image formation lenses and mirrors. object:p image: q focal length : f magnification :m real : light...
TRANSCRIPT
Object:pImage: qFocal length : fMagnification :MReal : light at imageVirtual : no light
1 1 2 2
1 2
sin sin
1 1 1( 1)
n n
nf R R
Mirrors Lenses
/
1 1 1
image
object
n c v
p q f
h qM
h p
1 2
2
Rf
+p
+q
+R
+f
Ray1- || to axis, then thru f
Ray2-thru f then ||
Ray3-thru R and back
(Ray-4- to center of mirror then reflect at same angle)
Ray1- || to axis, then thru f on back
Ray2-thru f on front then ||
Ray3-thru center of lens
1 2 1 2
1
2
1n n n n
p q R f
n qM
n p
Refraction at curved surface
+p
+q
+R
+f Convex-f Concave
Simple optical componentsImagine a spherical mirror
A horizontal ray gets reflected according to…
But if they area close to the axis they almost meet
The rays don’t really meet
Horizontal rays close to the axis focus
So, for• Optical components that Symmetric about an axis • With rays close to the axis• And not very thick when transparent
Horizontal rays are focused to a point
Axis of symmetry orprincipal axis
f
Focal pointRays form a point close to the axis are also focused
Image
Flat Mirrors Light rays appear to diverge from pointbehind mirror
QuickTime™ and aPhoto - JPEG decompressor
are needed to see this picture.
Law of Reflection: p =|q|independent of viewing angle
Image is Virtual; behind mirror, can’t project on screen, light rays don’t actually pass through image point
h h’
Principle axes
p
q
q
Magnification: M = h’/h = 1Image is Upright
p qp q
• In the overhead view of Figure 26.4, the image of the stone seen by observer 1 is at C. At which of the five points A, B, C, D, or E does observer 2 see the image?
’
x
x’
p
q
p-R
R-q
R
'
1sin sin x x
p R R
2
' 'sin sin '
x x
R q R
1 1
2 2
sin
sin
R q q
p R p
1 2p u q
u
1 1 2 1
p q R f
Just geometry…
Reflection by spherical mirrors
R
pq
Look at the mirror formula:
1 1 2 1
p q R f
q=R/2 when p is very large
All rays converge at R/2, the focal point
For concave mirror the image is in front of the mirror
The ray is reflected straight back because it move in the radial direction and so it is normal to the mirror
The image is inverted
h
h’
'h qM
h p
q
p
For convex mirror the image appears to be behind the mirror, it is virtual
The ray is reflected straight back because it move in the radial direction and so it is normal to the mirror
The image is not inverted
h
h’
'h qM
h p
qp
The mirror formula still holds, but with q<0 and R<0
The image is virtual The mirror is convex
• You wish to reflect sunlight from a mirror onto some paper under a pile of wood to start a fire. Which would be the best choice for the type of mirror? (a) flat (b) concave (c) convex
• Ray tracing for lenses (and mirrors)– ray parallel to the axis refracts (reflects) through
the focal point– ray through the focal point emerges parallel to
the axis– ray through the center of the lens (mirror)
• does not bend (lenses)• goes out at an equal angle (mirror)
Consequences of Snell’s law and the laws of reflection PLUS geometry
f’
f
p
q
h
-h’
26.2
• In a church choir loft, two parallel walls are 5.3 m apart. The singers stand against the north wall. The organist faces the south wall sitting 0.8 m away from it. To enable her to see the choir, a flat mirror 0.6 m wide is mounted on the south wall, straight in front of her. What is the width of the north wall she can see.
Exercise 26.2Sounds like ray tracingWill need the formula for reflected rays form mirrors
0.8m
organist
5.3m
choir0.6mL m 575.4m 6.02
m 9875.1 8.0
3.0
3.5
lL
ll
l
Exercise 26.11Sounds like an application of the formula
1 1 1
'
p q f
h pM
h q
/ 2 20 cm qpf
f Rp f
30 ( 20) 12) p 30 cm q 12 cm .4
30 20 30a M
60 ( 20) 15) p 60 cm q 15 cm 0.25
60 20 60b M
) 0 upright images in both casesc M
A spherical convex mirror has radius of curvature with magnitude 40 cm.Find q and M when p=30 cm and p=60 cm
http://www.phy.ntnu.edu.tw/java/Lens/lens_e.htmlcorrected apr 25
+ side - side
So, to make an image just do some ray-tracing
A ray trough a focal point is refracted to a horizontal ray
’
f’
f
A horizontal ray is refracted towards a focal point
p
q
h
-h’' ' '
' '
' ' '
'
' '
'
h f h f
p h h qθ
p f f q
p f f
f q f
Let f f
1 1 1
'
p q f
h qM
h p
The optical component may not be back-front symmetric so in general f≠f’
f=f’
Specific cases:
1 1 1
p q f
Spherical concave mirror of radius R
Spherical convex mirror of radius R
Spherical boundary of radius R between a medium of refraction index n1 (left) and refraction index n2 (right)
2
Rf
2
Rf
2112 /1'
1/ nn
Rf
nn
Rf
Thin lens of curvature radii Rleft, Rright and of material with refraction index n
left right
1 1 1( 1)n
f R R
1 2 2 1+ =n n n n
p q R
An object is:•REAL if light diverges from it•VIRTUAL if light converges toward it
Definitions and sign conventions
An image is:•VIRTUAL if light diverges from it•REAL if light converges toward it
f
f
p
qh h’
p (object) q (image) (object) ' (image) (image)
convex real real upright upright upright
concave virtual virtual inverted inverted inverted
f h h M
+p
+q
+R
+f Convex-f Concave
+p
+q
+R
+f
mirror
lens
For example:
p (object) q (image) (object) ' (image) (image)
convex real real upright upright upright
concave virtual virtual inverted inverted inverted
f h h M
An image image is:VIRTUAL if light diverges from itREAL if light converges toward it
An objectobject is:REAL if light diverges from itVIRTUAL if light converges toward it
+p
+q
+R
+f Convex-f Concave
1 1 1
'
p q f
h pM
h q
positive real image concave concave upright image
negative virtual image convex convex inverted image
q R f M
Sign conventions for mirrors:
+p
+q
+R
+f
Exercise Sounds like another application of the formulaThe tricky part are the signs:•M=+2 since the image is not inverted•The object and the image are on same sides of the lens so p and q have the different signs
1 1 1
qM
p q f p
2 q 2 84 cm
1 1 1 1 1
2.84 cm1
qM .
p
M M
f p q q q q
qf
M
Magnifying looks at a map. The image is virtual, of twice the size and |q|=2.84 cm. Find f
+p
+q
+R
+f Convex-f Concave
A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the transparent cornea. Assume that this outer surface has a 6mm surface of curvature and assume that the eyeball contains just one fluid with index of refraction 1.4. Where is the image? Does it fall on the retina? Describe it.
1 2 2 1n n n n
p q R
1.4(6 )21
1 .4
nR mmp so q mm
n
real, inverted
two lens system
Fig 36-32a, p.1151
first lens
1 1 1
10 20q
1 1 1
15 10q
q=30
1 1 1 qM
p q f p
second lens
p=40-30=10cm
q=-20 cm
M1=-30/15=-2
15 cm 40
http://www.hazelwood.k12.mo.us/~grichert/optics/intro.html
M=M1M2=-8
M2=-(-20)/(10)=4
real inverted
virtual, upright
Total
inverted
+p
+q
+R
f=2cmf=20 cm
f=2cmf=20 cm
virtual object?
first lens
1 1 1
15 2q
1 1 1
20q
q=20 cm
second lens
p=5-20=-15cmVIRTUAL OBJECT!
q=1.8 cm
double lens q=20 cm
1 1 1
p q f +p
+q
+R
Fig 36-41, p.1158
http://webphysics.davidson.edu/physlet_resources/dav_optics/Examples/eye_demo.html
• 6. If you cover the top half of the lens in Active Figure 26.24a with a piece of paper, which of the following happens to the appearance of the image of the object? (a) The bottom half disappears. (b) The top half disappears. (c) The entire image is visible but dimmer. (d) There is no change. (e) The entire image disappears.
http://www.phy.ntnu.edu.tw/java/Lens/lens_e.html
http://www.hazelwood.k12.mo.us/~grichert/optics/intro.html
http://webphysics.davidson.edu/physlet_resources/dav_optics/Examples/eye_demo.html
Rope with total mass m = 2 kg, L = 80 mand mass M = 20 kg at end
L = 80 m
M = 20 kg
Tension in rope F and wave speed v
2 kg
80 m0.025 kg /m Tension F Mg (20)(9.8) 196 N
v F
1960.025
88.5 m / s
If end of rope driven in SHM with f=0.056 Hz then wavelength of travelingwave up rope is
v
f
88.5
0.0561590 m L 80 m
If end of rope driven in SHM with f=0.056 Hz then wavelength of travelingwave up rope is ?
2
12
F=-kx x=Acos( t+ )
1
2spring
f fT
kSHO
m
gPendulum
L
PE kx Energy PE KE
total 0
0
2 2 2 20
2 20
sin( )
/
( / ) ( )
/tan
ma F kx bv F t
F mA
b m
b m
2
)2/(
2
)cos(
m
b
m
k
tAex tmb
/0
t mE E e where
b
( , ) sin( )
2
D x t A kx t
k vk
damped SHO
driven and dampled SHO
traveling waves
2 2
2 2 2
1d d
dt c dx
E E
8
0 0
13 10 m/sc
0
0
0 0
cos( )
cos( )
E kx t
B kx t
E cB
E x
B z
^
^
2av 0 0
1
2u E
0
1S E B
2
0 0
~ /EB E
Power areac
S
20
02avg
EPowerI S
Area c p=E/c
1 1 / 1 PowerP
F p E c I
A A t A t A c c
I=I0 cos2
Tv
LL S
S
v vf f
v v
positive direction is when one is moving toward the other
Object:pImage: qFocal length : fMagnification :MReal : light at imageVirtual : no light
1 1 2 2
1 2
sin sin
1 1 1( 1)
n n
nf R R
Mirrors Lenses
/
1 1 1
image
object
n c v
p q f
h qM
h p
1 2
2
Rf
+p
+q
+R
+f
Ray1- || to axis, then thru f
Ray2-thru f then ||
Ray3-thru R and back
(Ray-4- to center of mirror then reflect at same angle)
Ray1- || to axis, then thru f on back
Ray2-thru f on front then ||
Ray3-thru center of lens
1 2 1 2
1
2
1n n n n
p q R f
n qM
n p
Refraction at curved surface
+p
+q
+R
+f Convex-f Concave