Light Reflection and Refraction: Exploring the Behavior of Light Rays at Interfaces

1. When a light ray hits the interface between two different materials, some is reflected and some transmitted. Incident ray Ray reflected from top surface Ray transmitted into block Ray reflected from bottom surface Ray transmitted back into air Ray transmitted at bottom surface (very weak – not shown in text) Ray reflected into block at top surface (very weak and not shown in text – it splits into even weaker transmitted and reflected rays at side) Note that the transmitted ray changes direction: refraction. 7 6 Some of light passing through a material may be absorbed or scattered off in different directions. If almost all passes through without absorption or scattering: transparent If almost all is absorbed or scattered: opaque If some passes through: translucent Shiny objects (e.g. clean metals) reflect most of the light that hits them.

2. A beam of light is a collection of nearby rays. To see a beam, it must enter your eye!

3. A beam of light is a collection of nearby rays. To see a beam, it must enter your eye! Here, the beams of light are partly scattered by water drops (or dust of smog) in the air, so that they can be seen from the side.

4. Beam Scattered (or Reflected) by Particles Beam from a flash light The original beam serves as a light source. Particles that are in the beam (e.g. dust, water drops) may reflect the light. What we are seeing are the particles reflecting light to your eye. If there are a lot of them, they can outline the beam.

5. Reflection A rough surface will reflect each ray in a different direction. A smooth surface will reflect the rays in well defined directions. A mirror has a smooth, shiny surface. A movie screen is shiny (i.e. reflects light). Is it smooth or rough?

6. How to determine the direction of reflection from a smooth, shiny surface (e.g. a mirror)? If you know 1 (the direction of the incident ray with respect to the normal), how do you find 1’ (the direction of the reflected ray with respect to the normal)? Proof: Use Huygen’s Principle: All points on a wave front act as point sources for spherical waves (wavelets). The new position of the wave front is the surface connecting (tangent to) the wavelets.

7. Huygen’sPrinciple: • All points on a wave front act as point sources for spherical waves (wavelets). The new position of the wave front is the surface connecting (tangent to) the wavelets.

8. Reflection from a smooth surface Consider the beam with two parallel incident rays (1 and 2) which hit the surface at angle 1 and reflect at 1’. We want to find 1’. Ray 1 hits the surface t before ray 2,so when ray 2 hits the surface, wavelets from ray 1 will be on the sphere passing through D. Since the speed of the reflected and incident rays are equal, AD = BC. sin 1’ = cos ’ = AD/AC = BC/AC = cos  Therefore ’ =  and 1 = 1’.  angle of reflection = angle of incidence [angles measured with respect to normal and are always between 0o (normal incidence) and 90o(grazing incidence)]

9. angle of reflection (1’) = angle of incidence (1)

10. Consider two perpendicular mirrors: 1 = 1’  1 = 1’ But 2 = 90o - 1’  2 = 90o - 2= 1’ 2’ = 2 = 1’ = 1 Outgoing ray is parallel to incoming ray (true for any incoming 1 !!) 2’ 2’ 2 2 1 1’ 1 1’ If third perpendicular mirror (plane of screen), this would be true in 3 dimensions: Incoming light always reflected back to source, no matter what angle it comes in: retroreflection. (Retroreflector placed on moon to measure its distance from earth, on automobile tail lights, stop signs …)

11. Refraction: Change in direction of ray when passing between two materials with different light velocities [v =()-1/2: usually due to change in ] What determines the change in angle when light ray passes through a surface separating two materials, in which speeds are v1 and v2? V1 V2

12. Use Huygen’s principle: Consider two parallel rays. Suppose ray 1 hits the surface t before ray 2: BC = v1 t In the same time ray one will have advanced AD = v2 t, so AD/v2 = t = BC/v1 But AD = AC sin 2 and BC = AC sin 1. Therefore sin 2 / v2 = sin 1 / v1

13. sin 2 / v2 = sin 1 / v1 • Define index of refraction: • n  c/v • Snell’s law: • n2 sin 2 = n1 sin1 V1 = c/n1 V2 = c/n2 • Notes: • v  c, so n  1. • Angles defined with respect to normal • and are between 0o and 90o. • If normal incidence: 1 = 0, so 2 = 0 • Ray bends away from normal • (i.e. q2 > q1) when passing from larger • n to smaller (e.g. water, glass to air). • Ray bends toward normal • (i.e. q2< q1) when passing from • smaller n to larger (e.g. air to water, glass)

15. Refraction occurs because light travels with different speed in the two media. It is analogous to a marching band changing speed (e.g. going from pavement to mud) – when they hit the slower medium (mud), they must change their angle to stay in formation: refracted incident