Physics 141 Lecture 27 Lecture 27 - Mirrors...Fig. 1.7: Setup to observe the image from a concave...

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!


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


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


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


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


You will also get Images from Curved Mirrors:


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:


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:


object

image

SSʹ′

f

S > 2fimage is:

realinvertedsmaller

2f f


f > 0s > 0s’ > 0

SS’ f

object

image

S = 2f

2f

f

image is:real

invertedsame size


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


S’< 0f


f > S > 0

S

f

image is:virtualuprightbigger

f > 0s > 0s’ > 0


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


object image(virtual)

S > 0 Sʹ′< 0

f < 0

S > 0 image is:virtualuprightsmaller

f > 0s > 0s’ > 0


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)


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:


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


CheckPoints 4 & 5

0

20

40

60

80

1

0

20

40

60

80

1


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


f > S > 0 rays no longer meetin front of the mirror

but they do meet behind the mirror

S

f


0

18

35

53

70

1

CheckPoint 9

object image(virtual)

S > 0 Sʹ′ < 0

f < 0

S > 0 image is:virtualuprightsmaller


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)


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


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)


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)


Magnifica9on equa9on

Physics 141 Lecture 27 Lecture 27 - Mirrors...Fig. 1.7: Setup to observe the image from a concave...

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