Geometrical Optics
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UW ASTR 597
surface normal
incident ray
same
angle
exit ray
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“real” you
A
too long
shortest path;
equal angles
B
mirror only
needs to be half as
high as you are tall. Your
image will be twice as far from you
as the mirror.
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“image” you
4
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Geometrical Optics
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A
n1 = 1.0
n2 = 1.5
Snell’s Law:
n1sinθ1 = n2sinθ2
θ1
A
B
C
road
dirt
θ2
D (house)
Lecture 5
dist.
5
16
20
15
12
Δt@5
1
3.2
—
—
—
AD: 6.67 minutes
ABD: 6.0 minutes: the optimal path is a “refracted” one
ACD: 7.2 minutes
B
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leg
AB
AC
AD
BD
CD
7
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Δt@3
—
—
6.67
5
4
Note: both right triangles in figure are 3-4-5
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Geometrical Optics
01/31/2008
UW ASTR 597
incoming ray hugs surface
n1 = 1.0
n2 = 1.5
Note that the wheels
move faster (bigger space)
on the sidewalk, slower
(closer) in the grass
42°
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n1 = 1.5 n2 = 1.0
incoming ray
(100%)
96%
8% reflected in two
reflections (front & back)
Wheel that hits sidewalk starts to go faster,
which turns the axle, until the upper wheel
re-enters the grass and goes straight again
4%
92% transmitted
4%
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Lecture 5
image looks displaced
due to jog
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0.16%
12
3
Geometrical Optics
01/31/2008
UW ASTR 597
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pinhole
object
A lens, with front and back curved surfaces, bends
light twice, each diverting incoming ray towards
centerline.
Follows laws of refraction at each surface.
Parallel rays, coming, for instance from a specific
direction (like a distant bird) are focused by a convex
(positive) lens to a focal point.
image at
film plane
In a pinhole camera, the hole is so small that light hitting any particular point
on the film plane must have come from a particular direction outside the camera
object
image at
film plane
lens
Placing film at this point would record an image of
the distant bird at a very specific spot on the film.
Lenses map incoming angles into positions in the
focal plane.
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In a camera with a lens, the same applies: that a point on the film plane
more-or-less corresponds to a direction outside the camera. Lenses have
the important advantage of collecting more light than the pinhole admits
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real foci
Lecture 5
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virtual foci
s=∞
s’ = f
s=∞
s’ = -f
s=f
s’ = ∞
s = -f
s’ = ∞
s=∞
s’ = f
s=∞
s’ = -f
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Geometrical Optics
01/31/2008
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Take the parabola:
y = x2
Slope is y’ = 2x
Curvature is y’’ = 2
So R = 1/y’’ = 0.5
Slope is 1 (45°) at:
x = 0.5; y = 0.25
So focus is at 0.25:
f = R/2
Note that pathlength to focus is the same for depicted ray and one along x = 0
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f=D
D
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f = 4D
D
f/1 beam: “fast”
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f/4 beam: “slow”
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neg. S.A.
lens side
zero S.A.
pos. S.A.
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