WileyPLUS Assignment 3
Chapters 21, 22, 24, 25
Due Thursday, March 11 at 11 pm
Week of March 16-18
Tutorial and Test 3: chapters 22, 24-26
PHYS 1030 Final Exam
Friday, April 23, 1:30-4:30 pm
Frank Kennedy Brown Gym
30 questions, the whole course
Formula Sheet Provided
Wednesday, March 10, 2010
81
Conventions for the thin lens equation
Draw ray diagrams with rays travelling from left to right.
Normal situation:
Object
Real object to
left of lens,
object distance
do is positive
Lens
Image
Real image to right
of lens,
image distance
di is positive
Virtual object to right of lens, do is negative (2 or more lenses)
Virtual image to left of lens, di is negative
Converging (positive) lens, focal length f is positive
Diverging (negative) lens, focal length f is negative
Power of a lens, P = 1/f, (diopters): focal length in metres
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82
The Human Eye
ACCOMMODATION: the ability of the eye to focus on objects at
different distances.
Eye lens has its
Ciliary muscle,
relaxed
longest focal length
FAR POINT: greatest distance at which eye can focus
Normal value: infinity
Nearsighted eye: Far point < infinity
Eye lens compressed,
focal length decreased
NEAR POINT: closest distance at which eye can focus
Normal value: N = 25 cm
Farsighted eye: Near point > 25 cm
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83
Correcting near and farsightedness
Nearsighted:
• The corrective lens forms a virtual image of a distant object at the
person’s far point, or closer.
f<0
Farsighted:
• The corrective lens forms a virtual image of a nearby object at
the person’s near point, or further.
f>0
Wednesday, March 10, 2010
84
Prob. 26.115/107: A nearsighted person cannot read a sign that is
more than 5.2 m from his eyes. He wears contact lenses that do not
correct his vision completely, but do allow him to read signs located
up to distances of 12 m from his eyes.
What is the focal length of the contacts?
f = -9.2 m, P = -0.11 diopter
Wednesday, March 10, 2010
85
Prob. 26.75/67: A farsighted person has a near point that is at 67 cm.
She wears eyeglasses that are designed to enable her to read a
newspaper held at a distance of 25 cm from her eyes.
Find the focal length of the eyeglasses, assuming that they are worn
2.2 cm from the eyes.
f = 35.2 cm, P = 2.84 diopter
Wednesday, March 10, 2010
86
Angular size, magnification
(near point, closest at which eye can focus)
Angular size of the object: ! =
ho
N
(small angle, in radians, object
viewed by unaided eye at near
point, N)
Angular size of image formed
by the magnifying glass:
θ� =
ho
do
Angular magnification, M:
!� ho N
N
M= = × =
! do ho do
The magnifying glass lets the user view the object closer than the near point
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87
Magnifying Glass
!� N
Angular magnification: M = =
! do
di = – !
Two cases:
�
1) Final image is at infinity: so do = f
Then: M =
N N
=
do
f
1
1
1
= −
do f −∞
�
⇐ minimum magnification
2) Final image is at the near point, so di = – N
Thin lens equation:
Then, M =
Wednesday, March 10, 2010
1
1
1
f +N
= −
=
do
f −N
fN
N
f +N
N
=
= 1+
do
f
f
⇐ maximum magnification
88
Magnification Markings
Lenses are sometimes marked with the magnification they
produce when an image is formed at infinity.
For example, “10!”.
This means that,
M=
N
= 10,
f
using N = 25 cm, the normal near point.
So, f =
N
25
=
= 2.5 cm
10
10
Wednesday, March 10, 2010
89
Prob. 26.88/112: A stamp collector is viewing a stamp with a
magnifying glass held next to her eye. Her near point is 25 cm from
her eye.
a) What is the refractive power of a magnifying glass that has an
angular magnification of 6 when the image of the stamp is located at
the near point?
b) What is the angular magnification when the image of the stamp is
45 cm from the eye?
a) P = 20 diopter
b) M = 5.6
Wednesday, March 10, 2010
90
Astronomical Telescope
fo
fe
di1 ! fo
do1 very large
Formation of intermediate image
by the objective.
(! > 0, hi < 0, image inverted)
The eyepiece acts as a
magnifying glass to produce a
magnified final image.
M=
do2 ! fe
!�
hi fo
fo
�− × =−
!
fe hi
fe (exact when object and final image are at infinity)
Wednesday, March 10, 2010
91
Astronomical Telescope
di1 = fo
Object and final image at infinity: di1 = fo, do2 = fe
Distance between lenses, L = fo + fe
do2 = fe
Angular magnification:
M = -fo/fe
Wednesday, March 10, 2010
92
Prob. 26.98/92: An astronomical telescope has an angular
magnification of –132 and uses an objective with a refractive
power of 1.5 diopters.
What is the refractive power of the eyepiece?
Angular magnification, M = −
fo
= −132
fe
1
f
Refractive power,
P =
so,
M =−
Pe
= −132 and Po = 1.5 diopters
Po
Therefore, Pe = 132 " Po = 132 " 1.5 = 198 diopters.
fe =
1
= 0.00505 m = 5.1 mm
Pe
Wednesday, March 10, 2010
93
Opera Glasses
Like an astronomical telescope but with an eyepiece that is a
diverging (negative) lens.
The distance between the lenses is still:
L = fo + fe and the angular magnification is
M = – fo/fe
But, as fe < 0:
• the length, L = fo + fe, is less than an astronomical telescope of
the same magnification
• the image is the right way up (M > 0)
Wednesday, March 10, 2010
94
Compound Microscope
do1 " fo
!�
fo
do2 " fe fe
Distance between lenses = L
Magnification: compare angular size of final image,
θ’, to angular size, θ, of object at near point viewed
with the unaided eye.
With:
• object just outside the focal point of the objective, so do1 " fo
• first image at focal point of eyepiece (! final image at infinity)
θ�
(L − fe)N
Angular magnification, M = � −
θ
fo fe
N = near point
Wednesday, March 10, 2010
95
Prob. 26.93/86: A microscope for viewing blood cells has an
objective with a focal length of 0.5 cm and an eyepiece with a
focal length of 2.5 cm.
The distance between the two lenses is 14 cm.
If a blood cell subtends an angle of 2.1 ! 10-5 rad when viewed
with the naked eye at a near point of 25 cm, what angle does it
subtend when viewed through the microscope?
0.280 or 4.8x10-3 rad
Wednesday, March 10, 2010
96
Summary of Chapter 26
• Snell’s Law:
n1 sin !1 = n2 sin !2,
• Apparent depth:
d = d n1/n2
• Total internal reflection: n1 sin !c = n2,
v = c/n
!c = critical angle for total
• Lens equation:
1/do + 1/di = 1/f
• Linear magnification:
Two lenses:
m = – di/do
m = m1 m2
internal reflection
• Angular magnification: M = ! /!
• Magnifying glass:
M = N/f
M = N/f + 1
(image at infinity, N = near point)
(image at near point)
• Compound microscope: M = – (L - fe)N/fofe
Wednesday, March 10, 2010
Chapter 27:
Interference and
the Wave Nature
of Light
97
Owls and the
Rayleigh
Criterion
• Principle of linear superposition
• Interference and diffraction of light
• Resolving power
• Omit 27.9, x-ray diffraction
Wednesday, March 10, 2010
98
Principle of Linear Superposition
When two waves overlap, the resultant is the sum of the two.
Light waves: add the electric fields.
If the waves are of the same wavelength and are in phase, the amplitude
of the combined wave is increased. This is constructive interference.
Waves
in phase
again
l
E
Waves
in phase
Intensity ~ (Amplitude)2
Constructive
Interference
v
l
Time
v
l2 − l1 = (3.25 − 2.25)! = ! → waves in phase
Wednesday, March 10, 2010
99
If the waves are of the same wavelength and out of phase, the
amplitude of the combined wave is decreased, even zero if the two
waves have the same amplitude. This is destructive interference.
l
Waves
out of
phase
Destructive
Interference
v
Waves
in phase
l
v
Time
l2 − l1 = (3.25 − 2.75)! = 0.5! → waves out of phase
Wednesday, March 10, 2010
100
Coherent and incoherent light sources
Coherence: the waves maintain a constant phase relative to one another.
Constructive and destructive interference can then occur, depending on
the difference in path length. Example, light from a laser.
Constructive interference: the difference in path length is an integer
number of wavelengths:
l2 – l1 = m", m = 0, 1, 2, 3... (m = “order”)
Destructive interference occurs if:
l2 – l1 = (m + 1/2)", m = 0, 1, 2, 3...
If the waves emitted by the sources do not maintain a constant phase
relationship, the sources are “incoherent.” Example, light from a lamp.
Wednesday, March 10, 2010
101
Young’s double slit
experiment
Light passes through a single slit.
The light wave from the slit
spreads out to pass through two
slits which act as coherent
sources of light.
The light waves from the two
slits overlap on a screen.
Constructive and destructive
interference is seen as a series
of bright and dark bands – not
as images of the two slits.
Wednesday, March 10, 2010
102
Young’s double slit experiment
Intensity
Waves arrive
at the slits
in phase
0o
!
Wednesday, March 10, 2010
103
Constructive and destructive interference
l2 – l1 = 0
l1
Time
l2
Screen
Wednesday, March 10, 2010
104
Constructive and destructive interference
l1
l2 (from S2 to
point on screen)
Time
Screen
l2 − l1 = !
Wednesday, March 10, 2010
105
Constructive and destructive interference
l1
l2 − l1 =
!
2
l2 (from S2 to
point on screen)
Time
Screen
Wednesday, March 10, 2010
106