Phys-272 Lecture 24
Geometric Optics
Mirrors—Plane and Spherical
+
Hubble Space Telescope
Plane (Flat) Mirrors
Already know physics:
θR = θI
Source: Object – emits rays
in all directions.
Eye intercepts cone of rays.
Plane Mirrors
SIGN RULE convention for s and s′;
• IF object & incoming ray same side, s pos., otherwise neg.
• IF image & outgoing ray same side, s′ pos., otherwise neg.
• IF curvature center & outgoing ray on same side, pos.,
otherwise neg.
For mirror example, we have s = -s′
Plane Mirrors
Flat Mirror
(1) Draw first ray perpendicular to mirror 0 = θi = θr
(2) Draw second ray at angle. θi = θr
(3) Lines appear to intersect a distance d behind
mirror. This is the image location.
Virtual Image:
No light actually gets here
θr
θi
d
d
Clicker
A woman is looking at her reflection in a flat vertical 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 reflection she can see in the mirror.
A. Above her knee
B. Her knee
C. Below her knee
CD
Clicker
A woman is looking at her reflection in a flat vertical 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 reflection 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 cant see it
Concave 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 often
sections of spheres
(assumed in this class).
f
For such “spherical”
mirrors, we assume all
angles are small even
though we draw them
larger to easily see and
understand…
Aside:
For a spherical mirror, R = 2f
R
2f
center of sphere
sometimes labeled “C”
f
Recipe for finding image:
1) Draw ray parallel to axis reflection goes through focus
2) Draw ray through focus reflection is parallel to axis
3) Draw ray from object through radius = 2f, reflects back
through 2f
object
2f
f
image
You now know the position of the same point on the image
Ray tracing for a concave mirror
R
h
R-s’
h’
s-R
s
θ
θ
s’
Spherical Mirror
Concave Mirror
Point C, center of curvature.
Suppose object at P, then out
Going ray passes through P’.
α + θ = φ ; φ + θ = β ⇒ α + β = 2φ
h
h
h
tan α =
; tan β =
; tan φ =
s −δ
s '−δ
R −δ
Assume, angles small, δ ≅ 0,
paraxial rays, approx. parallel
to axis. Note s and s′ positive.
h
h
h
α≅ ; β≅ ; φ≅
s
s'
R
Mirror Equation:
1
1
2
+
=
s
s'
R
Spherical Mirror
Equation
Infinite distance, s= ∞, s′=R/2
image appears at ½ radius of
curvature.
1
1
2
+
=
s
s'
R
1
1
2
+
=
∞
s'
R
Nomenclature; focal length,
f=R/2, is image length when
object is at infinity.
1
1
1
+
=
s
s'
f
Real image – formed by converging rays. Can be viewed on screen.
Virtual image – rays do not go through. Cannot be shown on screen.
Concave mirror: can have real or virtual images.
Camera Region
image is:
real
inverted
smaller
S >2f
object
1 1 1
+ =
S S′ f
S′
M =−
S
f
2f
image
S
S’
f
Vanity Region
f >S>0
rays no-longer meet
in front of the mirror
object
f
f
S
but they do meet
behind the mirror
image
(virtual)
Lateral Magnification; spherical mirror
y | y′ |
θ= =
s
s′
Magnification =
y'
s'
=−
y
s
Remarks; note that y’=-|y’|, since the arrow is upside down and
s, s’, and y are positive. So the magnification by this definition
is negative, since y’ is upside down.
Examples of ray tracing for concave mirrors
s>R
image
Inverted
m=s’/s
s=R
m=-1
s=R/2=f
m=∞
∞
s<R/2
m=s’/s>1
Cosmetic/”vanity” mirror
More examples
s>R
image
Inverted
m=s’/s
s=R/2=f
m=∞
∞
R
= 10 cm
2
1 1 1
fs
+ = ; s′ =
s s′ f
s− f
(10)(30)
− s′
1
s = 30 cm; s ′ =
= 15 cm; m =
=−
30 − 10
s
2
R = 20 cm;
f =
1 1 1
fs
+ = ; s′ =
s s′ f
s− f
(10)(10)
− s′
′
s = 10 cm; s =
= ∞; m =
= −∞
10 − 10
s
Even more examples
1 1 1
fs
+ = ; s′ =
s s′ f
s− f
(10)(20)
− s′
s = 20 cm; s ′ =
= 20 cm; m =
= −1
20 − 10
s
1 1 1
fs
+ = ; s′ =
s s′ f
s− f
(10)(5)
= −10 cm;
s = 5 cm; s ′ =
5 − 10
− s′
(−10)
=−
= +2
m=
s
5
s=R
m=-1
s<R/2
m=s’/s>1
cosmetic mirror
Convex
mirror
For a convex mirror, the radius of curvature is inside the
mirror. We have another RULE, that when the curvature
center C is on the same side of the outgoing ray, then the
radius is positive. For the convex mirror we have NEGATIVE
radius. Also from our rule for s’, we observe this is negative in
the above drawing. Using same recipe as we used for concave
mirrors, we would find same formula:
1
1
2
+
=
s
s'
R
Again, note s’ and R are negative
y'
s'
=−
Magnification =
s
(this is positive) y
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:
f
Christmas Tree Ornament, Mirrors in 7-11’s
Convex mirror, R & s’ negative
Santa is 75cm from convex mirror of radius 3.6cm. What is image position and magnification?
s ' = −1.76cm
s'
− 1.76
1 1 2 1 1
2
m=− =−
= .0234
+ = = + =
s s ' R 75 s ' − 3.6
s
75
Another striking example:
Quick Clicker
2) The diagram below shows
three light rays reflected off
of a concave mirror. Which
ray is NOT correct?
R
A)
C)
B)
A)
B)
C)
f
Multi-part clicker
I). The image produced by a concave mirror of a real object is
a) always real.
b) always virtual.
c) sometimes real
and sometimes virtual.
II). The image produced by a concave mirror of a real object is
a) always upright.
b) always inverted.
c) sometimes upright
and sometimes inverted.
Is image of a real object from a concave mirror
real or virtual?
It depends on the position of the object relative to the focal point!
• Draw Rays or..
• 1/s’ = 1/f – 1/s (concave implies f > 0)
Is image of a real object from a concave mirror
upright or inverted?
Once again, it depends on position relative to focal point!
• If s’ > 0, real and inverted
• If s’ < 0, virtual and upright
I). The image produced by a convex mirror of a real object is
a) always real.
b) always virtual.
c) sometimes real
and sometimes virtual.
II). The image produced by a convex mirror of a real object is
a) always upright.
b) always inverted.
c) sometimes upright
and sometimes inverted.
BTW has the side mirror
on your car ever produced
an upside-down image of
the world when your car
was still upright? If so,
get a new car.
Is image of a real object from a convex mirror
real or virtual?
Same PROCEDURE as for concave mirrors we did earlier!
• Draw Rays or..
• 1/s’ = 1/f – 1/s (convex implies f < 0)
Is image of a real object from a convex mirror
upright or inverted?
Once again, depends on sign of s’ (real/inverted or virtual/upright)
• Here s’ < 0 ALWAYS… therefore virtual and upright
Clicker problem
• In order for a real object to create a real, inverted enlarged image,
a) we must use a concave mirror.
b) we must use a convex mirror.
c) neither a concave nor a convex mirror can produce this image.
Clicker problem
• In order for a real object to create a real, inverted enlarged image,
a) we must use a concave mirror.
b) we must use a convex mirror.
c) neither a concave nor a convex mirror can produce this image.
• A convex mirror can only produce a virtual image since all reflected
rays will diverge. Therefore, b) is false.
• To create a real image with a concave mirror, the object must be
outside the focal point.
• The example we just did gave a real,
inverted reduced image.
• Is it possible to choose the parameters
θ
R
θ
h
such that the image is enlarged??
R-s’
The easy (but clever) answer:
• h’ is a real image.
• Therefore consider the OBJECT to be
h’. The IMAGE will be h !!!
• Therefore it certainly IS POSSIBLE!!
equations follow…
h’
s-R
s
s’
Clicker problem
• In order for a real object to create a real, inverted enlarged image,
a) we must use a concave mirror.
b) we must use a convex mirror.
c) neither a concave nor a convex mirror can produce this image.
• The example we just did gave a real,
inverted reduced image.
• Is it possible to choose the parameters
h
such that the image is enlarged??
1 1
1
fs
+
=
s′ =
s s′
f
s− f
M =−
s′
s
M =−
R
R-s’
h’
s-R
s
f
s− f
Thus to get a real image we need: s > f
Thus to get an inverted enlarged image we need:
s<2f
Thus to get a real, inverted enlarged image we need:
f <s <2f
θ
θ
s’
Keck telescope on the
top of Mauna Kea
(13,600 ft)
Two 10m (33 ft)
diameter mirrors;
focal length 17.5
m
Mirror for the Hubble space telescope (f =57.6 m)
Lecture End
BACKUP SLIDES
Avoid the distortion and
bandwidth limitations of
the earth’s atomsphere.
Artificat of
CCD camera
Eagle Nebula viewed by the Hubble
Gravitational
lensing
A field of galaxies viewed by the Hubble
Executive Summary - Mirrors:
S > 2f
2f > S > f
f >S>0
S >0
real
inverted
smaller
real
inverted
bigger
virtual
upright
bigger
virtual
upright
smaller
concave
(converging)
f
1 1 1
+ =
S S′ f
S′
M =−
S
convex
(diverging)
f
S = 2f
object
1 1 1
+ =
S S′ f
S′
M =−
S
image is:
real
inverted
same size
f
image
2f
S’
S
f
1 1 1
+ =
S S′ f
S′
M =−
S
image is:
real
inverted
bigger
2f > S > f
object
2f
f
image
f
S’
S
image is:
virtual
upright
bigger
f >S>0
1 1 1
+ =
S S′ f
S′
M =−
S
object
f
image
(virtual)
f
S
S’<0
convex
f
The size is the only thing that changes
with a convex mirror, it is always
virtual and upright.
An arrow is located in front of a convex
spherical mirror of radius R = 50cm.
The tip of the arrow is located
at (-20cm,-15cm).
y
R=5
0
(-20,20,-15)
Where is the tip of the arrow’s image?
• Conceptual Analysis
• Mirror Equation: 1/s + 1/s’ = 1/f
• Magnification: M = -s’/s
• Strategic Analysis
•Use mirror equation to figure out the x coordinate of the image
•Use the magnification equation to figure out the y coordinate of the
tip of the image
x
An arrow is located in front of a convex
spherical mirror of radius R = 50cm.
The tip of the arrow is located
at (-20cm,-15cm).
y
R=5
0
x
(-20,20,-15)
What is the focal length of the mirror?
A) f =50cm
B) f = 25cm
C) f = -50cm
D) f = -25cm
For a spherical mirror | f | = R/2 = 25cm.
Rule for sign: Positive on side of mirror where light goes after hitting mirro
mirr
y
R
f <0
f = - 25 cm
An arrow is located in front of a convex
spherical mirror of radius R = 50cm.
The tip of the arrow is located
at (-20cm,-15cm).
y
R=5
0
f = -25 cm
(-20,20,-15)
What is the x coordinate of the image?
A) 11.1 cm
B) 22.5 cm
Mirror
equation
C) -11.1 cm
D) -22.5cm
1 1 1
= −
s′ f s
fs
s = 20 cm
s− f
f = -25 cm
(−25)(20)
s′ =
= -11.1 cm
20 + 25
s′ =
Since s’ < 0 the image is virtual (on the “other” side of the mirror)
x
An arrow is located in front of a convex
spherical mirror of radius R = 50cm.
The tip of the arrow is located
at (-20cm,-15cm).
y
R=5
0
x = 11.1 cm
f = -25 cm
(-20,20,-15)
What is the y coordinate of the tip of the image?
A) -11.1 cm
B) -10.7 cm
Magnification
equation
M =−
C) -9.1 cm
S′
S
s = 20 cm
s’ = -11.1 cm
D) -8.3cm
M = 0.556
yimage = 0.55 yobject = 0.556*(-15 cm) = -8.34 cm
x
Ray Tracing or Graphical techniques
Useful to help describe image.
To find the image position in
Practice we need to find the
Intersection of at least
two rays. Useful rays are
1) Focal point F, #2
2) Ray through C, #3
3) Vertex V, #4
4) Parallel ray reflects
through F, #1
S >0
1 1 1
+ =
S S′ f
image is:
virtual
upright
smaller
image
(virtual)
object
f<0
S>0
S’<0