Physics 1C
Lecture 25B
"There are two ways of spreading light: to be
the candle or the mirror that reflects it."
--Edith Wharton
Refraction of Light
When light passes from one medium to another, it
is refracted because the speed of light is
different in the two media.
The index of refraction, n, of a medium can be
defined:
n is a unitless ratio
For a vacuum, n=1 (exactly).
For all other media, n>1.
Indices of Refraction
Frequency Between Media
As light travels from one medium to another, its
frequency does not change.
Both the wave speed
and the wavelength do
change.
The wavefronts do not
pile up, nor are created
nor destroyed at the
boundary, so the
frequency must stay the
same.
λ1 n 2
=
λ2 n1
Snell’s Law
“Fast to slow, light bends towards the normal.”
Or think of a car axle: one wheel stuck in mud
(slow), other wheel on pavement (fast) pivots
around.
“Slow to fast,
light bends
away from the
normal.”
This is the
basis for
Snell’s Law.
Refraction of Light
The angle of refraction depends upon the
material (by index of refraction) and the angle of
incidence:
n1 sin θ1 = n 2 sin θ 2
where θ1 is the angle of
incidence (30o in the
diagram).
θ2 is the angle of refraction.
n1, n2 are the indices of
refraction of the first and
second media, respectively.
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Snell’s Law
For example, for this specific case (air to glass
with an incident angle of 30.0o):
n1 sin θ1 = n 2 sin θ 2
n1
sin θ 2 = sin θ1
n2
⎛
⎞
−1 n1
θ 2 = sin ⎜ sinθ1 ⎟
⎝ n 2
⎠
⎛ 1.0003
⎞
θ 2 = sin ⎜
sin 30°⎟
⎝ 1.52
⎠
−1
θ 2 = sin−1 (0.329) = 19.2°
Refraction of Light
What if instead of going from air to glass
with an incident angle of 30.0o, it went
from glass to air with the same incident
angle:
n1 sin θ1 = n 2 sin θ 2
⎛
⎞
n
θ 2 = sin−1⎜ 1 sinθ1 ⎟
⎝ n 2
⎠
€
n1
sin θ 2 = sin θ1
n2
⎛ 1.52
⎞
θ 2 = sin ⎜
sin 30°⎟
⎝ 1.0003
⎠
−1
θ 2 = sin−1 (0.760) = 49.4°
€
Here the angle increased
(slow to fast).
€
Refraction of Light
What if instead having an incident angle of 30.0o
(from glass to air) it had an incident angle of
60.0o (still glass to air):
n1 sin θ1 = n 2 sin θ 2
n1
sin θ 2 = sin θ1
n2
⎛
⎞
n
−1
1
⎞
−1⎛ 1.52
θ 2 = sin ⎜ sinθ1 ⎟
θ 2 = sin ⎜
sin60°⎟
⎝ n 2
⎠
⎝ 1.0003
⎠
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−1
θ 2 = sin (1.32) = ?????
Here the angle doesn’t exist. The angle was so
€
great that it refracted
the incident ray back to
the first medium (reflected?).
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Total Internal Reflection
This is known as Total Internal Reflection (TIR).
It can only occur if you move from a slow
medium to a fast medium such that the refracted
ray is bent away from the normal compared to
the incident ray.
Here you can see light
ray 5 undergoing total
internal reflection.
This means that at
angle that light ray 5
hits at, no light enters
the second medium.
Total Internal Reflection
We define the critical angle as a particular angle
of incidence that will result in an angle of
refraction of 90o.
For angles of
incidence greater than
the critical angle, the
beam is entirely
reflected at the
boundary (TIR).
This ray will obey the
Law of Reflection at
the surface boundary.
Conceptual Question 25B-1
A light ray travels from medium 1 to medium 3 as
shown in the figure below. What can we say about
the relationship between the index of refraction
for medium 1 (n1) and the index of refraction for
medium 3 (n3)?
A) n3 > n1.
B) n3 = n1.
C) n3 < n1.
D) We can’t
compare n1 to
n3 without
knowing n2.
Conceptual Question Answer
A light ray travels from medium 1 to medium 3 as
shown in the figure below. What can we say about
the relationship between the index of refraction
for medium 1 (n1) and the index of refraction for
medium 3 (n3)?
Use Snell’s Law twice:
Equation the two n2:
30o
Prisms
Example
An incident ray in air is headed straight towards
an equilateral plastic prism (n=1.50). The ray is
parallel to the bottom of the prism. Use Snell’s
Law to find the angle (with respect to the normal)
that the light ray exits the prism on the right.
Answer
The diagram is given but it is up to you to draw the
normal(s) and path of the ray.
Prisms
Answer
θ 2 = 19.47°
Start with:
30o
o
60
€
θ2
60o
Thus, the incident angle is 30o from air to plastic. Using
Snell’s Law we find:
n1 sin θ1 = n 2 sin θ 2
⎛ n1
⎞
θ 2 = sin ⎜ sinθ1 ⎟
⎝ n 2
⎠
−1
€
⎛ 1
⎞
θ 2 = sin ⎜ sin 30°⎟ = sin−1 (0.333)
⎝ 1.5
⎠
−1
€
Prisms
Answer
Now, we need to look at
the second boundary.
We can now examine the
small top triangle created
by the ray in the prism.
The bottom left angle on
this triangle will be:
70.53o
49.47o
60o
θ2
90° −19.47° = 70.53°
This means that the bottom right angle of the triangle
will be:
180° − 70.53° − 60° = 49.47°
€
Prisms
Answer
Is the 49.47o, the angle we
will use in Snell’s Law?
No, it is not with respect to
the normal.
θ 3 = 40.53°
We draw the normal and
find:
49.47o
60o
θ3
90° − 49.47° = 40.53°
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At the second boundary, will this ray be refracted or
totally internally reflected?
Check by calculating the critical angle.
n1
sin θ c =
n2
⎛ 1 ⎞
θ c = sin ⎜ ⎟ = sin−1 (0.667)
⎝ 1.5 ⎠
−1
refracted,
barely
θ c = 41.81°
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Prisms
Answer
Now we need to calculate
the outgoing ray with Snell’s
Law again:
θ 3 = 40.53°
n 2 sin θ 3 = n1 sin θ 4
⎛
⎞
n
θ 4 = sin−1⎜ 2 sinθ 3 ⎟
⎝ n1
⎠
θ3
€
⎛ 1.5
⎞
−1
θ 4 = sin ⎜ sin 40.53°⎟ = sin (0.975)
⎝ 1
⎠
−1
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θ 4 = 77.10°
This is the outgoing angle with
respect to the normal
θ4
For Next Time (FNT)
Start reading Chapter 26
Finish working on the homework
for Chapter 25