Refraction through glass block

Two refractions, one straight run. A ray in air meets the first face of the slab at angle of incidence $i_1$ and bends towards the normal (air is rarer, glass is denser), continuing inside the glass at angle of refraction $r_1$:

where: $\mu$ is the refractive index of the glass relative to air, $i_1$ is measured from the normal at the first face, and $r_1$ is measured from the same normal, on the glass side.

Why the second surface repeats the trick in reverse. Because the two faces of a rectangular slab are parallel, their normals are parallel too. So the ray travelling in a straight line inside the glass strikes the second face at exactly the same angle $r_1$ it left the first face with (angles between parallel lines and a common transversal are equal). At the second face the ray goes from glass to air — denser to rarer — so it bends away from the normal, emerging at angle of emergence $e$. Snell's law applied at this surface, worked backwards, forces $e = i_1$: the emergent ray leaves at precisely the angle the original ray arrived at, just at a different point on a different, parallel face.

Parallel but displaced. Since the incident and emergent rays make the same angle with parallel normals, they are themselves parallel lines — but not the same line. The perpendicular gap between them is the lateral displacement, and it grows with the slab's thickness $t$, with $i_1$, and with $\mu$; it vanishes only when $i_1 = 0$, since a ray straight along the normal never bends at all.

A ray strikes a glass slab of refractive index $1.5$ and thickness $6\ \text{cm}$ at an angle of incidence of $50°$. Find the lateral displacement of the emergent ray.

Step 1. Find $r_1$ using Snell's law at the first surface, exactly as before: $\sin r_1 = \dfrac{\sin 50°}{1.5} = \dfrac{0.766}{1.5} = 0.511$, so $r_1 = \sin^{-1}(0.511) \approx 30.7°$.

Step 2. Compute the bending angle used in the displacement formula: $i_1 - r_1 = 50° - 30.7° = 19.3°$, so $\sin(i_1 - r_1) = \sin 19.3° \approx 0.331$.

Step 3. Compute $\cos r_1$: $\cos 30.7° \approx 0.860$.

Step 4. Substitute into $d = \dfrac{t}{\cos r_1}\sin(i_1 - r_1)$: $d = \dfrac{6}{0.860} \times 0.331 = 6.98 \times 0.331 \approx 2.31\ \text{cm}$. The lateral displacement is about $2.31\ \text{cm}$, and the emergent ray leaves the far face at $e = 50°$, parallel to the original incident ray, just shifted sideways by this amount.

Competitive stretch

If the same slab is doubled in thickness to $12\ \text{cm}$ while $i_1$ and $\mu$ stay the same, $r_1$ is unchanged (it depends only on $i_1$ and $\mu$, not $t$), so the displacement simply scales with $t$: $d = \dfrac{12}{0.860}\times 0.331 \approx 4.62\ \text{cm}$ — exactly double, confirming that $d$ is directly proportional to thickness for fixed $i_1$ and $\mu$.

• Thinking the emergent ray changes direction, not just position — it is parallel to the incident ray ($e = i_1$ always, for parallel faces), only laterally shifted.
• Using $t$ itself as the length of the path inside the glass instead of $t/\cos r_1$ — the ray travels along the slanted path, not straight across the thickness.
• Plugging $i_1 - r_1$ into the displacement formula in degrees while leaving $\cos r_1$ or $\sin(i_1-r_1)$ computed with a calculator in radian mode, giving a wrong numeric answer despite the correct formula.
• Assuming displacement depends on thickness alone — it depends on $t$, $i_1$, and $\mu$ together (through $r_1$), and is zero at normal incidence regardless of thickness.

Ask the tutor: Want to try the displacement formula with a different thickness or angle of incidence, or see how it changes for a slab of a different refractive index?