted by θ_r, the angle of refraction. The ratio of the sines of these angles is a constant, n, called the refractive index of glass in air so that

sinθ_i/sinθ_r = n.                    (1)

We can explain everything in this paragraph if light is propagated as a wave – see post 19.21.

So, a ray passing from air into glass follows the path ABC; conversely, a ray passing from glass to air (initially along the line CB) would follow the path CBA. On emerging into air from glass, the ray is bent away from the normal to the interface.

Now let’s return to a ray passing from air into glass but with an angle of incidence of 90^o, as shown in the picture above. For a flat glass surface, this ray passes along the interface. Now, the angle of refraction is called the critical angle, θ_c. Substituting these values for the angles of incidence and refraction into equation 1 gives

n = sin90^o/sinθ_c. = 1/sinθ_c.                   (2)

since sin90^o = 1 (see post 16.50). The ray passes along the path A’BC’, in the picture above. A ray passing from C’ to B then follows the path C’BA’.

A ray passing through glass, towards the interface, that makes an angle of greater than θ_c to the normal can’t pass into the air because it would need to make an angle of 90^o with the normal. This is another way of saying that a ray making an angle of greater than θ_c to the normal can’t be propagated across the interface – so it is reflected, as explained in post 22.4. This type of reflection, arising at an interface with a medium of lower refractive index, is called total internal reflection. The ray follows the path C’’BA’’, in the picture above, where the angle of incidence, θ_i (θ_i > θ_c) is equal to the angle of reflection, θ_r, as explained in post 22.4. (Be careful, I have used the same symbol, θ_r, for two different ideas in this post for consistency with posts 19.21 and 22.24).

Total internal reflection in a prism is sometimes used to bend the path of light in optical instruments. What is a prism? A prism is a three-dimensional shape with two identical, parallel ends and flat side, as shown in the picture above. The prisms in the picture are called right prisms because the two identical ends are perpendicular to the flat sides; a prism that is not a right prism is called an oblique prism.

In optics, the word “prism” usually means a glass right triangular prism; the way is which it can bend the path of light is shown in the picture above, where the triangle is a section of the prism. The ray AB meets the lower face of the prism at right angles and so is not deflected, see appendix 1. It meets the next interface at B, making an angle of θ with the normal to the surface. If θ > θ_c, there is total internal reflection and the ray meets the interface at C. In the picture, I have chosen a prism whose shape means that BC is perpendicular to the face of the prism at C, so that the ray passes straight though. When designing an optical instrument, it is usual to make the entrance and exit angles 90^o, for the reason explained in appendix 2.

Related posts

22.4 Reflection
19.21 Refraction

Appendix 1

Why does a ray that is perpendicular to an air-glass interface pass straight through without being deflected?

Look at the third picture in post 19.21. Now B and A meet the interface at the same time. Then AD and BC are both zero. As a result, there is no difference in speed between the two paths, at any instant in time, so no refraction occurs.

Appendix 2

Why are optical instruments designed to have entrance and exit angles of the ray passing through a prism equal to 90^o? The reason is that the refractive index of light depends on its colour. This dependence of refractive index on colour is called dispersion. Red light (longer wavelength, post 19.9) is bent less than blue light (shorter wavelength, post 19.9).

When dispersion occurs in an optical instrument, white light is split into its different colours that appear in the image. This effect is called chromatic aberration. When the entrance and exit angles from a prism are both 90^o, no refraction occurs (see appendix 1). So any image formed by the instrument is fee from chromatic aberration.

Why does dispersion occur? Since violet light (shorter wavelength) is bent more than red light (longer wavelength), it must be that shorter wavelength light travels more slowly through dense media than longer wavelength light since

c_air/c_glass = sinθ_i/sinθ_r = n

where c_air is the speed in air and c_glass is the speed in glass, as shown in post 19.21.