Transcript F - 101physics

Optics
Mirrors and Lenses
Regular vs. Diffuse Reflection
 Smooth, shiny surfaces have a regular reflection:
Rough, dull surfaces have a diffuse reflection.
Diffuse reflection is when light is scattered in different
directions
Reflection
 We describe the path of light as straight-line rays
 Reflection off a flat surface follows a simple rule:
 angle in (incidence) equals angle out (reflection)
 angles measured from surface “normal” (perpendicular)
Laws of reflection
1 )The incident ray, the reflected ray and the normal all lie in the same plane.
normal
incident ray
mirror
2)The incident angel = the reflected angel

ˊ

reflected ray
24-1
mirrors
An object viewed using a flat mirror appears to be located behind the
mirror, because to the observer the diverging rays from the source appear to
come from behind the mirror
The image distance S behind the mirror equals the object distance S from the mirror
The image height h’ equals the object height h so that the lateral magnification
The image has an apparent left-right reversal
The image is virtual, not real!
1-Virtual images - light rays do not meet and the image is always upright or
right-side-up“ and also it cannot be projected Image only seems to be there
Real images - always upside down and are formed when light rays actually meet
example
• If the angle of incidence of a ray of light is 42owhat is each
of the following?
A-The angle of reflection
(42o)
B-The angle the incident ray makes with the mirror (48o)
C-The angle between the incident ray and the reflected
(90o)
Now you look into a mirror and see the image
of yourself.
a) In front of the mirror.
b) On the surface of the mirror.
C)Behind the mirror.
7
Example
A girl can just see her feet at the bottom edge of
the mirror.
Her eyes are 10 cm below
the top of her head.
(a)
What is the distance
between the girl and
her image in the mirror?
150 m
150 m
Distance = 150  2 = 300 cm
8
T . Norah Ali Almoneef

Signs: Image size and magnification
images can be upright (positive image size
h’) or inverted (negative image size h’)
Define magnification m = h’/h
Positive magnification: image orientation
unchanged relative to object
Negative magnification: image inverted
relative to object
l m l < 1 if image is smaller than object
l m l > 1 if image is bigger than object
l m l = 1 if image is same size as object
24-2 Thin Lenses
A lens is a transparent material made of glass or plastic that refracts light rays
and focuses (or appear to focus) them at a point
A converging lens will bend incoming light that
is parallel to the principal axis toward the
principal axis.Any lens that is thicker at its
center than at its edges is a converging lens
with positive f.
A diverging lens will bend incoming light that is
parallel to the principal axis away from the
principal axis.Any lens that is thicker at its
edges than at its center is a diverging lens with
negative f
Rules For Converging Lenses
1)
2)
3)
Any incident ray traveling parallel to the
principal axis of a converging lens will
refract through the lens and travel
through the focal point on the opposite
side of the lens.
Any incident ray traveling through the
focal point on the way to the lens will
refract through the lens and travel
parallel to the principal axis.
An incident ray which passes through the
center of the lens will in effect continue
in the same direction that it had when it
entered the lens.
Ray Diagram for Converging Lens, S > f
 The image is real
 The image is inverted
 The image is on the back side of the lens
1 1 1
  
f s s
SS
s
S
S-
S-
Ray Diagram for Converging Lens, S < f
 The image is virtual
 The image is upright
 The image is larger than the object
 The image is on the front side of the lens
Object Outside 2F
F
2F
2F
Real; inverted;
diminished
F
1. The image is inverted, i.e., opposite to the
object orientation.
3. The image is diminished in size, i.e., smaller than
the object.
2. The image is real, i.e., formed by actual light
on the opposite side of the lens.
Image is located between F and 2F
Example 3. A magnifying glass consists of a converging
lens of focal length 25 cm. A bug is 8 mm long and
placed 15 cm from the lens. What are the nature, size,
and location of image.
S = 15 cm; f = 25 cm
F
F
1 1 1
 
s s' f
pf
(15 cm)(25 cm)
q

p f
15 cm - 25 cm
S-= -37.5 cm
The fact that S- is negative means that the image is virtual (on same side as object).
Object Between 2F and F
F
2F
2F
Real; inverted;
enlarged
F
1. The image is inverted, i.e., opposite to the
object orientation.
3. The image is enlarged in size, i.e., larger than the
object.
2. The image is real; formed by actual light
rays on opposite side
Image is located beyond 2F
Object at Focal Length F
F
2F
F
2F
Parallel rays;
no image
formed
When the object is located at the focal length, the rays of
light are parallel. The lines never cross, and no image is
formed.
.
Example
Where must an object be placed to have unit magnification (
M = 1.00) (a) for a converging lens of focal length 12.0
cm ? (b) for a diverging lens of focal length 12.0 cm ?
a
b
1 1 1
 
f s s
1 1 1
 
12 s s
1 2

12 s
s  24cm
1 1 1
  
f s s
1 1 1
  
12 s s
1 2
 
12 s
s  24cm
example
A person uses a converging lens that has a focal length of 12.5 cm to
inspect a gem. The lens forms a virtual image 30.0 cm away.
Determine the magnification. Is the image upright or inverted?
solution
1 1 1
 
f s s
1
1 1
 
12.5 s 30
s  8.82
Since M  0, the image is upright.
 (30)
M
 3.4
8.82
example
A ray that starts from the top of an object and runs
parallel to the axis of the lens, would then pass
through the
a)principal focus of the
lens
b)center of the
lens
C)secondary focus of the lens
Example 5:
Derive an expression for calculating the magnification of
a lens when the object distance and focal length are
given.
1 1 1
 
s s' f
sf
s 
s f
From last equation:
h  s 
M 
h
s
s = -s M
Substituting for q in second equation gives . . .
sf
sM 
s f
Thus, . . .
M 
f
s f
Diverging Thin Lens
Incoming parallel rays DIVERGE
from a common point FOCAL
We still call this the point
Same f on both sides of lens
Negative focal length
Thinner in center
Ray Diagrams for Thin Lenses – Diverging
 For a diverging lens, the following three rays are drawn:
 Ray 1 is drawn parallel to the principal axis and emerges
directed away from the focal point on the front side of the
lens
 Ray 2 is drawn through the center of the lens and continues
in a straight line
 Ray 3 is drawn in the direction toward the focal point on the
back side of the lens and emerges from the lens parallel to the
principal axis
Ray Diagram for Diverging Lens
 The image is virtual
 The image is upright
 The image is smaller
 The image is on the front side of the lens
Sign Conventions for Thin Lenses
Quantity
Positive When
Negative When
Object locatio (s)
Object is in front of
the lens
Object is in back of
the lens
Image location (sˊ)
Image is in back of
the lens
Image is in front of
the lens
Image height (h’)
Image is upright
Image is inverted
R1 and R2
Center of curvature is
in back of the lens
Center of curvature is
in front of the lens
Focal length (f)
Converging lens
Diverging lens
The power of lens
The reciprocal of the focal length = the power of lens
1
P
f ( m)
If the focal length
f
is measured in meters then ;p measured in diopters
if two lenses with focal length f1 and f2 placed next to each other are
equivalent to a single lens with a focal length f satisfying
1 1 1
 
f
f1 f 2
P  P1  P2
Spherical Aberration
 Results from the focal
points of light rays far
from the principle axis are
different from the focal
points of rays passing near
the axis
 For a mirror, parabolic
shapes can be used to
correct for spherical
aberration
Spherical Aberration
With SA
37
SA free
Chromatic Aberration
 Different wavelengths of light refracted
by a lens focus at different points
 Violet rays are refracted more than
red rays
 The focal length for red light is
greater than the focal length
 for violet light
 Chromatic aberration can be minimized
by the use of a combination of
converging and diverging lenses
Multiple lenses can be used
to improve aberrations
Spherical Aberration
Chromatic Aberration
Lens Aberrations
Chromatic aberration can be improved by combining two or
more lenses that tend to cancel each other’s aberrations. This
only works perfectly for a single wavelength, however.
An object is placed 6.0 cm in front of a convex thin
lens of focal length 4.0 cm. Where is the image
formed and what is its magnification and
power?
f = 4.0 cm
s = 6.0 cm
1
1
1
+
=
s’
s
f
1
1
1
- 6 = s’
4
1
P=
0.04 m
1
f
_
1
=
s
s’
s’ = 12 cm
= 25.0 D
M = - 12 / 6 = -2
Negative means real, inverted image
41
1
T . Norah Ali Almoneef
1
1
1
+
=
s’
s
f
42
T . Norah Ali Almoneef
43
T . Norah Ali Almoneef
Example 1. A glass meniscus lens (n = 1.5) has a
concave surface of radius –40 cm and a convex surface
whose radius is +20 cm. What is the focal length of the
lens.
+20 cm
R1 = 20 cm, R2 = -40 cm
1 1 
1
 (n  1)   
f
 R1 R2 
-40 cm
n = 1.5
 1
  2 1 
1
1
 (1.5  1) 



f
 20 cm (40 cm   40 cm 
f = 80.0 cm
Converging (+) lens.
Example: What must be the radius of the curved surface
in a plano-convex lens in order that the focal length be 25
cm?
R1 = , f= 25 cm
1 0 1 
1
 ( n  1)   
f
  R2 
 1  0.500
1
 (1.5  1)   
25 cm
R2
 R2 
R2 = 12.5 cm
R2=?
R1=
f=?
R2 = 0.5(25 cm)
Convex (+) surface.
Example : What is the magnification of a diverging lens (f = -20
cm) the object is located 35 cm from the center of the lens?
First we find q . . . then M
F
s 
1 1 1
 
s s' f
sf
35 cm  20 cm

 12 .7cm
s  f 35 cm  ( 20 cm)
M
s'  ( 12 .7cm)

 0.364
s
35 cm
h'
s'
M  
h
s
s = +12.7 cm
M = +0.364
Example
An object is placed 20 cm in front of a converging lens of
focal length 10 cm. Where is the image? Is it upright
or inverted? Real or virtual? What is the magnification
of the image?
s  20 cm
f  10 cm
1 1 1
1
1
  

s f s 10 cm 20 cm
1
2
1
1



s 20 cm 20 cm 20 cm
s  20 cm
47
Real image,
magnification
= 1
T . Norah Ali Almoneef
Example
An object is placed 8 cm in front of a diverging lens
of focal length 4 cm. Where is the image? Is it
upright or inverted? Real or virtual? What is the
magnification of the image?
f  4 cm (concave)
1 1 1
 
s s f
s  4 cm
1 1 1
1
1
  

s f
s  4 cm 4 cm
s  2cm  0
m   s / s  0.5  0
48
T . Norah Ali Almoneef
Example
24(b). Given a lens with a focal length f = 5 cm
and object distance p = +10 cm, find the
following: i and m. Is the image real or virtual?
Upright or inverted? Draw 3 rays.
1
1
1


s
s
f
1 1 1
1
 

s 5 10
10
49
s  10 cm
h
s
m 
h
s
T . Norah Ali Almoneef
10
m    1
10
Image is real,
inverted.
24(e). Given a lens with the properties (lengths in cm) R1 =
+30, R2 = +30, s = +10, and n = 1.5, find the following: f, s
and m. Is the image real or virtual? Upright or inverted?
Draw 3 rays.
Real side
.
R1
Virtual side
F1
R2
p
F2
1
1 1
1
 1.5  1    
f
 30 30  30
50
h
s
m 
h
s
1 1 1
 
s f s
1 1
1
 n  1   
f
 R1 R2 
f  30cm
.
1 1 1
1
  
s 30 10 15
s  15 cm
T . Norah Ali Almoneef

m
15
 1.5
10
Image is virtual,
upright.
Example
An object is placed 5 cm in front of a converging lens of
focal length 10 cm. Where is the image? Is it upright
or inverted? Real or virtual? What is the magnification
of the image?
s  5cm
f  10 cm
1 1 1
1
1
  

s
f
s 10 cm 5cm
1
1
2
1



s 10 cm 10 cm 10 cm
s  10 cm
1 1 1
 
s s f
Virtual image, as viewed from the right, the light appears to be coming
from the (virtual) image, and not the object.
T . Norah Ali Almoneef
Magnification
= +2
51
51
52
T . Norah Ali Almoneef