The Refraction
of Light, the
Index of
Refraction, and
Snell’s Law

The bending or change in direction
of light when it travels from one
medium to another.




Refraction is due to changes in the speed of
light.
Different media slow down light by different
amounts.
The more that light slows down the more the
light is refracted.
When light from the spoon
passes from the water into
the air the light rays are
bent.




1) The incident ray, refracted ray & normal all
lie in the same plane
2) When a ray of light passes into a slower,
more optically dense medium, the ray of light
bends towards the normal. (example air 
glass)
3) When a ray of light passes into a faster, less
optically dense medium, the ray of light bends
away from the normal. (example glass 
water)
4) When the angle of incidence is zero, there is
no refraction. (No change in direction, yes a
change in speed)
4
Rule #1: The incident ray, normal and
refracted ray are all in the same plane.
Rule #2: Light bends towards the normal
when its speed decreases in a material.
Medium 2 will have a
slower speed of light
Bends Towards
the Normal
6
Rule #3: Light bends away from the
normal when the speed increases
material.
Medium 2 will have
a greater speed of light
Bends away from
the Normal
7

Rule #4: When the angle of incidence is zero,
there is no refraction. (No change in direction,
yes a change in speed)
8
Recall: The Index of Refraction
• The refractive index of a medium is determined by comparing the
speed of light in the medium with the speed of light in a vacuum
Index of refraction of material =
speed of light in vacuum
speed of light in medium
Where:
• n is the index of refraction
• c is the speed of light (vacuum) (always 3.00x108 m/s)
• v is the speed of light in a given medium.
The index of refraction has no units because both m/s speeds cancel out.
The speed of light in glass is 1.52 x 108 m/s.
What is the Index of Refraction for glass??
Given:
c = 3.00 x 108 m/s
v = 1.52 x 108 m/s
n=?
n = 3.00 x 108 m/s
1.52 x 108 m/s
n = 1.96
Therefore, the index of refraction of glass is 1.96.
Snell’s Law
• A formula that uses values for the index of
refraction to calculate the new angle that a ray
will take as a beam of light strikes the
interface between two media.
11
Snell’s Law
• If you call the indices of refraction of the two
media n1 and n2 and call the angles of
incidence and the angle of refraction θ1 and θ2
then the formula for Snell’s law is:
n1sinθ1 = n2sinθ2
12
Getting a value for sin using your
Calculator
• Different calculators may require different
inputs for calculating sin
• Option 1 (for most):
– Press sin then the number corresponding to the
angle (i.e. for sin 30° you would type sin then 30)
• Option 2:
– Other calculators will require you to type the
number first then punch in sin (i.e. for sin 30° you
would type 30 then sin)
13
Calculating an Unknown Angle
• When using Snell’s law, you may end up in a situation where
you are looking for the angle only:
0.65 = sinθwater
• In such a case, you will have to isolate the angle from the sin
symbol in front of it.
• To do this you will have to move the sin over to the other side
of the equation where it will become sin -1 or the inverse of
sin
• To calculate this, on most calculators, you will need to press
second function and then the sin symbol (look for sin -1)
• As for finding sin, depending on your calculator you may have
to press sin -1 1st or you may need to punch in the angle first.
14
Example 1
• In an experiment, a block of cubic zirconia is placed in water. A laser
beam is passed from the water through the cubic zirconia. The
angle of incidence is 50o, and the angle of refraction is 27o. What is
the index of refraction of cubic zirconia?
• Given:
– nwater = 1.33
– Θwater= 50o
– Θzirc= 27o
• Required:
Analysis:
n1sinθ1 = n2sinθ2
nwatersinθwater = nzircsinθzirc
– nzirc= ?
15
Example 1
• Solve:
Paraphrase:
nwatersinθwater = nzircsinθzirc
1.33 X Sin 50 = nzirc sin 27°
nzirc = 1.33 X sin 50
sin 27°
nzirc = 1.33 X 0.76
0.45
nzirc = 2.25
The index of refraction for cubic
zirconia is 2.25
16
Example 2
• When light passes from air into water at an angle
of 60o from the normal, what is the angle of
refraction?
• Given:
– nair= 1.0003
– nwater = 1.33
– Θair = 60o
• Required:
Analysis:
n1sinθ1 = n2sinθ2
nairsinθair = nwatersinθwater
– Θwater = ?
17
• Solve:
Example 2
nairsinθair = nwatersinθwater
1.0003 X sin60° = 1.33 sinθwater
1.0003 X 0.87 = 1.33 sinθwater
0.87 = 1.33 sinθwater
0.87 = sinθwater
1.33
0.65 = sinθwater
sin-1 0.65 = θwater
θwater = 40.5°
Paraphrase: The angle of refraction is 40.5 °
18
Now…
• 1) Answer Snell’s Law Problems on worksheet
• 2) Determine the unknown substance!
• 3) Hand in at end of class