Today 2/3
Total Internal Reflection
No HW due Today
Lab: Refractive index with Pins and Blocks
Start reading chapter 26.6-8 on Lenses
HW:
2/3 “Refraction” Due Thursday, 2/6
Exam I Thursday, Feb 13
This Week’s Lab
Pins
“Images” will not line up
when the “objects” do
This Week’s Lab
Pins
“Objects” will not line up
when the “Images” do
Refraction-Bending Light
Air - index of
refraction, nA = 1.0
Think of rows in a
marching band that slow
down in a mud puddle.
Wave crests are
closer together in
higher index
materials because
the waves travel
slower.
Water - index of refraction, nW = 1.33
Refraction-Bending Light
Air - index of
refraction, nA = 1.0
Think of rows in a
marching band that slow
down in mud puddle.
When light enters at
an angle, it “bends.”
Water - index of refraction, nW = 1.33
Ray direction always
perpendicular to the
waves.
Ray diagram example
Air - index of
refraction nA = 1.0
Refraction makes it
seem as though its
closer
Image
Object
Water - index of refraction nW = 1.33
Ray diagram example
Air - index of
refraction nA = 1.0
Image
Object
Water - index of refraction nW = 1.33
Refraction makes it
seem as though its
farther
Example:
n1sin 1 = n2sin 2
Note notation: light from n1 to n2
When looking into a glass of water, where
(at what angle, 2) should I place my eye
so that I can just barely see the red dot?
2
n = 1.0
n = 1.33
d
tan 1 = 4/7 1 = 30°
1.33 sin 30° = 1.0 sin2 2 = 41.7
1
7cm tall
4cm
How deep does the water appear to be?
tan 2 = 4/d
d = 4/tan41.7 = 4.5cm
Total internal reflection
Air - index of
refraction nA = 1.0
nAsin A = nWsin W
As W increases, so does A
A
until A becomes 90°. C is
called the “critical angle”. If
W is greater that C no light
will escape the water and all
will be “internally reflected.”
Total internal reflection
C only happens when light
nAsin 90 = nWsin C or
goes from high n to low n
nA = nWsin C
W
and it depends on both n’s!
Water - index of refraction nW = 1.33
Example:
n2 = n1sin C
What is the critical angle for light
traveling from water into air?
n2 = 1.0
sin C = n2/n1 = 0.752
C = 48.8°
C
Note that there is no critical angle for
light from low index to higher index.
n1 = 1.33
Example:
n2 = n1sin C
What happens if the angle of incidence
is greater than 48.8°?
n2 = 1.0
50°
n1 = 1.33
1 = 50°
n2 = n1sin C
Example: n2sin 2 = n1sin 1
What happens if I place a sheet of glass on the water, ng = 1.5?
No internal reflection possible-low to high n!
n2 = 1.0
1.5sin g = 1.33sin50°
g = 42.8°
Now what happens?
g = 42.8
C = Arcsin1/1.5 = 41.8°
ng = 1.5
1 = 50
n1 = 1.33
The light reflected
back into the glass.