Lecture 24 Chapter 34
Monday 13 March 2006
Optics
• Last time: Reflection and refraction
• Today: Image formation by lenses and mirrors
• Quiz Thursday: Chs. 33, 34
Optics Quiz on Thursday
• Ch. 33, Light waves.
– Review waves: fλ = v, y = ym sinωt
– Refraction: n = c/v = λ0/λ
– Snell’s Law: n1 sinθ1 = n2 sinθ2
• Ch. 34, Thin lenses, spherical mirrors.
–
–
–
–
IMAGES
Real and virtual images
Principal rays
Image location: 1/p + 1/i = 1/f
Magnification: m = -i/p
Spherical Mirrors
REAL IMAGE:
The light is really concentrated,
such as when you start a fire using sunlight and a lens.
• Focal point
In formulas the image distance i is positive.
• Focal distance
VIRTUAL IMAGE: The light only appears to
come from it, as when you seem to see your face behind
the bathroom mirror.
• Concave mirror is
converging: f > 0.
In formulas the image distance i is negative.
Images in Concave
Mirrors
• Convex mirror is
diverging: f < 0.
Locating the Images
Principal Rays for concave mirror
Let p = object distance,
let i = image distance.
Let p,i be positive for
real, negative for virtual.
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Lecture 24 Chapter 34
Monday 13 March 2006
Locating the Images
Principal Rays for convex mirror
Equations for the Images
** Small-angle Approximation **
Summarize Results:
1 1 1
+ =
p i f
Also: let magnification m be defined as
image size divided by object size. Then:
m=−
f = 1/ p + 1/ i
f = +6 cm
p = +15 cm
Negative m means
inverted image!
f = +6 cm
Example 1 solution
Spherical mirror example 1
An object is placed 15 cm in 1 /
front of a concave mirror
with focal length 6 cm.
Describe the resulting image.
i
p
p = +15 cm
Method 1: Principal Rays
O
•Real
•Reduced
•Inverted
F
I
6 cm
15 cm
O
F
6 cm
Method 2: 1 = 1 − 1 = 5 − 2 = 1 so i = +10 cm
i 6 15 30 30 10
m = − i / p = −10 / 15 = −2 / 3
15 cm
Example 2 Solution
Spherical mirror example 2
An object is placed 5 cm in
front of a concave mirror
with focal length 10 cm.
Describe the resulting image.
Method 1: Principal Rays
f = +10 cm
p = +5 cm
F
O
5 cm
F
O
5 cm
10 cm
•Virtual
•Enlarged
•Erect
I
10 cm
1
1
Method 2: = − 1 = 1 − 1 = − 1 so i = −10 cm
i f p 10 5
10
m = − i / p = − ( −10) / 5 = +2
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Lecture 24 Chapter 34
Monday 13 March 2006
Images in Thin Lenses
Thin Lenses
• Focal point
• Focal distance
• Convex lens is
converging: f > 0.
Let p = object distance,
let i = image distance.
• Concave lens is
diverging: f < 0.
Let p,i be positive for
real, negative for virtual.
Finding Images: Principal Rays
Locating the
Images
1. In parallel to axis: out through F.
Just as with the
mirrors, follow the
principal rays to
locate image
graphically.
Finding Images: Principal Rays
O
F
F
O
F
F
Finding Images: Principal Rays
3. Straight through center
2. In through F, out parallel
F
O
F
F
O
F
F
O
F
F
O
F
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Lecture 24 Chapter 34
Monday 13 March 2006
Equations for the Images
Thin-Lens Example 1
Given f=+20cm, p=+40cm, find i, m
Results in small-angle approximation:
I
1 1 1
+ =
p i f
i
m=−
p
O
F
F
1 1 1
1
1
1
= − =
−
=
i f p 20 40 40
Same equations as for spherical mirrors!
m=−
i
40
=−
= −1
p
40
†
Example 2
so i = +40cm
Image is:
• Real
• Inverted
• Same size as object
Example 3
Given f=+20cm, p=+10cm, find i, m
Given f=-30cm, p=+20cm, find i, m
I
F
O
F
1 1 1
1
1
1
= − =
− =−
i f p 20 10
20
so i = −20cm
†
Given f=-10cm, p=-4cm, find i, m
O
1 1 1
1
1
3
= − =
−
=+
i f p ( −10) ( −4)
20
i
20 / 3
20
m=− =−
=+
= +1.7
p
( −4)
12
I
1 1 1
1
1
5
= − =
−
=−
i f p ( −30) 20
60
i
( −12)
m=− =−
= +0.6
p
20
F
so i = −12cm
Image is:
• Virtual
• Erect
• Reduced
Recap: Formulas for spherical mirrors and
thin lenses in the small angle approximation
Example 4
F
I
†
Image is:
• Virtual
• Erect
• Enlarged
i
( −20)
m=− =−
= +2
p
10
O
F
1 1 1
+ =
p i f
F
so i = +6.7cm
Image is:
• Real
• Erect
• Enlarged
m=−
i
p
Definitions and sign conventions:
• f = focal length: + = converging, – = diverging
• p = object distance: + = real, – = virtual
• i = image distance: + = real, – = virtual
• m = magnification: + = erect, – = inverted
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