Download Chapter 26: Refraction, Lenses, Optical Instruments Refraction of light

Chapter 26: Refraction, Lenses, Optical Instruments
• Refraction of light, Snell’s law. Apparent depth
• Polarization of light on reflection
• Dispersion of light. Prisms, rainbows, sun dogs
• Formation of image by lenses, thin lens equation, magnification
• Combination of lenses
• The human eye – correction for near and farsightedness
• Optical instruments
• Omit 26.14, lens aberrations
Friday, March 9, 2007
1
Refraction of light
When light travels from one transparent medium to another, it is in general
deflected from its original direction –!this is refraction.
The amount by which
the light is deflected
depends on the
refractive index of
each medium.
Friday, March 9, 2007
2
Refractive index – a measure
of the speed of light in a
transparent medium
Refractive index =
n=
Speed of light in vacuum
Speed of light in medium
c
c
and v = = f λ
v
n
Prob. 26.6: Light has wavelength 340 nm and
frequency 5.403"1014 Hz when travelling
through some substance. What substance is it?
v = c/n = f ! = (5.403"1014 Hz)"(340"10-9m)
#
= 1.837"108 m/s
n = 3"108/1.837"108 = 1.633 – carbon disulphide
Friday, March 9, 2007
3
Prob. 26.8: A flat sheet of ice has a thickness of 2 cm. It is on top of a flat
sheet of quartz that has a thickness of 1.1 cm. Light strikes the ice
perpendicularly and travels through it and then through the quartz.
In the time it takes the light to travel through the two sheets, how far
would it have travelled in vacuum?
Ice: # t1 = l1/v1 = l1 n1/c as v1 = c/n1
Quartz:# t2 = l2/v2 = l2 n2/c
Ice
n1 = 1.309
l1 = 2 cm
Quartz
n2 = 1.544
l2 = 1.1 cm
In time (t1 + t2), light would travel a distance L = c(t1 + t2) in vacuum.
t1 = (0.020 m) " 1.309/c = 0.02618/c
t2 = (0.011 m) " 1.544/c = 0.01698/c
t1 + t2 = 0.04316/c
So L = c(t1 + t2) = 0.0432 m = 4.32 cm
Friday, March 9, 2007
4
Refraction and Reflection
Snell’s law: n1 sin !1 = n2 sin !2
Light incident from above:
If !1 = 30◦
“External
reflection”
n1 sin !1
n2
1 × sin 30◦
=
1.33
sin !2 =
!2 = 22.1◦
Friday, March 9, 2007
5
Snell’s law: n1 sin !1 = n2 sin !2
Light incident from below:
If !1 = 30◦
n1 sin !1
n2
1.33 × sin 30◦
=
1
sin !2 =
!2 = 41.7◦
“Internal
reflection”
(inside medium of higher refractive index)
Friday, March 9, 2007
6
Prob. 26.C8: Two rays of light converge to a point on a screen.
A plane-parallel plate of glass is placed in the path of this converging
light and the glass plate is parallel to the screen.
Will the point of convergence remain on the screen?
If not, will the point move toward the glass of away from it?
As n2 > n1, the ray is refracted
toward the normal to the surface
when it enters the glass.
glass
n1 = 1
n2 > n1
Friday, March 9, 2007
The converging rays meet
farther to the right, behind the
screen.
Friday, March 9, 2007
7
n1 = 1
n2 > n1
8
Rear view mirror
Day setting:
$ 100% reflection
Night setting:
$ 10% reflection
Friday, March 9, 2007
9
Apparent Depth
At what angle should
the searchlight be aimed
to illuminate the chest?
tan !2 =
2
3.3
!2 = 31.22◦
Snell’s Law: n1 sin !1 = n2 sin !2
sin !1 =
1.33 sin 31.22◦
= 0.6894
1
!1 = 43.6◦
The searchlight is aimed above the chest – the apparent depth of the chest is
less than its actual depth.
Friday, March 9, 2007
10
Apparent depth – formation of image
Apparent position
of chest
Friday, March 9, 2007
11
Prob. 26.17/19: Find the relationship between true and apparent depth for
small angles of incidence.
n1
!1
n2
x
x
d!
x
tan !2 =
d
tan !1 =
!2
d´
!1
d
x
apparent
depth
So, x = d ! tan !1 = d tan !2
Snell’s Law: n1 sin !1 = n2 sin !2
true depth
For small angles: sin ! ! tan ! ! ! radians
Therefore, n1!1 ! n2!2 and d "!1 ! d!2
So,
d ! !2 n1
" "
d !1 n2
Friday, March 9, 2007
→ d" #
dn1
n2
Apparent depth
12
Apparent depth (height)
Viewed from below the surface
of the water, the object appears
further above the surface than it
actually is.
n2
The same formula holds as for
apparent depth:
n1
d! "
dn1
n2
As n1 > n2, d´ > d
Friday, March 9, 2007
13
Prob. 26.16: A silver medallion is sealed within a transparent block of
plastic. An observer in the air, viewing the medallion from directly
above, see the medallion at an apparent depth of 1.6 cm beneath the top
surface of the block.
How far below the top surface would the medallion appear if the
observer (not wearing goggles) and the block were under water?
Friday, March 9, 2007
14
Prob. 26.20/18: A man in a boat is looking straight down at a fish in the
water directly beneath him. The fish is looking right back. They are the
same distance from the air-water interface.
To the man, the fish appears to be 2 m beneath his eyes.
To the fish, how far above its eyes does the man appear to be?
Friday, March 9, 2007
15
Chapter 26 so far...
Refractive index:
n = c/v
Snell’s law:
n1 sin%1 = n2 sin%2
Apparent depth:
d! = dn1/n2
Friday, March 9, 2007
16
Displacement of path of ray of light
n1 sin θ1 = n2 sin θ2 = n1 sin θ3
So, θ3 = θ1
Rotating the glass plate changes the
amount of displacement – used in some
optical instruments to line up an image
with reference lines (cross-hairs)
Friday, March 9, 2007
17
Prob. 26.19/17:
If "1 = 30o, the glass
plate is 6 mm thick
and n2 = 1.52, what
x
is the displacement?
dt = 6 mm
n2 = 1.52
Snell: n1 sin !1 = n2 sin !2 = n3 sin !3
= 30º
As n1 = n3, !1 = !3
That is, emerging ray is parallel with incident ray.
n1 sin !1 1 × sin 30◦
sin !2 =
=
→ !2 = 19.2◦
n2
1.52
Friday, March 9, 2007
18
26.19 contd
θ2 = 19.2
◦
A
B
90º
x
t = 6 mm
BCD:
x = L sin(!1 − !2)
ABD:
L = t/ cos !2
So, x =
t sin(!1 − !2)
cos !2
C
x=
L
6 sin(30◦ − 19.2◦)
cos 19.2◦
Displacement, x = 1.19 mm
!1 − !2
D
Friday, March 9, 2007
19
Snell’s Law: a light wave hitting a boundary
For any wave: v = f !
For light: v = c/n = f !
So, λ =
Rays are perpendicular
to wavefronts
c
1
, proportional to
nf
n
The incident wave
turns into the
refracted wave and
matches onto it.
The wavefronts
crumple and tilt to
adjust to the new
wavelength # the
ray is refracted.
Friday, March 9, 2007
(small n)
(large n)
20
Snell’s Law: a light wave hitting a boundary
Rays are perpendicular
to wavefronts
B
90º – "1
Wavefronts
90º
A
C
n1
! = c/(nf) for a light wave
c
= h sin "1
ABC: !1 =
n1 f
c
= h sin "2
ACD: !2 =
n2 f
c
So,
= n1 sin !1 = n2 sin !2
hf
Snell’s Law!
n2
90º
D
Wavefronts
Friday, March 9, 2007
21
Total internal
reflection
Internal
reflection
The normal case: both
reflected and refracted rays.
If the angle of incidence
increases, so does the angle
of refraction, until...
Refraction at the “critical
angle for total internal
reflection” – the refracted
ray is at 90o.
Possible only if n2 < n1
Friday, March 9, 2007
n1 sin !c = n2 sin 90◦
sin !c = n2/n1
22
n1 sin θc = n2 sin 90◦
n2 < n1
!1 > !c
If the angle of incidence is greater than the critical angle:
– according to Snell’s law, the sine of the angle of refraction is greater
# than 1.0, so there is no refracted ray
– the light undergoes “total internal reflection”
Example: n1 = 1.33 (water), n2 = 1 (air): sin "c = 1/1.33, "c = 48.8o
26.13
Friday, March 9, 2007
23
Prob. 26.105/13: A ray of light travels from the coin to the surface of the
liquid and is refracted as it enters the air. A person sees the ray as it
skims just above the surface of the liquid. How fast is the light travelling
in the liquid?
• What are the angles of incidence, refraction? Apply Snell’s law, find
# the index of refraction of the liquid.
90º
"
n2 = 1
n1 = ?
tan θ = 5/6, θ = 39.8◦
n1 sin " = n2 sin 90º
n1 = 1/sin " = 1.56
"
Friday, March 9, 2007
v = c/n1 = 1.92"108 m/s
24
Total internal reflection around the bend
Friday, March 9, 2007
25
Optical fibre – total internal reflection at the
walls steers the light around bends
Applications:
• Medicine – flexible optical fibres used to look inside the body.
# “Keyhole” surgery – add surgical instrument, laser beam to vaporize tissue.
• Communications – transmit telephone, radio, TV, internet signals on a
# laser beam inside a fibre optic cable – no external interference, much
# greater amount of information can be transmitted than with copper cable.
26.29
Friday, March 9, 2007
26
Prob. 26.-/29: The optical fibre shown consists of a core made of flint
glass surrounded by a cladding made of crown glass.
A beam of light enters the fibre from air at an angle "1 with respect to the
normal. What is "1 if the light strikes the core-cladding interface at the
critical angle "c?
B
n1 = 1
A
θ2 = 90◦ − θc
n3 = 1.523
n2 = 1.667
Friday, March 9, 2007
27
Total internal reflection in a prism
45◦ > !c
Glass: n $ 1.5
!c ! sin−1(1/1.5) = 42◦
Prisms “fold” the light
path to make the
binoculars shorter. Each
arm acts as a longer
telescope.
Friday, March 9, 2007
28
Polarization of light by reflection
Light reflected from a surface is in general partially polarized.
The reflected light is 100% polarized parallel to the surface when reflection
occurs at the “Brewster angle” "B, corresponding to a 90o angle between
reflected and refracted rays.
θB + θ2 + 90 = 180
◦
◦
!2
!B + !2 = 90◦
90º
!B
From the triangle: sin !2 = cos !B
Snell: n1 sin !B = n2 sin !2 = n2 cos !B
So, tan !B =
n2
n1
Brewster angle
Friday, March 9, 2007
29
Polarization of light by reflection
• Sunlight reflected from water – polarized horizontally. "B = 53º
• Polaroid type sun glasses reduce glare from reflected sunlight by
# filtering out horizontally polarized light.
• Digital watches – emitted light is polarized vertically (top to bottom
# in the display).
• Display turns dark if rotated by 90o when viewed through Polaroid sun
# glasses.
Friday, March 9, 2007
30
Dispersion by a prism
lower n, less refraction
Violet is refracted more than red
higher n, greater refraction
Friday, March 9, 2007
26.38
31
Prob. 26.38: A ray of sunlight is passing from diamond into crown glass;
the angle of incidence is 35º.
Indices of refraction for red and blue light:
Blue: ndiamond = 2.444# #
ncrownglass = 1.531
Red: ndiamond = 2.410# #
ncrownglass = 1.520
Determine the angle between the refracted red and blue rays in the crown
glass.
Friday, March 9, 2007
32
Dispersion by rain drops – rainbows
Secondary
rainbow
Primary
rainbow
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Friday, March 9, 2007
33
Dispersion by a raindrop – primary rainbow
The refractive index for violet is larger than
for red.
# violet is refracted through a larger angle
# than red
Friday, March 9, 2007
34
Dispersion – formation of a rainbow
The colours of the rainbow come from
raindrops at different height, red from
higher up, violet from lower down.
Friday, March 9, 2007
35
Violet is refracted through a larger angle than red
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Friday, March 9, 2007
36
Sun Dogs
Friday, March 9, 2007
37
Sun Dogs
Refraction by hexagonal
ice crystals
Violet is refracted through a larger angle
than red
http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/halo22.html#c3
Friday, March 9, 2007
38
Lenses
A positive (converging, convex) lens
Focal point
Focal point
A negative (diverging, concave) lens
Friday, March 9, 2007
39
Formation of image by thin lenses
Parallel to axis, passes through
focal point on right
Passes through focal point on
left, emerges parallel to axis
Passes through centre of lens
in a straight line
Parallel to axis, ray traced back
to axis passes through focal
point on left
Heads toward focal point on
right, emerges parallel to axis
Passes through centre of lens
in a straight line
Friday, March 9, 2007
40
Formation of a real image by a converging lens
Light appears
to originate
from image
Object is outside the
focal point
A “real” image – can be
seen on a screen placed at
the position of the image.
Image is inverted.
Friday, March 9, 2007
41
Formation of a virtual image by a converging lens
A “virtual” image – cannot be formed on a
screen. Image is upright, and magnified.
Light appears
to originate
from image
Object is inside
the focal point
Friday, March 9, 2007
42
Formation of virtual image by
a diverging lens
Light appears
to originate
from image
Image is virtual, upright, diminished
Friday, March 9, 2007
43
Thin lens equation
1
3
!
ho
!
!
!
1
3
f
Object distance
tan ! =
So,
ho hi
=
do di
hi di di − f
= =
ho do
f
That is:
Friday, March 9, 2007
1 1 1
+ =
do di f
Image distance
ho
hi
tan ! = =
f
di − f
Divide by di :
Thin lens equation
1
1 1
= −
do f di
Applicable to diverging
lenses too
44
Linear magnification
!
ho
!
!
!
f
Linear magnification, m = (height of image)/(height of object)
hi
hi di
(or from tan !)
m=
Similar triangles: =
ho
ho do
Sign convention, image is inverted, so: m = −
di
do
Friday, March 9, 2007
45
Conventions for the thin lens equation
Draw ray diagrams with rays travelling from left to right.
Normal situation:
Object # Lens # Image
Real object to left
of lens,
object distance,
do is positive
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
Ex8
Friday, March 9, 2007
46
Converging lens, f > 0
f>0
do (> 0)
di (> 0)
Object # Lens # Image
Diverging lens, f < 0
do (> 0)
f<0
di (< 0)
Friday, March 9, 2007
47
Example: A 1.7 m tall person stands 2.5 m in front of a camera. The
focal length of the lens is 0.05 m.
a) Find the image distance
b) Find the magnification and the height of the image on the film.
Friday, March 9, 2007
48
Prob. 26.50/104: To focus a camera on objects at different distances, the
converging lens is moved toward or away from the film, so a sharp
image always falls on the film.
A camera with a lens of focal length f = 200 mm is to be focussed on an
object located at a distance of 3.5 m and then at 50 m.
Over what distance must the lens be movable?
Friday, March 9, 2007
49
Thin lens equation
1
ho
3
!
!
!
!
1
3
f
Object distance
Thin lens equation:
Image distance
1 1 1
+ =
do di f
Linear magnification: m =
Friday, March 9, 2007
hi
di
=−
ho
do
50
Combinations of lenses – microscope
• Find the location of the image formed by the first lens as if the second
# lens did not exist.
• Use that image as an object (source of light) for the second lens using the
# sign convention for real and virtual objects.
1
3´
2
3
2
1´
Lens 2: light appears to come
from intermediate image
A microscope producing a virtual, inverted and magnified
final image. The eyepiece acts as a magnifying glass.
Friday, March 9, 2007
51
Prob. 26.59/60: Two identical diverging lenses are separated by 16 cm.
The focal length of each lens is –8 cm. An object is located 4 cm to the
left of the lens that is on the left. Determine the final image distance
relative to the lens on the right.
Friday, March 9, 2007
52
Thin lens equation
1
ho
3
!
!
!
!
1
3
f
Object distance
Image distance
1 1 1
Thin lens equation: + =
do di f
Linear magnification: m =
hi
di
=−
ho
do
Object to left: do > 0
Image to right: di > 0
Positive lens: f > 0
(converging, convex)
Negative lens: f < 0
(diverging, concave)
Friday, March 9, 2007
53
Combinations of lenses – microscope
• Find the location of the image formed by the first lens as if the second
# lens did not exist.
• Use that image as an object (source of light) for the second lens using the
# sign convention for real and virtual objects.
1
3´
2
3
2
1´
Lens 2: light appears to come
from intermediate image
A microscope producing a virtual, inverted and magnified
final image. The eyepiece acts as a magnifying glass.
Friday, March 9, 2007
54
26.66: Two converging lenses (f1 = 9 cm, f2 = 6 cm) are separated by 18
cm. The lens on the left has the longer focal length. An object stands 12
cm to the left of the left-hand lens.
a) Locate the final image relative to the lens on the right.
b) Obtain the overall magnification.
c) Is the final image real or virtual, upright or inverted, larger or smaller
than the object?
Friday, March 9, 2007
55
Prob. 26.64: A coin is located 20 cm to the left of a converging lens (f =
16 cm). A second, identical, lens is placed to the right of the first lens,
such that the image formed by the combination has the same size and
orientation as the original coin. Find the separation between the lenses.
Friday, March 9, 2007
56
The Human Eye
n = 1.33-1.34
n = 1.38
Most of the
refraction occurs at
the cornea
n = 1.41-1.45
n = 1.34
Sharpest image,
best colour
discrimination
Biomedical Applications of Introductory
Physics, Tuszynski & Dixon
Friday, March 9, 2007
57
The human eye
The eye focuses an image onto the retina by adjusting the focal length of
the eye lens. This is known as accommodation.
Eye lens has its
longest focal length
Ciliary muscle,
relaxed
Greatest distance at which eye can focus: “far point”
Normal value: infinity
Eye lens compressed,
focal length decreased
Closest distance at which eye can focus: “near point”
Normal value: N = 25 cm
Friday, March 9, 2007
58
Near and far points
Near point: closest distance at which unaided eye can focus,
normal value, N = 25 cm
Far point: greatest distance at which unaided eye can focus,
normal value infinity
Accommodation: the ability of the eye to adjust its focal length to focus on
objects at different distances.
Nearsighted eye: far point less than infinity " distant objects are blurred
Farsighted eye: near point greater than 25 cm " objects close by are
blurred
Friday, March 9, 2007
59
Nearsightedness
Objects in focus
The eye lens forms an image of a distant object in front of the
retina # blurred image on the retina
Correction is with a diverging lens that moves the image back
onto the retina. The corrective lens forms a virtual image in front
of the eye that is close enough for the eye to focus on.
Friday, March 9, 2007
60
Farsightedness
Objects in focus
The eye lens forms an image of a nearby object behind
the retina # blurred image on the retina
Correct with a converging lens that forms a virtual
image far enough away for the eye to focus on.
Friday, March 9, 2007
61
Correction of near and farsightedness
Use a corrective lens to form a virtual image at a distance at which
the eye can focus.
Nearsighted:
• The corrective lens forms an image of a distant object at the person’s far
point, or closer.
Farsighted:
• The corrective lens forms an image of an object at the person’s near
point, or further.
Power of a lens:
• Power is 1/f, focal length in metres, power in diopters.
# Example, f = –10 cm, power = 1/(– 0.1) = –10 diopters.
Friday, March 9, 2007
62
Prob. 26.107/71: 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?
Friday, March 9, 2007
63
Prob. 26.67: A farsighted person has a near point that is 67 cm from
her eyes. 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 –
a) 2.2 cm from the eyes,
b) 3.3 cm from the eyes.
Friday, March 9, 2007
64
Angular size, magnification
ho
Angular
size
of
the
object:
!
=
(near point, closest eye can focus at)
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
= × =
! do ho do
The magnifying glass lets the user view the object closer than the near point
M=
Friday, March 9, 2007
65
Magnifying Glass
Angular magnification: M =
!! N
=
! 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 =
Friday, March 9, 2007
1
1
1
f +N
= −
=
do f −N
fN
N
f +N
N
=
= 1+
do
f
f
$ maximum usable magnification
66
Magnification Markings
Lenses are sometimes marked with the magnification they produce
when an image is formed at infinity.
For example, “10"”.
This means that,
"
N
= 10,
" M=
f
with N = 25 cm, the normal near point.
So, f =
N
25
=
= 2.5 cm
10
10
Friday, March 9, 2007
67
Prob. 26.112: 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?
Friday, March 9, 2007
68
Astronomical Telescope
fo
fe
di1
Formation of intermediate image
by the objective.
hi
hi
(small angles)
!=
!
di1 fo
The eyepiece acts as a magnifying
do2
glass to produce a magnified final
image.
hi
hi
!! = −
#−
do2
fe
!
!
hi fo
fo
M= "− × =−
!
fe hi
fe (exact when object and final image are at infinity)
Friday, March 9, 2007
69
Astronomical Telescope
di1
Object at infinity:
• first image at focal point of objective
# di1 = fo
Final image at infinity:
• first image at focal point of eyepiece as well
# do2 = fe
# distance between lenses is:
" L = fo + fe
do2
and
di1
fo
M =−
=−
exactly
do2
fe
#
Friday, March 9, 2007
70
Prob. 26.92/90: 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 =−
Refractive power,
P =
so,
M =−
1
f
fo
= −132
fe
Pe
= −132
Po
Therefore, Pe = 132 # Po = 132 # 1.5 = 198 diopters.
fe =
1
= 0.00505 m = 5.1 mm
Pe
Friday, March 9, 2007
71
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 is less than an astronomical telescope of the same
# magnification
• the image is the right way up (M > 0)
Friday, March 9, 2007
72
Compound Microscope
!!
fo
fe
Distance between lenses = L
Magnification: compare angular size of final image, "%,
to angular size, ", of object at near point viewed with
the naked 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)
Angular magnification, M =
θ!
(L − fe)N
"−
θ
fo fe
N = near point
Friday, March 9, 2007
73
Compound Microscope
ho1
do1 ! fo
do2 ! fe
di1
hi1
fo
!!
Final image
fe
!
"
! "
L − fe
di1
× ho1 # −
× ho1
hi1 = m1ho1 = −
d01
fo
hi1 hi1
"
do2
f
"
!e
L − fe
!
× ho1
So, ! " −
fo fe
!! =
ho1
With the object at the near point and no microscope: ! =
N
"
!
!
!
L − fe
×N
Therefore, M = " −
!
fo fe
Friday, March 9, 2007
74
Prob. 26.86/98: 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?
Friday, March 9, 2007
75
Summary of Chapter 26
• Snell’s Law: ##
#
#
n1 sin "1 = n2 sin "2, # v = c/n
• Apparent depth: ##
#
d! = d n1/n2
• Total internal reflection: #n1 sin "c = n2, ""
"
"
" " " " " " " " "
• Lens equation: # # # 1/do + 1/di = 1/f
"c = critical angle for total
" internal reflection
• Linear magnification: # m = – di/do
" Two lenses:# # # # m = m1 m2
• Angular magnification:# M = "!/"
• Magnifying glass:# #
#
#
# # # #
Friday, March 9, 2007
M = N/f # # (image at infinity, N = near point)
M = N/f + 1# (image at near point)
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