Refraction
Last Time…
Total internal reflection
•  Refraction occurs when light moves into
medium with different index of refraction.
•  Result: light direction bends according to
Snell’s law
n1 sin θ1 = n 2 sin θ 2
θi,1
θr
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Reflection and
refraction
n1
n2
θ2
Angle of refraction
Why Snell’s law?
Refraction angle
n1
n1
Reflected
ray
v2<v1
n2 >n1
•  Light moves at different speed in media with different
index of refraction.
θi
Reflected
ray
n1
v2>v1
n2 < n1
slower in medium 2
faster in medium 2
n2>n1
Refracted ray bent
toward normal
n2<n1
Refracted ray bent
away from normal
n2>n1
Which of these fluids has the
smallest index of refraction
(highest light speed)?
A
B
A.  Fluid A
C.  Fluid C
C
v2<v1
θ2
Numerical Example
Quick quiz
B.  Fluid B
θr
A beam of light is traveling underwater, aimed up at the
surface at 45˚ away from the surface normal. Part of
it is reflected back into the water, and part is
transmitted into the air.
Air
n2=1.00
Water
n1=1.33
θ2
n1 sin θ1 = n 2 sin θ 2
n1
sin θ1 = 0.94
n2
n

θ 2 = arcsin 1 sin θ1  = 70˚
 n2

sin θ 2 =
θ1=45˚
D.  All equal
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Quick quiz
A trout looks up through
the surface at the
setting sun, and at the
moon directly
overhead.
He sees
Total Internal Reflection
n2=1.0
•  Is possible when light is directed from n1 > n2
⇒ refracted rays bend away from the normal
•  Critical angle: angle of
incidence that will result in
an angle of refraction of
90° (sinθ2 = 1)
n1=1.33
A.  Moon directly overhead, sun ~ parallel to water surface
B.  Moon directly overhead, sun ~ 40˚ above water surface
C.  Moon ~ 40˚ from vertical, sun ~ parallel to water surface
For water:
D.  Moon and sun aligned at 40˚ from vertical.
sin θ c =
1
= 0.75 ⇒ θ c = 48.75˚
1.333
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Optical Fibers
The cladding has a
lower n than the core
Light rays and images
•  Plastic or glass light pipes
•  Applications:
–  Medicine: endoscope (light can be
directed even if bent and the
surgeon can view areas in the body
using a camera.)
–  Telecommunications



Each point on object
reflects light
Light propagates out,
represented by rays
perpendicular to
wavefront.
Lens in our eye does
some ‘imaging’ so
that we identify origin
of light rays.
Mon. Feb. 4, 2008
Question
10
How do you ‘see’ this?
Does the fish appear
Object
A.  Closer than actual
n=1.00
B.  Farther than actual
Lens
C.  Same as actual
Image
n=1.33


Virtual
image
Mon. Feb. 4, 2008
Physics 208, Lecture 4
Physics 208, Lecture 4
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Lens bends (refracts) light rays— forms image on retina
Sensors on retina report to brain.
  Color information, intensity information
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Physics 208, Lecture 4
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How a lens works
Making a real image

Position surfaces to bend light rays in just the right way
Spherical surfaces are very close to the right ones.
Lens: Refracts light so that rays originating from
a point are focused to a point on the other side.
n1
n1
n2>n1
Optical Axis
F
This is a
real image
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Physics 208, Lecture 4
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Thin-lens approximation:
Ray tracing
F
Object
Optical
Axis
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Physics 208, Lecture 4
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Different object positions
Image
Image (real, inverted)
Object
F
1) Rays parallel to optical axis pass through focal point.
2) Rays through center of lens are not refracted.
Image (real, inverted)
3) Rays through F emerge parallel to optical axis.
Here image is real, inverted, enlarged
Image
(virtual, upright)
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Physics 208, Lecture 4
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These rays seem to originate
from tip of a ‘virtual’ arrow.
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Physics 208, Lecture 4
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Making an image
Quick Quiz
I project a focused image onto a screen 2
meters away. I now want to make the image
bigger without changing the lens. I should
A.  Move screen farther away only
Object distance
Image distance
s
s’
Object
B.  Move screen closer only
Image
C.  Move screen closer and object toward lens
D.  Move screen farther and object toward lens
f
E.  Move screen farther and object away from lens
f
How are all these related?
1 1 1
+ =
s s′ f
focal length
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Physics 208, Lecture 4
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€
3
Image size vs object size:
Magnification
Your eye can change focal length
Object
1 1 1
+ =
s s′ f

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Object
height
Lens
Image distance
Object
dist.
Image
What is range of focal lengths if it can focus from
near point (25 cm) to inf. onto retina 1.7 cm away?
Image height
image height s′ image distance
= =
object height s object distance = M = Magnification
1
1
1
+
= ⇒ f = 1.7cm
∞ 1.7cm f
1
1
1
+
= ⇒ f = 1.59cm
Object at near point:
25cm 1.7cm f
Object at infinity:
€
Very limited range
€
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Physics 208, Lecture 4
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Different object positions
Question

At what object distance does image size
equal object size (magnification=1)?
A.  Object distance = f
Image (real, inverted)
Object
1 1 1
+ =
s s′ f
B.  Object distance = 2f
C.  Object distance = f/2
D.  Object distance = infinity
E.  Object distance =0
Image (real, inverted)
€
Image
(virtual, upright)
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Physics 208, Lecture 4
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Mon. Feb. 4, 2008
These rays seem to originate
from tip of a ‘virtual’ arrow.
Physics 208, Lecture 4
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Virtual images
objects closer to a converging lens than
the focal length form a virtual image
Image
(virtual, upright)



Do these rays come from real image, a
virtual image, or an object?
These rays seem to originate
from tip of a ‘virtual’ arrow.
Virtual image


Can’t tell. Rays are exactly equivalent, and can be
imaged by a lens in exactly the same way.
can’t be recorded on film,
Can’t be seen on a screen.
But rays can be focused by another lens


e.g. lens in your eye (focus on retina)
e.g. lens in a camera (focus on film plane)
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