Refraction through a rectangular glass slab

Hold a thick glass slab over a printed page and the letters underneath look nudged sideways, not blurred or tilted — a strange, specific kind of shift. That shift is not a trick of the glass; it is what happens when a ray of light crosses two parallel surfaces, bending once going in and bending back by exactly the same amount coming out. Because a rectangular slab has two flat faces parallel to each other, the ray that emerges is always parallel to the ray that went in — only displaced sideways from its original path. This article puts a number on that sideways shift, called lateral displacement, and shows how it follows directly from the same law of refraction (Snell's law) you already use at a single surface, applied twice, once at each face of the slab.

The core insight

A ray crossing a rectangular glass slab bends toward the normal entering and away from the normal leaving, so the emergent ray is parallel to the incident ray but laterally displaced by an amount that grows with the slab's thickness, the angle of incidence, and the glass's refractive index.

The core idea

A rectangular glass slab has two flat surfaces that are parallel to each other. A light ray from air meeting the first surface at angle of incidence $i_1$ bends towards the normal (air to denser glass), travelling inside at the smaller angle of refraction $r_1$ . Since Snell's law governs this bend at every point of entry, the ratio of the sines of the two angles is fixed by the material:

n = \frac{\sin i_1}{\sin r_1}

where: $n$ is the refractive index of the glass with respect to air (a pure number, no units), $i_1$ is the angle of incidence in air, and $r_1$ is the angle of refraction inside the glass, both measured from the normal at the first surface. Inside the slab the ray travels in a straight line (one uniform medium) until it reaches the second surface, the glass-to-air boundary, where it refracts again — this time away from the normal, emerging at the angle of emergence $e$ . Because the two surfaces are parallel, so are their two normals, and simple geometry of alternate angles forces

e = i_1

so the emergent ray leaves parallel to the original incident ray. It does not leave along the same line, though: the sideways offset between the incoming ray's line and the outgoing ray's line, measured perpendicular to that line, is called the lateral displacement, usually written $d$ . It grows with the slab's thickness $t$ , with the angle of incidence, and with the refractive index of the glass — thicker slab or larger angle, bigger shift; look straight through the slab at $i_1 = 0°$ and $d = 0$ , which is why a slab held flat in front of the eye does not distort a view seen dead ahead.

Figure

Ray diagram of light entering a glass slab bending toward the normal then leaving parallel to the incident ray but shifted sideways by the lateral displacement

Derivation ALateral displacement through a glass slab

The goal is to derive an exact formula for the lateral displacement $d$ of a ray passing through a rectangular glass slab of thickness $t$ , in terms of the angle of incidence $i$ , the angle of refraction $r$ inside the glass, and $t$ itself, using only the triangle formed by the refracted ray inside the slab and the perpendicular distance to the undeviated straight-line path — no result beyond plane geometry and the sine ratio already used above.

Step 1. Picture the ray if the slab were not there: it would carry straight on along the original incident direction. The actual ray instead bends at the first surface and travels along the refracted path inside the glass, at angle $r$ to the normal, until it reaches the second surface after crossing the slab's thickness $t$ .

Step 2. Along the normal direction, the refracted ray covers the slab's thickness $t$ while, in the direction along the surface, it drifts sideways from where an undeviated ray would be. The path length of the refracted ray inside the glass, measured along its own direction, is $t/\cos r$ , since $t$ is the perpendicular (normal) distance across the slab and $r$ is the angle the ray makes with that normal.

Step 3. The angle between the actual refracted ray and the undeviated straight-line path is $(i - r)$ , because the undeviated ray would continue at angle $i$ to the normal while the refracted ray bends to angle $r$ . The lateral displacement $d$ is the perpendicular distance from the emergent ray's line back to the undeviated line, and it is the side of a right triangle whose hypotenuse is the path length found in Step 2 and whose included angle is $(i-r)$ :

d = t\,\frac{\sin(i-r)}{\cos r}

Step 4. Sanity-check the limits. At normal incidence, $i = 0°$ and $r = 0°$ , so $\sin(i - r) = 0$ and $d = 0$ — no shift, matching the everyday observation that looking straight through a slab shows no displacement. For a fixed glass ( $r$ tied to $i$ by Snell's law), increasing $t$ or increasing $i$ increases $\sin(i - r)$ and hence $d$ increases — matching the qualitative rule that thicker slabs and larger angles of incidence give bigger sideways shifts.

Worked example

A ray of light strikes a rectangular glass slab of thickness $t = 3\ \text{cm}$ at an angle of incidence $i_1 = 50°$ , refracting inside the glass at $r_1 = 30°$ (the same angles as before). Find (a) the refractive index of the glass, and (b) the lateral displacement of the emergent ray.

Step 1. Refractive index from Snell's law: $n = \dfrac{\sin i_1}{\sin r_1} = \dfrac{\sin 50°}{\sin 30°} = \dfrac{0.766}{0.500} = 1.53$ .

Step 2. Angle between the refracted ray and the undeviated path: $i_1 - r_1 = 50° - 30° = 20°$ , so $\sin(i_1 - r_1) = \sin 20° = 0.342$ .

Step 3. Apply the lateral displacement formula with $t = 3\ \text{cm}$ and $\cos r_1 = \cos 30° = 0.866$ :

d = t\,\frac{\sin(i_1-r_1)}{\cos r_1} = 3 \times \frac{0.342}{0.866} = 3 \times 0.395 = 1.18\ \text{cm}

Step 4. Check the direction of reasoning: the emergent ray leaves the second face at $e = i_1 = 50°$ (angle of emergence equals angle of incidence, as established earlier), confirming it is parallel to the incident ray, and this worked-out $d = 1.18\ \text{cm}$ is exactly the sideways gap between the two parallel lines. The glass has refractive index $n \approx 1.53$ , and the emergent ray is shifted sideways by $d \approx 1.18\ \text{cm}$ from where an undeviated ray would land.

Competitive stretch

If the same ray struck a slab of the same glass but only $t = 1\ \text{cm}$ thick, at the same $50°$ angle of incidence, the refraction angle $r_1$ (fixed by $n$ and Snell's law) would stay $30°$ , so only $t$ changes in the formula: $d = 1 \times 0.395 = 0.395\ \text{cm}$ , about a third of the earlier shift — confirming that displacement scales directly, and linearly, with slab thickness for a fixed glass and fixed angle of incidence.

Common traps

Assuming the emergent ray exits along the same line as the incident ray. It is only parallel, offset by lateral displacement $d$ ; the two lines coincide only at $i_1 = 0°$ .
Forgetting to apply the law of refraction at both surfaces. The angle of refraction $r_1$ is found at the first (air-to-glass) surface; the lateral displacement formula also needs this same $r_1$ , not a fresh angle at the second surface.
Treating lateral displacement as depending on angle of incidence alone. The formula $d = t\sin(i-r)/\cos r$ shows it depends jointly on thickness $t$ , on $i$ , and on $r$ (which itself depends on $n$ ) — changing any one changes $d$ .
Mixing up the angle of refraction inside the glass with the angle of emergence. Only $i_1$ and $e$ are equal for a rectangular slab; the internal angle $r_1$ is smaller than both and appears only in tracing the path and in the displacement formula.

Ask the tutor: If tracing the ray path or plugging numbers into the lateral displacement formula feels confusing, ask the AI tutor to walk through the geometry with you, step by step.

Why does the emergent ray end up parallel to the incident ray instead of just going straight through?

The ray does bend, twice — once entering the denser glass, bending toward the normal, and once leaving back into air, bending away from the normal by the same law applied in reverse. Because the two surfaces of a rectangular slab are parallel, their normals are parallel too, and the geometry of the two equal-and-opposite bends forces the final direction to match the original one exactly, so $e = i_1$ . What it does not undo is the sideways position: the ray has travelled some extra distance sideways while inside the glass, and that gap survives as the lateral displacement.

Does lateral displacement depend on the colour of the light?

In principle, yes, though this is beyond what the formula alone shows at this stage. The refractive index $n$ of glass is very slightly different for different colours (this is what causes dispersion in a prism), so $r_1$ for red light differs slightly from $r_1$ for violet light at the same $i_1$ , which in turn changes $d$ slightly through the formula $d = t\sin(i-r)/\cos r$ . For a single rectangular slab this effect is small and is not part of the standard Class 10 treatment, where a single refractive index is assumed for the glass.

Is there a simple way to remember which way the ray bends at each surface?

Yes: bending is always toward the normal when light enters a denser medium (like air into glass) and away from the normal when it leaves into a rarer medium (like glass into air). At the first face of the slab, air-to-glass, the ray bends toward the normal, so $r_1$ is smaller than $i_1$ . At the second face, glass-to-air, it bends away from the normal by exactly the reverse amount, so the emergence angle grows back to equal $i_1$ . Track "denser bends toward, rarer bends away" and the whole ray path through any slab follows.

Refraction through a rectangular glass slab

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads