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Ray Model of Light
•Treats light as a straight ray
Geometric (Ray) Optics
Angle of incidence = Angle of Reflection
Chapter 23
Rough vs. Smooth Surfaces
Rough surfaces
Smooth Surfaces
Appear opaque
Can see at various points
Mirror-like
Can see reflection only at qr
Specular reflection
Plane Mirror
• Image distance = object distance
• Virtual image
• Real Images
– Light does pass through image point
– Can photograph the image at image position
– Image appears to come from behind mirror
– Cannot photograph the image at that position
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Flat Mirror: Ex 1
A woman 1.60 m tall stands in front of a mirror.
What must height must the mirror be, and how
high must it be off the floor? Assume that her
eyes are 10.0 cm from the top of her head
Top of head
• Top of mirror (Pt F) can be 5 cm below the top of
her head = 5 cm
Feet
• To see feet = ½ eyes to floor
• ½(160cm – 10cm) = 75 cm
Total Length of Mirror
• 75 + 5 = 80 cm
From Floor
• 75 cm
Mirror Equations
f=r
2
1 + 1 = 1
do di
f
m = h i = - di
ho
do
Concave Mirror
Convex Mirror
Magnifies
Shaving/Make-up mirror
Outside focal length=
inverted, real image
Inside focal length =
upright, virtual image
Demagnifies
Truck/shoplifting mirrors
Always produces a virtual,
upright image
Concave: Outside focal pt
(real, inverted)
(r = radius of curvature)
Concave: Inside focal pt
(virtual, upright)
Convex:
Virtual, upright
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Concave Mirror
di = +60 cm (real, same side as object)
m = -2
(inverted)
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di = infinity
di = -60 cm (virtual image, opposite side of mirror)
m = +4
(upright)
Convex Mirror
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di = -9.4 cm (virtual image, opposite side of mirror)
m = +0.375 (upright)
di = -7.5 cm (virtual image, opposite side of mirror)
m = +0.5
(upright)
Sign Conventions for Mirrors
f and r
+ Concave
f and r
- Convex
do
+ (must be in front of surface)
di
+ In front of silver surface (real)
di
- In back of silver surface (virtual)
m
+ Upright
m
- Inverted
di = -6.0 cm (virtual image, opposite side of mirror)
m = +0.6
(upright)
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Concave: Ex 1
A 1.50 cm high olive is placed 20.0 cm from a
concave mirror (r= 30.0 cm). Find the position of
the image and its size.
f = r/2 = 30.0 cm/2 = 15.0 cm
1 + 1 = 1
do di
f
Concave: Ex 2
A 1.00 cm object is placed 10.0 cm from a concave
mirror (r=30.0 cm) What is the position and
magnification of the object?
Convex: Ex 1
A convex rearview mirror has a radius of -40.0 cm.
What is the location and magnification of a
vehicle that is 10.0 m from the mirror?
1
+
20.0 cm
1
=
di
1
15.0 cm
di = 60.0 cm
m = -di
=
-60.0 cm =
-3
do
20.0 cm
m = hi
ho
hi = mho = (-3)(1.50 cm) = -4.5 cm (real, inverted)
f = r/2 = 30.0 cm/2 = 15.0 cm
1 + 1 = 1
do
di
f
1
+
10.0 cm
1 = 1
di
30.0 cm
di = -30.0 cm
(virtual image behind mirror)
m = m = -di =
do
-(-30.0 cm)
10.0 cm
m = +3
(upright, enlarged image)
f = r/2 = 0.400 m/2 = -0.200 m
1 +
do
1 =
di
1
+
10.0 m
1
f
1 =
di
1
-0.200 m
di = -0.196 m (19.6 cm behind mirror)
m = -di =
do
-(-0.196 m)
10.0 m
m = 0.0196
(upright, demagnified image)
(this is about 1/51st )
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Index of Refraction
•
•
•
•
Light slows when passing through a substance
Must be absorbed and re-emitted
Eyes slow light by ~30%
Bose-Einstein condensate (50 nanokelvins) v = 38
mph
v = speed in material
n = index of refraction
Refraction: Ex 1
Calculate the speed of light in water
Snell’s Law
n1sinq1 = n2sinq2
if n2 > n1
bends toward normal
v = c/n
v = (3.00 X 108 m/s)(1.33) = 2.26 X 108 m/s
if n1 > n2
bends away from normal
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If you shine a flashlight into a pool at 60.0o to the
normal, what angle will it take in the water?
Snell’s Law: Graphing Activity
A laser is shined through a liquid, and the angle of
refraction was recorded. Using a graph,
determine the index of refraction.
Approach:
1. Graph q1 vs. q2 and see if there is a linear
relationship.
2. Graph sinq1 vs. sinq2 and see if there is a linear
relationship.
q2= 40.6
Angle of Incidence
(degrees)
0.00
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
n1sinq1 = n2sinq2
Angle of Refraction
(degrees)
0.00
7.5
14.9
22.1
28.9
35.2
40.6
45.0
47.8
Light strikes a pane of glass at 60.0o. What is the
angle of refraction in the glass (n = 1.50)?
sinq1 = n2sinq2
n1
y
=mx
+ b
m = slope = n2/n1
m ~ 1.3
qA = 35.3o
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What angle does it leave the glass, qB?
Total Internal Reflection
• Critical Angle
– Angle at which no light leaves the material
– Refracted angle is 90o
90o
qB = 60.0o
Angle is the same but beam is shifted
• Used in fiber optic cables
qC
• Can see edges of pond/pool from underwater
• Used for binoculars and camera optics
n1sinq1 = n2sinq2
n1sinq1 = n2sin(90o)
sinqC = n2
n1
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What is the critical angle for light leaving water
going into air?
Thin Lenses
• Converging
• Diverging
• Focal length is the same on
both sides, even if curvature
is different (eyeglasses)
qC = 49o
Converging Lens
• Focal Pt and
Focal Plane are
at same distance
Diverging lens
• pushes light rays apart
• Still has a focal point
Converging Lenses
• Forms a real, inverted image
• Used for farsightedness
• Camera lenses
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• Can project on a screen or put your eye at same
place as the screen
Diverging Lenses
• Forms a virtual, upright image
• Used for nearsightedness
Why is it virtual?
• Rays that form the image do not come from the
object itself
• Seem to come from the focal point.
diverging lens
converging lens
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Lens Equations
f=r
2
Sign Conventions for Lenses
(r = radius of curvature)
1 + 1 = 1
do di
f
m = h i = - di
ho
do
Converging Lens: Example 1
A 7.60 cm flower is placed 100.0 cm from a 5.00 cm
focal length lens. Calculate the image distance,
magnification and size of the image.
f and r
+ Converging
f and r
- Diverging
do
+ On side from which light comes
do
- On opposite side
di
+ Behind lens (real)
di
- In front of lens(virtual)
m
+ Upright
m
- Inverted
1 + 1 = 1
do di
f
1
+
100 cm
1 = 1
di
5.00 cm
di = 5.26 cm behind the lens (real)
m = -di/do = - 5.26/100 = -0.0526 (inverted)
hi = mho = (-0.0526)(7.6 cm) = -0.40 cm
Converging Lens: Example 2
Suppose the flower is now placed 2.00 cm from the
5.00 cm focal length lens. Calculate the image
distance and magnification.
2.00 cm
1 + 1 = 1
do di
f
1
+
2.00 cm
1 = 1
di
5.00 cm
di = -3.33 cm in front of the lens (virtual)
m = -di/do = - 3.33/2 = +1.67 (upright)
Acts as magnifier
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What is the focal length of a converging lens that
produces an image 1.6 cm of Mr. Saba riding his
narwhal very far away?
0 + 1
=
1.6cm
1
f
f = 1.6 cm
This is how the focal length of lens is measured,
using distant objects like buldings, trees, stars.
Diverging Lens: Ex 1
An insect is placed 100 cm from a -25 cm focal
length diverging lens. Calculate the image
distance and magnification.
object
distance
image
distance
m = --20/100 = +0.2
di = -20 cm
(virtual image)
Combinations of Lenses
• Microscopes, telescopes, zoom lenses
• Image distance of one lens becomes the object
distance of the second
• Total magnification = m1m2
Combination of Lenses: Ex 1
Two converging lenses (f1=20.0 cm, f2 = 25.0 cm) are
placed 80.0 cm apart. An object is placed 60.0 cm
in front of the first. What is the position and
magnification of the final image?
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1 + 1 = 1
do di
f
1
+
1 = 1
60.0cm
di
20.0cm
di1 = 30.0 cm
(real)
m1 = -30/60 = -0.5 (inverted)
do2 must equal 80 cm – 30 cm = 50 cm
1
+
1 = 1
50.0cm di
25.0cm
di2 = 50.0 cm
(real)
m2 = -50/50 = -1.00 (inverted from di1)
Combination of Lenses: Ex 2
A microscope is designed with two converging
lenses (f1=0.28 cm, f2 = 2.50 cm) that are placed
17.0 cm apart. An object is placed 0.290 cm in
front of the first. What is the position and
magnification of the final image?
mtotal = m1m2 = (-0.5)(-1) = +0.5
The final image is upright
1 + 1 = 1
do di
f
1
+
1 = 1
0.29cm di
0.28cm
di1 = 8.12 cm
m1 = -8.12/.29 = -28.0
do2 must equal 17.0 cm – 8.12 cm = 8.88 cm
(real)
(inverted)
1
+
1 = 1
8.88cm
di
2.50cm
di2 = 3.48 cm
m2 = -3.48/8.88 = -0.392
(real)
(inverted from di1)
mtotal = m1m2 = (-28)(-0.392) = 11.0
The final image is upright
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A diverging lens (f1=-15.0 cm) is placed in front of a
converging lens (f2 = +15.0 cm). The distance
between the lenses is 30.0 cm. An object is
placed 10.0 cm in front of the first lens. Calculate
the position and magnification of the final image?
A diverging lens (f1=-20.0 cm) is placed in front of a
converging lens (f2 = +25.0 cm). The distance
between the lenses is 30.0 cm. An object is
placed 30.0 cm in front of the first lens. Calculate
the position and magnification of the final image.
[ans: 25.7 cm, -0.43]
[ans: 61.8 cm, -0.59]
Combination of Lenses: Ex 3
1 + 1 = 1
do1 di1 f1
To measure the focal length of a diverging lens, a
converging lens is placed in contact with it. The
focal length of the converging lens is 16.0 cm. The
overall image is a 28.5 cm. The sun is used as the
object.
1 + 1
=
∞
di1
di1 = 16.0 cm
1 + 1 = 1
do2 di2 f2
1 + 1 = 1
-16 28.5 f2
1
16.0 cm
Lensmaker’s Equation
(the do becomes negative since it is
in back of the lens)
1 = (n-1) 1
f
R1
+
1
R2
R1 and R2 are positive for convex (converging) lenses
f2 = -36.4 cm
R1 and R2 are negative for concave (diverging) lense
R = ∞ for flat surfaces
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Lensmakers: Ex 1
1 = (n-1) 1
f
R1
+
1
R2
A lens is made from glass (n=1.50). One surface is
convex with a radius of curvature of 22.4 cm. The
concave surface has an r of 46.2 cm. What is the
focal length?
1 = (1.50-1)
f
1
22.4 cm
1
46.2
f = 87 cm
Lensmakers: Ex 2
Where will the lens focus an object 2.00 m away?
A lens is made from Lucite (n=1.51). One surface is
concave with a radius of curvature of -18.4 cm.
The other is flat. What is the focal length?
1 + 1 = 1
do di
f
1
+
2.00m
1 = 1
di
0.87 m
di = 1.53 m
1 = (1.51-1)
f
f = -36 cm
1
+
-18.4
0
1 = (n-1) 1
f
R1
+
1
R2
1 = (1.51-1)
f
1
+
-18.4
1
∞
Diopters
• Power of a lens (optometrists)
P=1
f
• Power is in diopeters (focal length must be in
meters)
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1.Calculate the power in diopters
of a 30.0 cm focal length lens.
(3.33 diopters)
2.Calculate the power in diopters
of a -25.0 cm focal length lens.
(-4 diopters)
3.Calculate the focal length of a
2.00 diopter lens. (50 cm)
4.Calculate the focal length of a
-4.00 diopter lens (-25 cm)
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L = 3.9 X 10-4 m
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Ch 23 Test Rev
+
13 m
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