Light – Reflection and Refraction · Lesson 9 of 15
Refraction through a Rectangular Glass Slab
“The ray leaves parallel to where it started, just slightly off course.”
• Trace incident, refracted and emergent rays through a glass slab. • Explain bending toward and away from the normal. • Explain why the emergent ray is parallel to the incident ray. • Describe lateral displacement and normal incidence. • State and apply the laws of refraction and Snell's law.
A rectangular slab has two parallel faces. A ray bends when it enters the glass and bends again when it leaves. The second bending does not return the ray to its original line, but it does restore its original direction, producing a parallel sideways-shifted ray.
Tracing the Ray
At the air-glass surface, the ray travels from an optically rarer medium to a denser one and bends toward the normal. Inside the slab it travels straight. At the glass-air surface, it moves from denser to rarer and bends away from the normal. Label the first segment incident ray, the segment in glass refracted ray, and the final segment emergent ray.
The slab faces are parallel, so their normals are parallel. For the same surrounding medium on both sides, the deviation at the second face is equal and opposite to that at the first. Consequently, the emergent ray is parallel to the incident ray. It is displaced sideways; the perpendicular separation between the emergent ray and the incident ray's original direction is lateral displacement.
Laws of Refraction
- The incident ray, refracted ray and normal at the point of incidence lie in the same plane.
- For light of a given colour and a fixed pair of media, the ratio sin i/sin r is constant for 0° < i < 90°.
Reading Snell's Law
If a ray enters a denser medium from a rarer one, it bends toward the normal, so r < i and sin i/sin r is greater than one. In the reverse direction, the ray bends away, so r > i. The law must be applied to the stated direction of travel; reversing the media reverses the relative index.
Problem
A ray enters glass from air with i = 45° and r = 28°. Estimate the refractive-index ratio.
- 1.Use n = sin i/sin r for this direction.
- 2.sin 45° ≈ 0.707 and sin 28° ≈ 0.469.
- 3.n ≈ 0.707/0.469 ≈ 1.51.
- 4.A value above one is reasonable because the ray enters an optically denser medium and bends toward the normal.
Problem
A ray strikes a glass slab normally. Describe its complete path.
- 1.Here i = 0° at the first face.
- 2.The ray continues along the normal without directional deviation, though its speed changes.
- 3.It also meets the second parallel face normally and emerges undeviated.
- 4.There is no lateral displacement for this normal path.
Problem
Light travels from glass of refractive index 1.50 into air at an incidence angle of 30°. Estimate the refraction angle.
- 1.For glass to air, the relative index of air with respect to glass is 1.00/1.50 = 0.667.
- 2.Use sin i/sin r = 0.667, so sin r = sin 30°/0.667.
- 3.sin r = 0.500/0.667 ≈ 0.750, giving r ≈ 48.6°.
- 4.The refracted angle is larger than the incident angle, which agrees with bending away from the normal on entering the optically rarer air.
Activity
Fix paper to a board and trace the outline of a rectangular slab. Place two vertical pins along an oblique incident line. Looking through the opposite face, place two more pins so that all four appear collinear. Remove slab and pins, join the pin points, draw normals at both faces, and label the three ray segments. Compare i and r and verify that the emergent and incident directions are parallel.
Quiz
How does light bend from air into glass?
Why is the emergent ray parallel to the incident ray?
What is lateral displacement?
In Snell's law, angles are measured from what?
What happens at normal incidence?
Practice Problems
- Name the three ray segments through a slab. Solution: Incident ray before entry, refracted ray inside glass and emergent ray after exit.
- Explain the two bending directions. Solution: Air to glass slows light and bends it toward the normal; glass to air speeds it and bends it away.
- For i = 30° and r = 19.5°, estimate n. Solution: n = sin30°/sin19.5° ≈ 0.500/0.334 ≈ 1.50.
- Why is the emergent ray shifted even though it is parallel? Solution: The ray travels along a changed direction inside the finite-thickness slab before the second face restores its original direction.
- Predict the effect of increasing slab thickness at the same incidence angle. Solution: The refracted ray travels farther sideways inside glass, so lateral displacement generally increases.
Key Takeaways
• A ray bends toward the normal on entering an optically denser medium. • It bends away on entering an optically rarer medium. • Parallel slab faces produce a parallel emergent ray. • The emergent ray is laterally displaced from the original line. • Snell's law connects incidence and refraction angles for fixed media and colour. • All refraction angles are measured from the normal.