Download CHAPTER 32 OPTICAL IMAGES • Introduction to ray diagrams

Introduction to ray diagrams:
CHAPTER 32
OPTICAL IMAGES
• Introduction to ray diagrams
*
object
image
• Reflection from mirrors
• Plane mirrors
• Spherical mirrors
• Sign convention
• Lenses
• Converging Lenses
• Diverging Lenses
• Chromatic aberration
*
Rays emerge from all points on an object. Only some of
those from the top of the head are shown here. Note: the
image does not really exist ... you cannot display it on a
screen, for example! It is called a virtual image.
** Note that only 2 rays are required to locate
the image position **
Geometry of image formation (plane mirror):
s′
s
B
1
θ
A
Q
θ
θ
θ
θ
O
Question 32.1: You are 1.62 m tall and want to be able to
see your full image in a vertical plane mirror.
P
(a) What is the minimum height of the mirror that will
meet your needs?
2
(b) How far above the floor should the bottom of the
ΔOAB and ΔOPQ are congruent triangles,
i.e., OA = OP and AB = PQ
mirror in (a) be placed, assuming that the top of your
head is 14 cm above your eye level?
(c) If you move closer to the mirror, describe what
Lateral magnification m =
PQ
( = 1).
AB
Also,
you would see in the mirror.
Use ray diagrams to explain your answers.
object distance (OA) = image distance (OP)
i.e., s = s′ .
Image is virtual and upright.
1.62 m
A ray diagram showing rays from your feet and the top
of your head reaching your eyes is shown above.
(a) Therefore, the mirror must be half your height, i.e.,
0.81 m.
(c) You move closer to the mirror. A ray diagram
showing rays from your feet and the top of your head
(b) The top of the minimum height mirror must be
halfway between your eyes and the top of your head,
i.e., 1.55 m from the floor. Therefore, the bottom of the
mirror should be
1.55 m − 0.81 m = 0.74 m,
above the floor.
reaching your eyes is shown above, with the same size
mirror. Notice you can still see all your body; it makes
no difference where you stand!
Formation of an image with two plane mirrors.
P1
P2
1
Reflection in a plane mirror produces:
2
lateral inversion
z′
y′
z
y
x
RH set of axes
(object)
x′
LH set of axes
(image)
The rays from the object into the eye satisfy the law of
reflection. We see that the image at P1 due to mirror 1
acts as an object for mirror 2 . The final image appears
at P2. Note that both images are virtual.
IMPORTANT CONCEPT: Even though the image at
P1 is virtual, it can act as an object for mirror 2 .
Question 32.2: When two plane mirrors are parallel, such
as on opposite walls in a hairdressing salon, multiple
images arise because each image in one mirror serves as
Even though a virtual image does not actually exist, you
an object for the other mirror. An object is placed
can see it and photogragh it because it can act as an
between parallel mirrors separated by 3.00 m. The object
object for another lens or mirror and produce a real
is 1.00 m in front of the left mirror and 2.00 m in front of
image.
the right mirror.
PS: Ever done a selfie in a mirror?
(a) What is the distance from the left mirror to the
first four images in that mirror?
(b) What is the distance from the right mirror to the
first four images in that mirror?
3.0 m
1.0 m
5.0 m
2.0 m
4.0 m
See the notes I wrote on
7.0 m
8.0 m
“retroreflection”
10.0 m
11.0 m
available on the useful notes link on the web-site.
(a) For the left hand mirror (red) the images are
1.0 m; 5.0 m; 7.0 m; 11.0 m
behind the mirror.
(b) For the right hand mirror (blue) the images are
2.0 m; 4.0 m; 8.0 m; 10.0 m
behind the mirror.
REFLECTION FROM CURVED SURFACES
1. Concave mirror
optical
axis object
2
1
θ
α
φ
•C
β
ℓ
θ
γ
image
r
s′
optical
axis
object
•
image
r
s′
s
s
NOTE: The image produced in this example is real; it
Using simple geometry we have:
actually exists, i.e., we could see it on a screen.
β = α + θ and γ = α + 2θ.
Eliminate θ and we get:
α + γ = 2β
If α, β and γ are small then:
ℓ
ℓ
ℓ
α≈
β≈
γ≈
s
r
s′
1 1 2
i.e., + =
s s′ r
My convention is that if the rays forming the image are
solid lines, it is a real image. Dashed lines are rays that
do not actually exist ... they show paths that rays appear
to follow, for example, behind a mirror.
Beam parallel to the
optical axis (i.e., from
a very distant object)
•C
•
F
r
s= ∞
•
s′
C
1 1 2
+ =
s s′ r
r
If s = ∞ then s′ = = f ,
2
where F is the principal focus and f is the focal length.
1 1 1
∴ + = .
s s′ f
Note, it is a real focus.
Can you think of an example?
s= f
s′ = ∞
To produce a parallel beam of light put the source at
the principal focus.
Can you think of an example?
Image formation in a concave mirror
Tee, hee, hee ...
there’s a sign convention
for mirrors and lenses!
I. Object distance: s > r
s
object
y
•F y′ •C
image
•
light, s > 0.
Image distance ( s′ ): if the image is on the same side
of the reflecting or refracting surface as the outgoing
light, s′ > 0.
•
1
Object distance ( s): if the object is on the same side
of the reflecting or refracting surface as the incoming
•
2
Radius of curvature (r): if the center of curvature is
on the same side as the outgoing light, r > 0.
s′
s>0
s′ > 0
r>0
The lateral (or linear) magnification is defined as
m = y′ y .
But y s = − y ′ s′ .
∴m = − s′ s
(⇒ negative).
Definition of magnification.
Note in this case that m < 0 and m < 1. The image is
real, inverted and reduced.
Image formation in a concave mirror
Image formation in a concave mirror
II. Object distance: r 2 < s < r (i.e., f < s < r)
III. Object distance: s < f (i.e., s < r 2)
s
image
object
•F
y′
y
object
y
•F
•C
y′
s
•C
s′
image
1
2
s′
s>0
2
1
s′ > 0
r>0
s>0
s′ < 0
r>0
The lateral (or linear) magnification is
m = y ′ y = − s′ s (⇒ negative).
The lateral (or linear) magnification is
m = y ′ y = s′ s (⇒ positive).
Note that m < 0 and m > 1. The image is real, inverted
Note that m > 1. The image is virtual, upright and
and enlarged.
enlarged.
Here’s a summary!
For a concave mirror ...
F
C
r
f >0
m = − s′ s .
Question 32.3: A concave mirror has a radius of
curvature of 24.0 cm. Find the image position for
objects that are placed
• When s > r: then m < 0 and m < 1. The image is
real, inverted and reduced.
• When f < s < r: then m < 0 and m > 1. The image
is real, inverted and enlarged.
• When s < f : then m > 1. The image is virtual,
upright and enlarged.
positive m ( m > 0) ⇔ virtual image (upright)
negative m ( m < 0) ⇔ real image (inverted)
(a) 55.0 cm,
(b) 24.0 cm,
(c) 12.0 cm, and
(d) 8.00 cm from the mirror.
For each case give the magnification and state whether
the image is real or virtual; upright or inverted.
Given r = 0.24 m. Since the mirror is concave then r and
f ( = r 2 = 0.12 m) are both positive.
1 1 1
(a) For s = 0.55 m and using + = , we have:
s s′ f
1
1
1
=
−
= 6.52 m −1, i.e., s′ = 0.15 m.
s′ 0.12 m 0.55 m
∴m = − s′ s = −0.15 m 0.55 m = −0.27.
Since s′ > 0, m < 0 and m < 1, the image is on the same
side as the object, it is real, inverted and reduced.
(b) For s = 0.24 m:
1
1
1
=
−
= 0.417 m −1, i.e., s′ = 0.24 m.
s′ 0.12 m 0.24 m
∴m = − s′ s = −0.24 m 0.24 m = −1.00.
Since s′ > 0, m < 0 and m = 1, the image is on the same
side as the object, it is real, inverted and the same size
as the object.
(c) s = 0.12 m:
1
1
1
1
=
−
= 0, i.e., s′ = − = ∞.
s′ 0.12 m 0.12 m
0
So, the emerging rays are parallel and do not form an
image.
(d) s = 0.08m:
1
1
1
=
−
= −4.17 m −1, i.e., s′ = −0.24 m.
s′ 0.12 m 0.08 m
∴m = − s′ s = 0.24 m 0.08 m = 3.0.
Since s′ < 0, m > 0 and m > 1, the image is on the
opposite side to the object, it is virtual, upright and
enlarged.
2. Convex mirror:
2
2
θ
θ
θ
θ
1
γ
α
object
s
s′
θ
r
β
image
1
•C
θ = α + β and 2θ = α + γ
Eliminate θ and we get: γ − α = 2β
ℓ ℓ 2ℓ
1 1 2
i.e., − =
so, − = .
s′ s r
s′ s r
But, using the sign convention we find:
s′ < 0
1 1 2
∴ + = ,
s′ s r
s
s′
θ
r
β
image
•C
1 1 2
+ = .
s′ s r
Using simple geometry:
s>0
object
γ
α
r < 0,
which is the same result obtained for a concave mirror!
When s = ∞, i.e., an incoming parallel beam of light, we
1 2 1
have:
= =
s′ r f
r
i.e., f =
(virtual focus),
2
1 1 1
so + = .
s′ s f
** Note that by the sign convention, r < 0, and so
f < 0 for a convex lens **
Image formation in a convex mirror
1
•
object
2
image
s′
F
C
•
Question 32.4: A convex mirror has a radius of
curvature of 24.0 cm. Find the image position for
objects that are placed
s
(a) 55.0 cm,
By simple geometry the lateral magnification of a
(b) 24.0 cm,
convex mirror is:
(c) 12.0 cm, and
m = y ′ y = − s′ s ,
but s′ < 0, so m > 0.
(d) 8.00 cm from the mirror.
For each case give the magnification and state whether
the image is real or virtual; upright or inverted.
Note also, 0 < m < 1 always and so the image is upright,
virtual and reduced in size.
Can you think of an example ?
Given r = 0.24 m. Since the mirror is convex then r and
f ( = r 2 = 0.12 m) are negative.
(a) s = 0.55 m:
1
1
1
=−
−
= −10.2 m −1,
s′
0.12 m 0.55 m
i.e., s′ = −0.10 m.
∴m = − s′ s = 0.10 m 0.55 m = +0.18.
Since s′ < 0, m > 0 and m < 1, the image is on the
opposite side as the object, it is virtual, upright and
reduced.
(b) s = 0.24 m:
1
1
1
=−
−
= −12.5 m −1,
s′
0.12 m 0.24 m
i.e., s′ = −0.08 m.
∴m = − s′ s = 0.08 m 0.24 m = +0.33.
Since s′ < 0, m > 0 and m < 1, the image is on the
opposite side as the object, it is virtual, upright and
reduced.
(c) s = 0.12 m:
1
1
1
=−
−
= −16.7 m −1,
s′
0.12 m 0.12 m
i.e., s′ = −0.06 m.
∴m = − s′ s = 0.06 m 0.12 m = +0.50.
Since s′ < 0, m > 0 and m < 1, the image is on the
opposite as the object, it is virtual, upright and reduced.
(d) s = 0.08 m:
1
1
1
=−
−
= −20.83 m −1,
s′
0.12 m 0.08 m
i.e., s′ = −0.048 m.
∴m = − s′ s = 0.048 m 0.08 m = +0.60.
Since s′ < 0, m > 0 and m > 1, the image is on the
opposite side to the object, it is virtual, upright and
reduced.
(b) The mirror cannot be convex because a convex
mirror always produces a diminished virtual image.
Therefore, it must be concave.
s = 2.10 cm.
(a)
Question 32.5: A dentist wants a small mirror that will
s
produce an upright image that has a magnification of
s′
The image is
upright and so it is
virtual, i.e., behind
5.50 when the mirror is located 2.10 cm from a tooth.
the mirror, so s′ < 0.
(a) Should the mirror be concave or convex?
(b) What should the radius of curvature of the
mirror be?
The magnification: m = − s′ s = 5.50
∴ s′ = −5.50 × 2.10 cm = −11.6 cm.
1 1 1 2
1
1
∴ + = = =
−
= 0.39 cm−1
s s′ f r 2.10 cm 11.6 cm
2
i.e., r =
= 5.13 cm.
0.39 cm−1
∴f = r 2 = 2.56 cm,
so s < f .
Types of Lenses ...
Converging
Lenses
Planoconvex
Biconvex**
Positive Meniscus
See the notes on the web-site for chapter 32 on how to
solve mirror problems using a scale drawing ...
Diverging
Lenses
Planoconcave
Biconcave**
Negative Meniscus
Cross-sections of various types of lenses. The upper
group are converging lenses, the lower group are
diverging lenses.
Convex lens ...
Converging lens
(convex)
5.0 cm
Principal
focus
Parallel
beam
Real focus
Diverging lens
(concave)
Focal length
(f)
Lens-maker’s formula:
n1
Principal
focus
Parallel
beam
n2
r1
Real focus
r2
Focal length
(f)
1 ⎛ n2 ⎞ ⎛ 1 1 ⎞
= ⎜ − 1⎟ ⎜ − ⎟
f ⎝ n1 ⎠ ⎝ r1 r2 ⎠
Virtual focus
Optometrists who prescribe lenses and opticians who
Principal
focus
make them do not normally specify the focal length of a
Parallel
beam
lens but its refractive power, in units called diopters.
Refractive power =
Focal length
(f)
1
(diopters)
f (in meters)
Converging lenses have f > 0 but diverging lenses have
f < 0.
Sign convention: If the center of curvature is on the same
side as the outgoing light r > 0 .
Incoming light
•
•
r2
r1 > 0
Incoming light
r2 < 0
Question 32.6: Place the following in order of
increasing focal length, i.e., from shortest to longest:
r1
Outgoing light
r2
•
•
r1
Incoming light
Outgoing light
r1
r1 < 0
r2 > 0
Outgoing light
r1
•
•
r2
r1 > 0
r2 > 0
1
2
r1
r1
r1
r2
3
r2 > r1
r1
1
r1 > 0
r2 = ∞
r1
r1
r1
r2
2
both positive
3
r2 > r1
both positive
1
1
r
1:
= (n − 1) . ∴f1 = 1 .
f1
r1
(n − 1)
2:
3:
⎛ 1 1⎞
1
= (n − 1)⎜ − ⎟ = 0. ∴f 2 = ∞.
f2
⎝ r1 r1 ⎠
1
⎛ 1 1⎞
⎛r −r ⎞
= (n − 1)⎜ − ⎟ = (n − 1)⎜ 2 1 ⎟ .
f3
⎝ r1 r2 ⎠
⎝ r1r2 ⎠
r
r2
∴f 3 = 1
.
(n − 1) (r2 − r1 )
>1
∴f1 < f 3 < f 2 .
Note, if n1 = 1 (air), the lens maker’s formula becomes:
⎛ 1 1⎞
1
= (n 2 − 1)⎜ − ⎟ .
f
⎝ r1 r2 ⎠
But n 2 varies with wavelength (λ) ... dispersion.
Therefore, the focal length, f, varies with light color.
Since, for glass, n 2 increases for decreasing λ (i.e., from
red light to blue light ), the focal length for blue light is
less than for red light. So, blue light is “bent” more than
red light.
blurred focus
“chromatic aberration”
** See the articles on the web-site I wrote about
aberration **
** See also lens-maker’s formula
under “useful notes” **
Incoming light
Outgoing light
r1
•
•
r1
•
•
r2
Question 32.7: Find the focal length of a lens that has
an index of refraction of 1.62, a convex surface with a
radius of curvature of r1 = 0.40 m and a concave
surface with a radius of curvature of r2 = 1.00 m.
r2
Using the sign convention r1 > 0 and r2 > 0. We have
1 ⎛ n2 ⎞ ⎛ 1 1 ⎞
= ⎜ − 1⎟ ⎜ − ⎟ ,
f ⎝ n1 ⎠ ⎝ r1 r2 ⎠
where n 2 = 1.62 and n1 = 1.00.
1
1
1 ⎞
∴ = (1.62 −1)⎛
−
= 0.93 m −1,
⎝ 0.40 m 1.00 m ⎠
f
i.e., f = 1.08 m.
Convex lens ...
f
s′
s
f
F
F
s
y
y′
virtual
upright
image
Image
Object
f
f
If s > 2f then m < 1, i.e., an inverted and reduced image.
s′ < 0
0
For a convex lens we have (see “useful notes”):
1 1 1
+ =
s s′ f
and
m = y ′ y = − s′ s .
s′ − f
f
(Also, by substitution: m = −
=−
.)
f
s−f
f
s−f
s>0
f >0
m= −
real
inverted
image
s′ > 0
object distance (s)
s<f
s= f
s>f
When:
•
s>f
•
s = 2f then
m = 1 (real, inverted)
•
s=f
then
m= ∞
•
s<f
then
m > 1 (virtual, upright)
then
m < 0 (real, inverted)
Interesting case !!
s′
s
Object
Question 32.8: An object that is 0.03 m high is placed
Image
f
0.25 m in front of a thin lens that has a power equal to
f
10.0 D.
m= −
f
s− f
If s < f , then m > 1. Also,
1 1
1
since > , then < 0,
s f
s′
i.e., s′ < 0.
So, to use a convex lens as a magnifying glass make
s < f , i.e., object distance less than the focal length of
the lens. The image is on the same side as the object
and it is enlarged, upright and virtual.
(a) Find the position and the size of the image.
(b) Repeat part (a) if the object is placed 0.20 m in
front of the lens.
(c) Repeat part (a) if the object is placed 0.05 m in
front of the lens.
1
The power of a lens is given by P = .
f
∴f = P −1 = 0.10 m.
1 1 1
(a) With s = 0.25 m. The lens formula is + = .
s s′ f
1 1 1 s−f
fs
∴ = − =
, i.e., s′ =
.
s′ f s
fs
s−f
With s = 0.25 m and f = 0.10 m, we find
(0.10 m)(0.25 m)
s′ =
= 0.167 m.
(0.25 m − 0.10 m)
∴m = −
s′
0.167 m
=−
= −0.667.
s
0.25 m
The object height is y = 0.03 m, so the image height is
y ′ = my = (−0.667)(0.03 m) = −0.02 m.
Since s′ > 0 the image is real. Since y ′ < y and y ′ < 0,
the image is reduced and inverted.
∴m = −
s′
0.20 m
=−
= −1.00,
s
0.20 m
and the image height is
y ′ = my = (−1.00)(0.03 m) = −0.03 m.
Since s′ > 0 the image is real. Since y ′ = y and y ′ < 0,
the image is the same size as the object and inverted.
(c) With s = 0.05 m, we find
(0.10 m)(0.05 m)
s′ =
= −0.10 m.
(0.05 m − 0.10 m)
∴m = −
s′ 0.10 m
=
= +2.00,
s 0.05 m
and the image height is
y ′ = my = (2.00)(0.03 m) = 0.06 m
Since s′ < 0 the image is virtual. Since y ′ > y and
(b) With s = 0.20 m, we find
(0.10 m)(0.20 m)
s′ =
= 0.02 m.
(0.20 m − 0.10 m)
y ′ > 0, the image is enlarged and upright.
Remember Fermat’s principle? It states that a light ray
traveling between two points will take the path with the
least time. So, if one of
A
A
B
C
C
B
these light ray paths took
less time than any of the
others, by Fermat’s
principle they would all
take that path! But we
Question 32.9: If the rays all start from the object at the
same time, which arrives at the image first or do they
know that light travel along A, B and C ... so they must
arrive at the same time?
all take the same time (even though the distances
traveled are not the same)! Although path C is the
shortest, the ray travels further through glass where the
velocity is less.
Concave lens ...
s
Object
A
Image
F
A
Question 32.10: A double concave lens* that has an
s′
index of refraction of 1.435 has radii whose magnitudes
f
are 0.03 m and 0.25 m. An object is positioned 0.80 m
to the left of the lens. Find
Note for a concave lens: s > 0, s′ < 0, f < 0 and s′ < s .
Since m = − s′ s then 0 < m < 1, always.
(a) the focal length of the lens,
(b) the position of the image, and
(c) the magnification of the image.
(d) Is the image real or virtual, upright or inverted?
So, a diverging lens always produces a virtual image of a
real object no matter where it is positioned. The image is
upright and smaller relative to the object
ABCDEF
ABCDEF
* Both sides of the lens are concave.
Incoming light
r1
r2
Outgoing light
(c) m = −
s′
−0.22 m
=−
= +0.275.
s
0.80 m
(d) Since m > 0 the image is virtual and upright. Since
r1 = −0.30 m (opposite side of outgoing light).
r2 = 0.25 m (same side of outgoing light).
n1 = 1.00 and n 2 = 1.45
⎞⎛ 1 1 ⎞
1 ⎛n
(a) = ⎜ 2 − 1⎟ ⎜ − ⎟
f ⎝ n1 ⎠ ⎝ r1 r2 ⎠
1.45 ⎞ ⎛
1
1 ⎞
=⎛
−1
−
= −3.30 m −1
⎝ 1.00 ⎠ ⎝ −0.30 m 0.25 m ⎠
∴f = −0.303 m
m > 1 the image is diminished.
See useful notes on web-site to see how to solve this
problem using a scale drawing.
Object
(diverging lens).
1 1 1
(b) s = 0.80 m and + = .
s s′ f
1
1
1
∴
+ =
0.80 m s′ −0.303 m
1
1
1
i.e., =
−
= −4.55 m−1
s′ −0.303 m 0.80 m
∴ s′ = −0.22 m
(opposite side to outgoing light)
Image
Incoming light
Outgoing light
Note: the image is not necessarily on the same side as
the outgoing light!
FRESNEL LENS:
Only the radii of the
front and back surfaces
of a lens determine the
focal length, not the
thickness in between!
So, much weight can
be saved with this plano-convex lens by removing the
“unused” part of a lens!
Compared with regular lenses, Fresnel lenses produce a
poorer image quality, so they tend to be used only where
quality is not critical or where the bulk of a solid lens
would be prohibitive. Cheap Fresnel lenses can be
stamped or moulded out of transparent plastic and are
used in overhead projectors, projection televisions, and
hand-held sheet magnifying glasses. Fresnel lenses have
been used to increase the visual size of CRT displays in
pocket televisions, and they are also used in traffic lights.
Lighthouse lens