Reflection vs. Refraction When light strikes a mirror, it is reflected. The law of reflection tells us that the reflected angle is equal to the incident.

Reflection vs. Refraction• When light strikes a mirror, it is

reflected.• The law of reflection tells us that the

reflected angle is equal to the incident angle.

Reflection vs. Refraction• When light strikes the

interface between two transparent materials, the light divides into two parts.• Some is reflected (θr = θi)• The rest is transmitted

across the interface.

Reflection vs. Refraction• When a light ray enters

the second material, it changes direction – it is refracted.• The transmitted ray has a

different direction than the incident ray.

Index of Refraction• Light travels through a vacuum at a

speed of c = 3.00 x 108 m/s• As light travels through other

materials (air, water, glass, etc.) atoms in the material absorb, reemit, and scatter the light.• This results in light traveling through

the material at a speed less than c.

Index of Refraction• The index of refraction (n) describes

the extent to which the speed of light in a particular material differs from that in a vacuum.

Example 1

• = 1.24x108 m/s

Snell’s Law of Refraction• Light going from smaller n

to larger n, ray is bent towards the normal.• Light going from larger n

to smaller n, ray is bent away from the normal.

1 1 2 2sin sinn n Angle of incidence Angle of refraction

Example 2• A light ray strikes an air/water surface at an angle of 46° w/r/t the normal. The refractive index for water is 1.33. Find the angle of refraction when the direction of the ray is (a) from air to water and (b) from water to air.• A) n1 = 1.00 (incident ray is in the air)θ1 = 46 °n2 = 1.33 (refractive index of water)θ2 = ?

Example 2 part An1 = 1.00 (incident ray is in the air)θ1 = 46 °n2 = 1.33 (refractive index of water)θ2 = ?AirWater

46°

Ɵ2n1sinƟ1 = n2sinƟ2

Ɵ2=33°

Example 2 part An1 = 1.33 (incident ray is in the water)θ1 = 46 °n2 = 1.00 (refractive index of air)θ2 = ?WaterAir

46°

Ɵ2n1sinƟ1 = n2sinƟ2

Ɵ2=74°

Mini-lab: Pencil in water

• Fill the beaker halfway with water

• Hold the pencil vertically in the center of the beaker

• View the pencil from the side with the beaker at eye level

• Slowly move the pencil towards the side of the beaker.

Mini-lab: Pencil in water

• Fill the beaker halfway with water

• Hold the pencil vertically in the center of the beaker

• View the pencil from the side with the beaker at eye level

• Slowly move the pencil towards the side of the beaker.

Apparent Depth• An object lying under water

appears to be closer to the surface than it actually is because light is refracted away from the normal when it enter the air.

• The eye sees the virtual image at the point where the refracted rays are extended back into the water (dashed lines)

Apparent Depth Example

• A searchlight on a yacht is being used at night to illuminate a sunken chest. At what angle of incidence Ɵ1 should the light be aimed?

Example 2

d’ = d()A swimmer is treading water (n=1.33) at the surface of a pool 3.00 m deep. She sees a coin on the bottom directly below. How deep does the coin appear to be?

Displacement of light by a transparent slab

• When light strikes a transparent slab with parallel surfaces, the emergent ray is parallel to the incident ray, just displaced from it.• Apply Snell’s Law at each

interface to prove it !

Assignment

Focus p. 834 #1, 5Problems p. 835 #1-3, 9, 10, 12, 16