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mirrors and lenses
PHY232 – Spring 2007
Jon Pumplin
http://www.pa.msu.edu/~pumplin/PHY232
(Ppt courtesy of Remco Zegers)
we saw…
 that light can be reflected or refracted at boundaries
between material with a different index of refraction.
 by shaping the surfaces of the boundaries we can make
devices that can focus or otherwise alter an image.
 Here we focus on mirrors and lenses for which the
properties can be described well by a few equations.
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the plane (=flat) mirror
p
 in the previous chapter we
already studied flat mirrors.
 Distance from the object to the
mirror is the object distance p
 Distance from the image to the
mirror is the image distance -q
 in case of a flat mirror, an
observer sees a virtual image,
meaning that the rays do not
actually come from it.
 the image size (h’ ) is the same as
the object size (h), meaning that
the magnification h’/h=1
 the image is not inverted
-q
NOTE: a virtual image
cannot be projected on
a screen but is ‘visible’ by
the eye or another optical
instrument.
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question
 You are standing in front (say 1 m) of a mirror that is less high than
your height. Is there a chance that you can still see your complete
image?
 a) yes b) no
object
PHY232 - Pumplin - Mirrors and lenses
image
4
ray diagrams
 to understand the properties of optical elements we use
ray diagrams, in which we draw the most important
elements and parameters to understand the elements
h
h’
p
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-q
5
concave mirrors
M
F
C
C: center of mirror curvature
F: focal point
a light ray passing through the center of curvature will be
reflected back upon itself because it strikes the mirror
normally to the surface.
a light ray traveling parallel to the central axis of the mirror
will be reflected to the focal point F, with FM=CM/2
The distance FM is called the focal length f.
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concave mirrors: an object outside F
O
I
F
step 1: draw the ray from the top of the object parallel to the central
axis and its reflection (through F).
step 2: draw the ray from the top of the object through F and its
reflection (parallel to the central axis)
the image of the top of the object is located where the reflected
rays meet
step 3: note that a ray from the bottom of the object just reflects back.
construct the image I
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concave mirrors: an object outside F
O
I
F
The image is:
a) inverted (upside down)
b) real (light rays pass through it)
c) For this location, image is smaller than object
d) If interchange image and object, image would
be bigger than object.
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concave mirrors: object outside F
O
I
F
distance object-mirror: p
distance image-mirror: q
distance focal point-mirror: f
mirror equation: 1/p + 1/q = 1/f
given p,f this equation can be used to calculate q
magnification: M = -q/p
can be used to calculate magnification.
• if negative: the image is inverted
• if smaller than 1, object is demagnified
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example
 An object is placed 12 cm in front of a a concave mirror
with focal length 5 cm. What are:
 a) the location of the image
 b) the magnification
a) 1/p + 1/q = 1/f so 1/12 + 1/q = 1/5
1/q = 1/5 - 1/12 so q = 8.57 cm
b) M = -q/p = -8.57/12 = -0.71
this means that size of the image is only 71% of the
the size of the object and that it is inverted.
c) Image is real (whenever q>0, I.e., whenever M > 0)
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concave mirrors: an object inside F
the image is:
a) not inverted
b) virtual
c) magnified
F
O
I
step 1: draw the ray from the top of the object parallel to the central
axis and its reflection (through F).
step 2: draw the ray from the top of the object through F and its
reflection (parallel to the central axis)
the image of the top of the object is located where the reflected rays
meet: in this you must draw virtual rays on the other side of the lens
step 3: note that a ray from the bottom of the object just reflects back.
creates a magnified image (“shaving mirror”)
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concave mirrors: an object inside F
the image is:
a) not inverted
b) virtual
c) magnified
F
O
I
The lens equation and equation for magnification are still
valid. However, since the image is now on the other
side of the mirror, its sign should be negative
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example
 an object is placed 2 cm in front of a lens with a focal
length of 5 cm. What are the a) image distance and b) the
magnification?
a) 1/p + 1/q = 1/f so 1/2 + 1/q = 1/5
1/q = 1/5 - 1/2 so q = -3.3 cm (note the ‘-’ sign!)
b) M = -q/p = -(-3.3)/2 = +1.65
this means that size of the image is 65% larger than
the size of the object and that it is not inverted (+).
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demo: the virtual pig
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convex mirrors: (p>|f|or p<|f| doesn’t matter)
O
I
F
F is now located on the other
side of the mirror
step 1: draw the ray from the top of the object parallel to the central
axis and its reflection (through F).
step 2: draw the ray from the top of the object through F and its
reflection (parallel to the central axis)
the image of the top of the object is located where the reflected
rays meet
step 3: note that a ray from the bottom of the object just reflects back.
construct the image I (examples: garden ball, rearview mirrors that
warn “objects are closer than they appear”)
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convex mirrors
O
I
F
F is now located on the other
side of the mirror
the image is:
a) not inverted
b) virtual
c) demagnified
The lens/mirror equation and equation for magnification are
still valid. However, since the image and focal point are now
on the other side of the mirror, their signs should be negative
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example
 an object with a height of 3 cm is placed 6 cm in front of a
convex mirror, with f=-3 cm. What are a) the image
distance and b) the magnification?
answer a): 1/p+1/q=1/f with p=6 cm f=-3 cm
1/6 + 1/q = -1/3 so q=-2 cm
b) M=-q/p=-(-2)/6=1/3
the image is only 33% of the height of the object.
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Mirrors: an overview
type
p?
image
image
direction
M
q
f
concave p>f
real
inverted
|M|>0 M -
+
+
concave p<f
virtual
not
inverted
|M|>1 M +
-
+
convex
virtual
not
inverted
|M|<1 M +
-
-
any p
 mirror equation 1/p + 1/q = 1/f
 F = R/2 where R is the radius of the mirror
 magnification: M = -q/p
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Lenses
 Lenses function by refracting light at their surfaces
 Their action depends on
 radii of the curvatures of both surfaces
 the refractive index of the lens
 converging (positive lenses) have positive focal length
and are always thickest in the center
+
 diverging (negative lenses) have negative focal length
used in
and are thickest at the edges
-
drawings
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lensmakers equation
object
R2
1
2
R1
f: focal length of lens
n: refractive index of lens
R1 radius of front surface
R2 radius of back surface
R2 is negative if the center of the circle is on the left of
curvature 2 of the lens
R1 is positive if the center of the circle is on the right of
curvature 1 of the lens
if the lens is not in air then (nlens-nmedium)
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example
object
R2
1
2
R1
 Given R1=10 cm and R2=5
cm, what is the focal length?
The lens is made of glass
(n=1.5)
R1 is on the right of the curvature, so positive +10 cm
R2 is on the left of the curvature, so negative –5 cm
n=1.5
1/f=0.5(0.1-(-0.2))=0.15 f=+6.67 cm
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example 2
object
R1
1
2
R2
 Given R1=5 cm and R2=10
cm, what is the focal length?
The lens is made of glass
(n=1.5)
R1 is on the left of curvature 1 so R1=-5 cm
R2 is on the right of curvature 2 so R2=+10 cm
n=1.5
1/f=0.5(-0.2-0.1)=-0.15 f=-6.67 cm
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converging lens p>f
I
O
F
F
+
1) A ray parallel to the central axis will be bend through the focal point
2) A ray through the center of the lens will continue unperturbed
3) A ray through the focal point of the lens will be bend parallel to the
central axis
4) the image is located at the crossing of the above 3 rays (you need
just 2 of them).
A real inverted image is created. The magnification
depends on p: |M| can be <1, 1 or >1
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lens equation
I
O
F
F
+
The equation that connects object distance p, image
distance q and focal length f is (just like for mirrors):
1/p + 1/q = 1/f
Similarly for the magnification:
M=-q/p
q is positive if the image is on the opposite side of the lens as the object
NOTE THAT THIS IS DIFFERENT THAN THE CASE FOR MIRRORS. In either
case, q >0 means the image is on the side where the light from the lens
or mirror is.
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example
 an object is put 20 cm in front of a positive lens, with focal
length of 12 cm. a) What is the image distance q? b) What
is the magnification?
a) P = 20 cm, f = 12 cm use 1/p + 1/q = 1/f
1/20 + 1/q = 1/12 solve for q gives q=30 cm
b) M = -q/p = -30/20 = -1.5
The image is inverted (M negative).
The image is magnified (|M|>1)
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converging lens p<f
I
F
O
F
+
1) A ray parallel to the central axis will be bend through the focal point
2) A ray through the center of the lens will continue unperturbed
3) A ray through the focal point of the lens will be bend parallel to the
central axis
4) the image is located at the crossing of the above 3 rays (you need
just 2 of them).
A virtual non-inverted image is created.
Magnification >1
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example
 an object is put 2 cm in front of a positive lens, with focal
length of 3 cm. a) What is the image distance q? b) What
is the magnification?
a) p=2 cm, f=3 cm use 1/p + 1/q = 1/f
1/2 + 1/q = 1/3 solve for q gives q=-6 cm
NOTE q is negative which means it is on the same
side of the lens as the object
b) M=-q/p=-(-6)/2=+3
The image is not inverted (M positive).
The image is magnified (|M|>1)
The image is virtual
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question
 An object is placed in front of a converging (positive) lens with the
object distance larger than the focal distance. An image is created
on a screen on the other side of the lens. Then, the lower half of the
lens is covered with a piece of wood. Which of the following is true:
 a) the image on the screen will become less bright only
 b) half of the image on the screen will disappear only
 c) half of the image will disappear and the remainder of the image
will become less bright.
I
O
F
F
+
rays of light can still make it to any point on the image,
but there are less of them, so less bright (only)
PHY232 - Pumplin - Mirrors and lenses
screen
28
NOT CORRECT
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diverging lens p>|f|
O
F
F
I
-
1) A ray parallel to the central axis will be bend so that the ray passes
through the focal point IN FRONT of the lens
2) A ray through the center of the lens will continue unperturbed
3) A ray aimed at the focal point on the other side of the lens will be
bent parallel to the central axis
4) the image is located at the crossing of the above 3 rays (you need
just 2 of them).
A virtual non-inverted image is created.
The magnification |M|<1
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example
 an object is put 5 cm in front of a negative lens, with focal
length of -3 cm. a) What is the image distance q? b) What
is the magnification?
a) p=5 cm, f=-3 cm use 1/p + 1/q = 1/f
1/5 + 1/q = -1/3 solve for q gives q=-1.88 cm
b) M=-q/p=-(-1.88)/5=+0.375
The image is non-inverted (M positive).
The image is demagnified (|M|<1)
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diverging lens p<|f|
F
O
F
I
-
1) A ray parallel to the central axis will be bend so that the ray passes
through the focal point IN FRONT of the lens
2) A ray through the center of the lens will continue unperturbed
3) A ray aimed at the focal point on the other side of the lens will be
bent parallel to the central axis
4) the image is located at the crossing of the above 3 rays (you need
just 2 of them).
A virtual non-inverted image is created.
The magnification |M|<1 similar to case with p>|f|
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example
 an object is put 2 cm in front of a negative lens, with focal
length of -3 cm. a) What is the image distance q? b) What
is the magnification?
a) p = 2 cm, f = -3 cm use 1/p + 1/q = 1/f
1/2 + 1/q = -1/3 solve for q gives q = -1.2 cm
b) M = -q/p = -(-1.2)/2 = +0.6
The image is non-inverted (M positive).
The image is demagnified (|M|<1)
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lenses, an overview
type
p?
image
image
direction
M
q
f
converging
p>f
real
inverted
|M|>0 M -
+
+
converging
p<f
virtual
not
inverted
|M|>1 M +
-
+
diverging
any p
virtual
not
inverted
|M|<1 M +
-
-
 mirror or lens equation 1/p + 1/q = 1/f
 mirror or lens magnification: M = -q/p
 lens makers equation: 1/f=(n-1)(1/R1-1/R2)
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chromatic aberrations
Chromatic aberrations are due to light of different
wavelengths having a different index of refraction
Can be corrected by combining lenses/mirrors
If n varies with wavelength, the focal length f
changes with wavelength
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two lenses
 an object, 1 cm high, is placed 5 cm in front of a
converging mirror with a focal length of 3 cm. This setup is
placed in front of a diverging mirror with a focal length of
–5 cm. The distance between the two lenses is 10 cm.
Where is the image located, and what are its properties?
+
3cm
5 cm
5cm
15 cm
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answer
+
O
5cm
3cm
5 cm
I1
I2
5+7.5=
12.5 cm
15 cm
15-1.67=13.33 cm
consider the converging lens first:
1/p+1/q=1/f so 1/5+1/q=1/3 so q=7.5 cm
M=-q/p=-7.5/5=-1.5, so 1.5x1cm=1.5 cm high
A real inverted magnified image I1
consider the action of diverging lens on the image constructed above
1/p+1/q=1/f so 1/(2.5) + 1/q = 1/(-5) q=-1.67 cm
M=-q/p=-(-1.67)/2.5=0.67, so 0.67x 1.5=1 cm high
A virtual non-inverted demagnified image I2 relative to I1
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