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Physics 141Lecture 27
Today’s Concept:
A) Mirrors
Electricity & Magne9sm Lecture 27, Slide 1
Reflection
Angle of incidence = Angle of reflec9on
θi = θr
θi
θr
That’s all of the physics – everything else is just geometry!
Electricity & Magne9sm Lecture 27, Slide 4
Flat Mirror
All you see is what reaches your eyesYou think object’s loca9on is where rays appear to come from.
θrθi
Flat Mirror
Object
All rays origina9ng from peak will appear to come from same point behind mirror! Image
Electricity & Magne9sm Lecture 27, Slide 5
Flat Mirror
3) Lines appear to intersect a distance d behind mirror. This is the image loca9on.
Virtual Image: No light actually gets here
d d
1) Draw first ray perpendicular to mirror 0 = θi = θr
2) Draw second ray at angle. θi = θr
θrθi
Electricity & Magne9sm Lecture 27, Slide 6
Clicker Question
A woman is looking at her reflec9on in a flat ver9cal mirror. The lowest part of her body she can see is her knee.
If she stands closer to the mirror, what will be the lowest part of her reflec9on she can see in the mirror.
A) Above her knee
B) Her knee
C) Below her knee
Electricity & Magne9sm Lecture 27, Slide 7
Clicker Question
A woman is looking at her reflec9on in a flat ver9cal mirror. The lowest part of her body she can see is her knee. If she stands closer to the mirror, what will be the lowest part of her reflec9on she can see in the mirror.
A) Above her knee
B) Her knee
C) Below her knee
If the light doesn’t get to your eye then you can’t see it
Electricity & Magne9sm Lecture 27, Slide 8
You will also get Images from Curved Mirrors:
Electricity & Magne9sm Lecture 27, Slide 9
Concave: Consider the case where the shape of the mirror is such that light rays parallel to the axis of the mirror are all “focused” to a common spot a distance f in front of the mirror:
These mirrors are o_ensec9ons of spheres
(assumed in this class).
For such “spherical” mirrors, we assume all angles are
small even though we draw them big to make it easy to
see…
Note: analogous to “converging lens”Real object can produce real image
fElectricity & Magne9sm Lecture 27, Slide 10
2f f
For a spherical mirror, R = 2f
R
center of spheresome9mes labeled “C”
Aside:
Electricity & Magne9sm Lecture 27, Slide 11
object
1) Draw ray parallel to axis reflec9on goes through focus
2) Draw ray through focus reflec9on is parallel to axis
image
You now know the posi9on of the same point on the image
2f f
normal
normal
Note: any other ray from 9p of arrow will be reflected according to θi = θr and will intersect the two rays shown at the image point.
Recipe for Finding Image:
Electricity & Magne9sm Lecture 27, Slide 12
object
image
SSʹ′
f
S > 2fimage is:
realinvertedsmaller
2f f
Electricity & Magne9sm Lecture 27, Slide 13
f > 0s > 0s’ > 0
SS’ f
object
image
S = 2f
2f
f
image is:real
invertedsame size
Electricity & Magne9sm Lecture 27, Slide 14
SS’f
object
image
2f > S > f
2f f
image is:real
invertedbigger
Electricity & Magne9sm Lecture 27, Slide 15
f
objectimage(virtual)
f > S > 0 rays no longer meetin front of the mirror
but they do meet behind the mirror
S
f
Electricity & Magne9sm Lecture 27, Slide 16
S’< 0f
objectimage(virtual)
f > S > 0
S
f
image is:virtualuprightbigger
f > 0s > 0s’ > 0
Electricity & Magne9sm Lecture 27, Slide 17
f
Convex: Consider the case where the shape of the mirror is such that light rays parallel to the axis of the mirror are all “focused” to a common spot a distance f behind the mirror:
Note: analogous to “diverging lens”Real object will produce virtual image
Electricity & Magne9sm Lecture 27, Slide 18
object image(virtual)
S > 0 Sʹ′< 0
f < 0
S > 0 image is:virtualuprightsmaller
f > 0s > 0s’ > 0
Electricity & Magne9sm Lecture 27, Slide 19
S > 2freal
invertedsmaller
2f > S > f real
invertedbigger
f > S > 0 virtualuprightbigger
S > 0 virtualuprightsmaller
f > 0
f < 0
Executive Summary – Mirrors & Lenses:
fConverging
Diverging ffConvex
(diverging)
fConcave(Converging)
Electricity & Magne9sm Lecture 27, Slide 20
Lens sign conven9ons
S: posi9ve if object is “upstream” of lens S’ : posi9ve if image is “downstream” of lensf: posi9ve if converging lens
Mirrors sign conven9ons
S: posi9ve if object is “upstream” of mirror S’: posi9ve if image is “upstream” of mirrorf: posi9ve if converging mirror (concave)
s’ is posi9ve for a real imagef is posi9ve when it can produce a real image
You just have to keep the signs straight:
It’s Always the Same:
Electricity & Magne9sm Lecture 27, Slide 21
Happy Day!
Unit 29: One day only
Suggested ac9vi9es:
*Ac9vity 29-‐5: Ray Tracing to Locate the Virtual Image
*Ac9vity 29-‐7: Forming an Image with a Concave Mirror
No grades for this unit, par9cipa9on only.We’ll move on to Unit 30 on Friday
(c) Part of the law of reflection states that the incident ray, the reflected ray and the normal to the reflecting surface lie in the same plane. Is this adequately tested by this experiment? Explain.
Experiment OP1: Ray Optics, Reflection and Concave Mirrors
OP1.5
90°
90°
d1
d2
Image ofFilament
Fig. 1.5: Finding the Virtual Image of the Filament.
Part IV - Using Ray Tracing to Locate the Virtual Image (C)
In this part you will use the method developed in part II to find the position of
the virtual image of a filament formed by a plane mirror. The basic idea isillustrated in Figure 1.5. The (C) means that you may use your own ideas and
be creative in doing this part. Take notes clearly describing the procedure in
your lab notebook. Analyze the data and discuss the results. [Do not write on the
ray table.]
Part V - The Focal Length of a Concave Mirror
The image formed by a plane mirror is always virtual. A concave mirror can
form either a real image or a virtual image. A real image occurs if the rays come
from a distant source. If the source is closer than one focal length from the
mirror, then a virtual image is formed. If the source is exactly at the mirror’s
focal point, the rays emerge parallel from the mirror. This is the principle used in
flashlights, headlights and spotlights to produce a parallel beam of light.
Ideally, concave mirrors have a paraboloidal shape. Spherical shapes are
cheaper to manufacture than paraboloids and most inexpensive concave mirrors
are spherical. For most applications the loss in quality is insignificant,
particularly where the mirror’s f-number (focal length/diameter) is large. In the
following exercise you will use the cylindrical ray optics mirror to visualize its
image-forming properties and find the mirror’s focal length by producing a
parallel beam of rays.
1. Mount the slit plate and ray table. Place the slit plate directly on the light
source and move the ray table so its edge touches it. Use the ray table
surface with a centimetre grid and turn it so that the grid is aligned with the
optical bench axis.
Figure 29.6: Finding a virtual image.
Page 29-10 Studio Physics Activity Guide SFU
© 2008 by S. Johnson. Adapted from PHYS 131 Optics lab #1
Physics 131 Laboratory Manual
OP1.6
move source
image moves
y mm y' mm
Fig. 1.6: Rays
when source is at
focal point of the
mirror
Concave Mirror
Fig. 1.7: Setup to observe the image from a concave mirror
2. Observe the real image formed by a concave mirror. Place the ray
optics mirror on the ray table, turn on the light and observe the rays
reflected from the concave surface. When the mirror is at the far end of the
table you should see the reflected rays converging at the position of the real
image.
3. Rotate mirror to show aberration. If you turn the mirror so that the
image position is off the optical axis of the bench, you will notice that not
all rays converge at the same place. This is because the mirror’s shape is
circular instead of parabolic. This effect is known as spherical aberration
when it occurs in spherical mirrors.
4. Measure the source-mirror and image-mirror distances and
magnification. Orient the ray mirror so that the image is exactly on the
optic axis. Record the object-to-mirror distance, the image-to-mirror
distance. Slide the light source box a few mm away from the alignment rail
keeping its edge parallel to the
rail. Record the direction and
distance that the image is
displaced for a given light
source displacement. This
shows the magnification and
inversion of the image.
5. Find where the mirror gives
parallel reflected rays. Slide
the mirror closer to the light
source until the reflected rays
diverge. The rays don’t actually cross now, but their projections on the
dark side of the mirror will cross at the position of the virtual image. Move
the mirror back until the emerging rays are parallel. Measure the distance
between the light source and the mirror. Remove the mirror from the table
and have your partner repeat this measurement to get two independent
values for the focal length
VI - Forming an Image with a Concave Mirror.
If you cut a strip along any diameter of a spherical mirror you get something
close to a cylindrical mirror. It's not surprising then that we can understand
spherical mirrors from the properties of cylindrical mirrors studied in the last
part.
1. Arrange components. Set up the equipment as shown in Fig 1.7 with the
concave side of the mirror facing the light source. The viewing screen
should cover only half of the hole in the component holder so that light
from the filament reaches the mirror.
2. Find focal point of a distant object. Estimate the focal length of the
.
Figure 29.7: Set-up to observe the image from a concave mirror
(b) Choose five distances between 50.0 cm and 5.0 cm. These are object-mirror distances o for which you will measure the image-mirror distances i and the image heights h’. Record o, i, h and h’ in the table below. Calculate f from o and i "in each instance and record these values in the table below. Show at least one calculation of f in the space below.
o i h h’ f(calc)
(c) How do your five values of f agree?
Unit 29 – Reflection Page 29-15Author: Sarah Johnson
© 2008 by S. Johnson.
C is not correct as it does not go through the focal point.
0
15
30
45
60
1
CheckPoint 2
Electricity & Magne9sm Lecture 27, Slide 22
CheckPoints 4 & 5
0
20
40
60
80
1
0
20
40
60
80
1
Electricity & Magne9sm Lecture 27, Slide 23
0
13
25
38
50
1
If the object is behind the focal length it will reflect an inverted image.
If the object is in front of the focal length it will produce a virtual upright image.
CheckPoint 7
SS’
f
object
image
2f > S > f
2f f
image is:real
invertedbigger
f
objectimage(virtual)
f > S > 0 rays no longer meetin front of the mirror
but they do meet behind the mirror
S
f
Electricity & Magne9sm Lecture 27, Slide 24
0
18
35
53
70
1
CheckPoint 9
object image(virtual)
S > 0 Sʹ′ < 0
f < 0
S > 0 image is:virtualuprightsmaller
Electricity & Magne9sm Lecture 27, Slide 25
An arrow is located in front of a convex spherical mirror of radius R = 50cm.The 9p of the arrow is located at (−20cm,−15cm).
Where is the 9p of the arrow’s image?
Conceptual Analysis Mirror Equa9on: 1/s + 1/sʹ′ = 1/f Magnifica9on: M = −s’/s
Strategic Analysis Use mirror equa9on to figure out the x coordinate of the image Use the magnifica9on equa9on to figure out the y coordinate of the 9p of the image
Calculation
R = 50y
x(–20,–15)
Electricity & Magne9sm Lecture 27, Slide 26
What is the focal length of the mirror?
A) f = 50cm B) f = 25cm C) f = −50cm D) f = −25cm
Rule for sign: Positive on side of mirror where light goes after hitting mirror
f = − 25 cm
< 0
Calculation
R = 50y
x(-20,-15)
An arrow is located in front of a convex spherical mirror of radius R = 50cm.The 9p of the arrow is located at (−20cm,−15cm).
For a spherical mirror | f | = R/2 = 25cm.
f
Ry
Electricity & Magne9sm Lecture 27, Slide 27
What is the x coordinate of the image?
A) 11.1 cm B) 22.5 cm C) −11.1 cm D) −22.5cm
s = 20 cmf = −25 cm
= −11.1 cm
Since sʹ′ < 0 the image is virtual (on the “other” side of the mirror)
CalculationAn arrow is located in front of a convex spherical mirror of radius R = 50cm.The 9p of the arrow is located at (−20cm,−15cm).
f = −25 cm
R = 50y
x
(-20,-15)
Electricity & Magne9sm Lecture 27, Slide 28
Mirror equa9on
What is the y coordinate of the 9p of the image?
A) −11.1 cm B) −10.7 cm C) −9.1 cm D) −8.3cm
x = 11.1cm
s = 20 cmsʹ′= −11.1 cm
M = 0.556
yimage = 0.55 yobject = 0.556⨉(−15 cm) = −8.34 cm
CalculationAn arrow is located in front of a convex spherical mirror of radius R = 50cm.The 9p of the arrow is located at (−20cm,−15cm).
f = −25 cm
R = 50y
x
(-20,-15)
Electricity & Magne9sm Lecture 27, Slide 29
Magnifica9on equa9on