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1
32 Optical Images
• image formation
• reflection & refraction
• mirror & lens equations
• Human eye
• Spherical aberration
• Chromatic aberration
2
image formation
• real image: rays converge to a pointEx. sun’s rays focused by magnifier
• virtual image: apparent source of light divergence Ex. image seen in mirror
3
Plane Mirror Image
• distances measured from mirror & axis
• image height hi = object height ho
• object distance do = image distance di
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concave mirror images
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convex mirror images
• always “virtual”, “upright”, “diminished”
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lens images
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Lens & Mirror Equation
• relate do, di & f.
• lateral magnification (LM) = hi/ho.
you must learn the sign conventions to use these formulas
fdd io
111
o
i
o
i
d
d
h
hLM
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imbedded object
• image distance < object distance
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lens power
• power = 1/(focal-length(meters))
• unit: [D, diopters, 1/m]
• used for corrective lenses
• Ex. near-sighted person P = -5.0Df = 1/P = 1/(-5) = -0.2m = -20cm.
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human eye• average index of refraction ~ 1.4
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Far Sighted Eye
• correction requires converging lens
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near sighted eye
• correction requires diverging lens
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Spherical Aberration
• Spherical aberration: light striking near edge focus at different points than light striking near center
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ChromaticAberration
• Chromatic aberration: different colors focus at different points
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Summary
• Image formation by lenses & mirrors
• Real/virtual, orientation, magnification.
• Near & far-sighted problem & correction
• Spherical & chromatic aberration
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2. A document is sealed in a glass cube (n=1.5) at a depth of 10cm from one surface. Calculate s’ and m.
5.100.11
10
5.1
s
00.1)10(1
)67.6(5.1
2
1
sn
snm
Object is in n1=1.5. Viewer is in air n2=1.00. Radius of surface is r = infinity.
1/s’ = -0.15 s’ = -6.67cm.
.
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OI
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Question: Reverse the designations in the example above. Object is now the fish, so n1 = 1.33 (fish location). Let the fish be 10cm from bowl surface. Where is the image of the fish formed that the cat will see?
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33.100.100.1
10
33.1
s
197.1)10(00.1
)00.9(33.1
2
1
sn
snm
Answer: s = 10cm , s’ = ?, r = -15cm, n1 = 1.33, n2 = 1.00.
0.133 + 1/s’ = 0.022 1/s’ = -0.111 s’ = -9.00cm
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Calculating Focal Length for a Thin Lens
cmrr
nf
1215
1
10
1)15.1(
11)1(
1
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Example: A double convex, thin glass lens with n = 1.5 has radii of curvature of 10 and 15cm. Find its focal length.
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21
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r
nn
s
n
s
n 1221
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00.133.133.1
10
1
s
s, cat’s nose
n1 holds object, here equal to 1.00
0.1 +1.33/s’= 0.022
1.33/s’ = -0.078
s’ = -17.1cm
sn
snm
2
1
29.1)10(33.1
)1.17(00.1
m
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