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.
•𝑛=
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑎 𝑣𝑎𝑐𝑢𝑢𝑚
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
=
𝑐
𝑣
Example 1
• If the index of refraction for diamond is n =
2.419, what is the speed of light in diamond?
• 𝑛=
𝑐
𝑣
• 𝑛𝑣 = 𝑐
• 𝑣=
• 𝑣=
𝑐
𝑛
3.00𝑥108 𝑚/𝑠
=
2.419
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.
n1 sin  1  n 2 sin  2
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 A
n1 = 1.00 (incident
ray is in the air)
θ1 = 46 °
n2 = 1.33 (refractive
index of water)
θ2 = ?
n1sinƟ1 = n2sinƟ2
Ɵ2=33°
46°
Air
Water
Ɵ2
Example 2 part A
n1 = 1.33 (incident
ray is in the water)
θ1 = 46 °
n2 = 1.00 (refractive
index of air)
θ2 = ?
n1sinƟ1 = n2sinƟ2
Ɵ2=74°
46°
Water
Air
Ɵ2
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
𝑛2
d( )
𝑛1
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, 5
Problems p. 835 #1-3, 9, 10, 12,
16