Refraction at plane surfaces

We start from Snell's law and the principle of reversibility, and reach three results that let you convert freely between any pair of media.

Part (a): {}_1n_2 = n_2/n_1 = v_1/v_2

Step 1 — Snell's law in its two-medium form. For a ray passing from medium 1 (angle i) into medium 2 (angle r):

Step 2 — make \sin i/\sin r the subject. Divide both sides by n_1\sin r:

Step 3 — name the left-hand side. By definition, \sin i/\sin r for this pair of media is the refractive index of 2 with respect to 1:

Step 4 — bring in the speeds. Absolute index is defined as n = c/v, so n_1 = c/v_1 and n_2 = c/v_2. Substituting:

Step 5 — cancel c and flip the divided fraction:

Part (b): {}_1n_2 \times {}_2n_1 = 1

Step 6 — state the principle of reversibility. If the direction of a ray is exactly reversed, it retraces its own path. So send a ray back from medium 2 along the old refracted ray. It now strikes the boundary at angle r and must emerge along the old incident ray, at angle i.

Step 7 — apply the definition to this reversed journey. The ray now goes from 2 to 1, so the angle in medium 2 is the incidence angle and the angle in medium 1 is the refraction angle:

Step 8 — compare with {}_1n_2 = \sin i/\sin r from Step 3. The right-hand side of Step 7 is the reciprocal of the right-hand side of Step 3:

Step 9 — multiply both sides by {}_1n_2:

Part (c): the chain relation

Step 10 — write each relative index as a ratio of absolute indices, using Part (a):

Step 11 — cancel n_2, which appears once on top and once below:

Step 12 — recognise the right-hand side as {}_1n_3, again by Part (a):

Step 13 — multiply both sides by {}_3n_1 and use Part (b) on the right, since {}_1n_3 \times {}_3n_1 = 1:

Result.

What it means. You only ever need a table of absolute indices; any relative index is a ratio of two of them. Part (b) says glass-with-respect-to-water is the exact reciprocal of water-with-respect-to-glass, and Part (c) says a chain of media closing back on itself multiplies to one — like walking round a loop and returning to the same height.

Conditions used. A single wavelength (since n depends on colour), and media that are transparent and uniform.

We start from Snell's law applied to a ray leaving the water, and reach a way of measuring n with nothing but a ruler.

Setting up. A coin lies at O, a depth OA vertically below A on the flat surface of water of refractive index n. Take two rays from the coin: OA straight up along the normal, which passes out undeviated, and OQ making a small angle \theta with the normal.

Step 1 — apply Snell's law at Q. The ray goes from water into air, so the water term carries n and the air term carries 1. Let \phi be the angle in air:

Step 2 — make n the subject:

Step 3 — locate the image. The emergent ray bends away from the normal (air is rarer, so \phi > \theta). Produced backwards, it meets the normal OA at a point I above O. The eye receives both rays as if they came from I, so IA is the apparent depth.

Step 4 — use the right-angled triangle OAQ, right-angled at A, with \theta at O:

Step 5 — use the right-angled triangle IAQ, right-angled at A, with \phi at I (the angle at I equals \phi because IQ is the emergent ray and IA is parallel to the normal at Q):

Step 6 — restrict to a nearly normal view. If we look almost straight down, \theta and \phi are both small, and for small angles \sin x \approx \tan x:

Step 7 — replace the sines in Step 2 by tangents:

Step 8 — substitute Steps 4 and 5:

Step 9 — divide the fractions. AQ cancels, and dividing by AQ/OA means multiplying by OA/AQ:

Step 10 — name the two lengths:

Step 11 — how far the coin seems to rise. The shift is the difference of the two depths, and from Step 10, IA = OA/n:

Step 12 — take out the common factor OA:

Result.

What it means. Because n > 1, the apparent depth is always less than the real depth — a pond always looks shallower than it is, which is a genuine danger. Both lengths are measured with a ruler, so this gives a simple laboratory method for n.

Conditions used. The viewing must be nearly normal (Step 6). If you look at the water at a slant, the small-angle step fails and the simple ratio no longer holds. The surface must also be flat and the liquid uniform.

We start by showing the emergent ray is parallel to the incident ray, then find how far sideways it has been pushed.

Setting up. The slab has thickness t and refractive index n. The ray enters at A with angle of incidence i and angle of refraction r, travels inside along AB, and leaves at B with angle of incidence r (inside) and angle of emergence e.

Step 1 — Snell's law at the first face, air into glass:

Step 2 — why the angle inside at B is also r. The two faces are parallel, so their normals are parallel. AB cuts both normals, and alternate angles between parallel lines are equal. So the angle of incidence inside the glass at B equals r.

Step 3 — Snell's law at the second face, glass into air:

Step 4 — compare Steps 1 and 3. Both left-hand sides equal n, so

Step 5 — cancel \sin r and take the inverse sine:

So the emergent ray is parallel to the incident ray: the slab changes the ray's position but not its direction. That is the first part of the Group C question.

Now the shift itself.

Step 6 — define the shift. Produce the incident ray straight on, as if the glass were not there. Drop the perpendicular BN from B onto this undeviated line. Then

Step 7 — find the angle at B in triangle ABN. AN lies along the original direction, which makes angle i with the normal at A; AB makes angle r with the same normal. The angle between them is therefore their difference:

Step 8 — use the right-angled triangle ABN, right-angled at N:

Step 9 — make d the subject:

Step 10 — now find AB. Drop the perpendicular AM from A onto the second face; its length is the thickness of the slab, AM = t. In the right-angled triangle ABM the angle at A is r (both AM and the normal at A are perpendicular to the faces):

Step 11 — make AB the subject:

Step 12 — substitute this into Step 9:

Result.

What it means. A window pane does not distort the direction of what you see, only shifts it a little sideways — which is why you can read through a glass block at an angle without the scene tilting. The shift grows with the thickness t and with the angle of incidence.

Conditions used. The two faces must be parallel (Step 2 fails otherwise — that is why a prism does deviate light), and the same medium (air) lies on both sides.

We start from Snell's law at the one special angle where the refracted ray just grazes the boundary, and reach a relation between the critical angle and the refractive index.

Setting up. Light travels in a denser medium of absolute index n_1 and meets a rarer medium of index n_2. By the definition of the critical angle, when the angle of incidence is C the angle of refraction is exactly 90^{\circ}.

Step 1 — write Snell's law for this boundary:

Step 2 — substitute the critical condition i = C and r = 90^{\circ}:

Step 3 — evaluate \sin 90^{\circ} = 1:

Step 4 — divide both sides by n_1:

Step 5 — take the usual case of a denser medium in air. Then n_2 = 1 and we write n_1 = n:

Step 6 — invert to get n from a measured C:

Result.

What it means. A bigger n makes 1/n smaller, so a denser material has a smaller critical angle and traps light more easily — this is exactly why diamond sparkles. Step 4 also explains why total internal reflection can never happen from rarer to denser: then n_2/n_1 > 1, and no angle has a sine greater than 1, so no critical angle exists.

Conditions used. Light must be going from denser to rarer, and one wavelength at a time (since n, and hence C, depends slightly on colour).