Transcript Refraction of Light

Refraction of Light
Light travels in straight lines. However, when
light travels through different mediums, its
path is deviated. This deviation is called
refraction.
Each medium has its own level of refraction.
We can calculate the index of refraction, n, of a
medium using:
n = c/v
Where c is the speed of an electromagnetic
wave in a vacuum, 3.00 x 108 m/s
and v is the speed of light in that medium in m/s
Example
Using the table of values below, identify the
medium through which light travels when its
velocity is 8.57 x 107 m/s.
Medium
Water
Ethyl
alcohol
Plexiglass
Diamond
Gallium
phosphide
Index of
refraction
(n)
1.33
1.36
1.51
2.42
3.50
n = c/v
c= 3.00 x 108 m/s
v= 8.57 x 107 m/s
n = c/v
n= 3.00 x 108 m/s
8.57 x 107 m/s
n= 3.50, gallium phosphide
Note:
The greater the index of refraction of a medium,
the slower the speed at which light will travel
through it.
A relative index of refraction can also be
calculated when a light ray does not pass
through a vacuum.
The relative index of refraction is a ratio of the
index of refractions of the two mediums
involved.
Example:
Calculate the relative index of refraction of a
light ray that travels through water and then
through Plexiglass.
nR = n2 / n1
n1 : index of refraction of the first medium
n2: index of refraction of the second medium
nR = 1.51 / 1.33 = 1.14
Geometry of Refraction
Since light is travelling in
two different mediums,
the angle of incidence is
no longer equal to the
angel of refraction.
Laws of refraction:
1. The incident ray, the
normal and the
refracted ray are all on
the same plane.
Incident light ray
θi
n1
n2
θR
refracted light ray
2. The ratio of the sine of the angle of incidence
and refraction are equal to the ratio between
the index of refraction of the two mediums.
sin θi = n2
sin θR
n1
n1 sin θi = n2 sin θR
Example:
A light travel from a diamond (n=2.42) into
water (n=1.33). If the rays have an angle of
incidence of 30o, calculate the angle of
refraction.
n1 sin θi = n2 sin θR
2.42 x sin (30) = 1.33 x sin θR
θR = 65o
EXAMPLE
A light ray travelling through air encounters
a glass cup with an angle of incidence of
35o. The light ray then travels through
water and finally through ice. Determine
the final angle of refraction once the light
ray exits the ice.
glass
water
ice
Total Internal Reflection
When a light ray travels in a medium with a high
refractive index and encounters one with a
low refractive index, refraction does NOT
always occur; the light ray can be reflected
instead.
The angle of incidence will determine whether
the ray is reflected of refracted.
If the angle of incidence is greater than or equal
to the critical angle, then internal reflection
occurs. If not the ray will be refracted into the
second medium.
medium 2
medium 1
We can determine the value of the critical angle
by using
θC = sin-1 (n2/n1)
Where θC is the critical angle
n1 is the refractive index of medium 1
n1 is the refractive index of medium 2
Example
An aquarium has a built in
light source. The light rays
hit the aquarium’s glass
walls at an angle of
incidence of 65o.
a. Determine whether reflection or refraction
occurs.
b. At what angle will the light ray reflect/refract?