Transcript p>|f
mirrors and lenses
PHY232
Remco Zegers
[email protected]
Room W109 – cyclotron building
http://www.nscl.msu.edu/~zegers/phy232.html
an important point
objects do not emit rays of light that get ‘seen’ by your
eye. Light (from a bulb or the sun) gets reflected off the
object towards your eye.
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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 flat mirror
p
in the previous chapter we
already saw flat mirrors.
The distance from the object to
the mirror the object distance p
The 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
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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
q
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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) smaller than the object
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concave mirrors: an 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
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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.
create the image
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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?
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demo: the virtual pig
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convex mirrors: an object outside F (p>|f|)
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
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convex mirrors: an object outside F (p>|f|)
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?
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convex mirrors with p < |f|
the situation is exactly the same as for the situation with
p > |f|. The demagnification will be different though…
F
O
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F
19
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
p>|f|
virtual
not
inverted
|M|<1 M +
-
-
convex
p<|f|
virtual
not
inverted
|M|<1 M +
-
-
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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lon-capa
now do problems 7,8,11 of lon-capa 8
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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)
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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)
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example 3
object
R1
1
2
R2
Given R1=5 cm and R2=,
what is the focal length? The
lens is made of glass (n=1.5)
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question
A person is trying to make a lens but decides to make
both surfaces flat, resulting in essentially a flat piece of
glass on both sides. What is the focal length of this ‘lens’?
a) infinity
b) 0
c) cannot say, depends on the index of refraction n
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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
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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?
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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?
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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.
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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?
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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?
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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
p>|f|
virtual
not
inverted
|M|<1 M +
-
-
diverging
p<|f|
virtual
not
inverted
|M|<1 M +
-
-
mirror equation 1/p + 1/q = 1/f
magnification: M=-q/p
lens makers equation: 1/f=(n-1)(1/R1-1/R2)
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spherical aberrations: Hubble space telescope
spherical aberrations are due to the rays hitting the lens
at different locations have a different focal point
perfect
distorted
example: Hubble
before
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after
correction
40
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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lon-capa
now do problems 9,10,12 of lon-capa 8
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