Optics
description
Transcript of Optics
Mirrors and Lenses
Optics
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
ReflectionWe describe the path of light as straight-line raysReflection off a flat surface follows a simple rule:
angle in (incidence) equals angle out (reflection)angles measured from surface “normal” (perpendicular)
1 )The incident ray,
the reflected ray
and the normal all lie in the same plane.
Laws of reflection
normal
incident ray reflected ray
mirror
ˊ
2)The incident angel = the reflected angel
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 behind the mirror equals the object distance 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!
S S
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)
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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.
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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.
150 m
150 m
(a) What is the distance between the girl and her image in the mirror? Distance = 150 2 = 300 cm
Signs: Image size and magnification
images can be upright (positive image size h’) or inverted (negative image size h’)Define magnification m = h’/hPositive magnification: image orientation unchanged relative to objectNegative magnification: image inverted relative to objectl m l < 1 if image is smaller than objectl m l > 1 if image is bigger than objectl 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 Lenses1) 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.
2) 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.
3) 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, Ray Diagram for Converging Lens, S > S > ff
The image is realrealThe image is invertedinvertedThe image is on the back sideon the back side of the lens
S
S
S-
S-
S-
s
ssf
111
Ray Diagram for Converging Lens, Ray Diagram for Converging Lens, SS < < ff
The image is virtualvirtualThe image is uprightuprightThe image is largerlarger than the objectThe image is on the front sidethe front side of the lens
Object Outside 2F
11 . .The image is The image is inverted,inverted, i.e., i.e., opposite to the object opposite to the object orientationorientation..
22 . .The image is The image is real, real, i.e., formed i.e., formed by actual light on the opposite by actual light on the opposite
side of the lensside of the lens . .
33 . .The image is The image is diminished diminished in size, in size, i.e., smaller than the objecti.e., smaller than the object.. Image is located
between F and 2F
Image is located between F and 2F
Real; inverted; diminished
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
(15 cm)(25 cm)
15 cm - 25 cm
pfq
p f
S-= -37.5 cm
The fact that S- is negative means that the image is virtual (on same side as object).
The fact thatS- is negative means that the image isvirtual on) (same side as object.
fss
1
'
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Object Between 2F and F
FF
FF
2F2F
2F2F
Real; inverted; enlarged
11 . .The image is The image is invertedinverted, i.e., , i.e., opposite to the object orientationopposite to the object orientation.. 22 . .The image is The image is realreal; formed by ; formed by
actual light rays on opposite actual light rays on opposite sideside33 . .The image is The image is enlarged enlarged in size, i.e., in size, i.e.,
larger than the objectlarger than the object.. Image is located beyond 2F
Image is located beyond 2F
Object at Focal Length F
When the object is located at the focal length, the rays of light are parallel. The lines never cross, and no image is 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.
Parallel rays; no image formed
. ExampleExample 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 ?
cmss
ss
ssf
24
2
12
1
11
12
1
111
ba
cmss
ss
ssf
24
2
12
1
11
12
1
111
exampleexample
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?
4.382.8
)30(
82.830
11
5.12
1
111
M
ss
ssf
solutionsolutionSince 0M ,the image is upright .
exampleA 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.
From last equation: = -s MSubstituting for q in second equation gives. . .
Thus . . . ,
fss
1
'
11 s
s
h
hM
fs
sfs
fs
fM
fs
sfsM
s
Diverging Thin Lens
Incoming parallel rays DIVERGE from a common point FOCALWe still call this the pointSame f on both sides of lensNegative focal lengthThinner in center
Ray Diagrams for Thin Lenses – Ray Diagrams for Thin Lenses –
DivergingDiverging
For a diverging lensdiverging lens, the following three rays are drawn:Ray 1Ray 1 is drawn parallel to the principal parallel to the principal
axisaxis and emerges directed away from the away from the focal point on the front sidefocal point on the front side of the lens
Ray 2Ray 2 is drawn through the centerthrough the center of the lens and continues in a straight linecontinues in a straight line
Ray 3Ray 3 is drawn in the direction toward the direction toward the focal point on the back sidefocal point on the back side of the lens and emerges from the lens parallel to the parallel to the principal axisprincipal axis
Ray Diagram for Diverging LensRay Diagram for Diverging Lens
The image is virtualvirtualThe image is uprightuprightThe image is smallersmallerThe image is on the front sidethe front side of the lens
Sign Conventions for Thin Lenses
QuantityPositive 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 uprightImage is inverted
R1 and R2Center of curvature is in back of the lens
Center of curvature is in front of the lens
Focal length (f)Converging lensDiverging lens
The power of lens
The reciprocal of the focal length = the power of lens
)(
1
mfP
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
21
21
111
PPP
fff
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
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With SA SA free
Chromatic AberrationDifferent wavelengths of light refracted
by a lens focus at different pointsViolet rays are refracted more than
red raysThe 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.
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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 ?
s = 6.0 cm f = 4.0 cm
P = 1
0.04 m =25.0 D
1s
1s’
1f
+ =1
s1 1f
=
s’
_
16
1 14
=
s’
- s’ =12 cm
Negative means real, inverted image
M = - 12 / 6 = -2
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1s
1s’
1f
+ =
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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.
R1 = 20 cm, R2 = -40 cm
--4040 cmcm
++2020 cmcm
n = 1.5n = 1.51 2
1 1 1( 1)n
f R R
1 1 1 2 1(1.5 1)
20 cm ( 40 cm 40 cmf
f = 80.0 cmf = 80.0 cm Converging (+) lensConverging (+) 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
2
1 1 1( 1)n
f R
R1= R2?=
f ? =
00
2 2
1 1 0.500(1.5 1)
25 cm R R
R2 = 12.5 cmR2 = 12.5 cm Convex (+) surface.
R2 = 0.5(25 cm)
Example : What is the magnification of a diverging lens (f = -20 cm) the object is located 35 cm from the center of the lens?
FF
First we find q . . . then M
s = +12.7 cm
M = +0.364
fss
1
'
11
cmcmcmcmcm
fssf
s 7.12)20(35
2035
s
s
h
hM
''
364.035
)7.12('
cmcm
ss
M
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ExampleAn 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?
Real image ,magnification =
cmscmcmcms
cmcmsfs
cmf
cms
2020
120
120
2120
110
1111
10
20
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ExampleAn 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?
05.0/
0241
41111
4
111
(concave) 4
ssm
cmscmcmsfs
cmsfss
cmf
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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.
Image is real ,inverted.
m10
10 1
Example
fss111
101
101
511
s
cms 10
ss
hh
m
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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.
cmf 30
m 15
101.5
Image is virtual, upright.
Virtual side Real side
R1. .F1 F2
pR2
21
111
1RR
nf
301
301
301
15.11
f
ss
hh
m
sfs111
151
101
3011
s
cms 15
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ExampleAn 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?
Virtual image, as viewed from the right, the light appears to be coming from the (virtual) image, and not the object.
Magnification = +251
cmscmcmcms
cmcmsfs
cmf
cms
1010
110
210
115
110
1111
10
5
fss111
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