mirrors and lenses

1
mirrors and lenses

• PHY232
• Remco Zegers
• zegers_at_nscl.msu.edu
• Room W109 cyclotron building
• http//www.nscl.msu.edu/zegers/phy232.html

2
quiz (extra credit)

• a ray of light moves from air to a material with
  ngt1, at an initial angle of 300 with the normal.
  After reaching the material, the ray bends and
  has an outgoing angle of 200 with the normal. Now
  consider the reverse case a ray of light travels
  from the same material with ngt1 to air. The
  incident angle is 200. After reaching air, the
  angle with the normal will be
• a) 200
• b) between 200 and 300
• c) 300
• d) greater than 300

?0
300
200
200
3
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.

4
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.

5
the flat mirror

• 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/h1
• the image is not inverted

p
q
NOTE a virtual image cannot be projected on a
screen but is visible by the eye or another
optical instrument.
6
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

7
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
8
concave mirrors
F
M
C
C center of mirror curvature
F focal point
9
concave mirrors an object outside F
O
F
step 3 note that a ray from the bottom of the
object just reflects back.
10
concave mirrors an object outside F
O
F
I

• The image is
• inverted (upside down)
• real (light rays pass through it)
• smaller than the object

11
concave mirrors an object outside F
O
F
I
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

12
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

• 1/p1/q1/f so 1/121/q1/5
• 1/q1/5-1/12 so q8.57 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.

13
concave mirrors an object inside F
O
F
I

• the image is
• not inverted
• virtual
• magnified

step 3 note that a ray from the bottom of the
object just reflects back.
14
concave mirrors an object inside F
O
F
I

• the image is
• not inverted
• virtual
• magnified

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
15
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?

• 1/p1/q1/f so 1/21/q1/5
• 1/q1/5-1/2 so q-3.3 cm (note the -
  sign!)
• b) M-q/p-(-3.3)/21.65
• this means that size of the image is 65
  larger than
• the size of the object and that it is not
  inverted ().

16
demo the virtual pig
17
convex mirrors an object outside F (pgtf)
O
F
F is now located on the other side of the mirror
step 3 note that a ray from the bottom of the
object just reflects back.
18
convex mirrors an object outside F (pgtf)
O
F
I
F is now located on the other side of the mirror

• the image is
• not inverted
• virtual
• 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
19
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/p1/q1/f with p6 cm f-3 cm 1/6
1/q -1/3 so q-2 cm b) M-q/p-(-2)/61/3 th
e image is only 33 of the height of the object.
20
convex mirrors with p lt f

• the situation is exactly the same as for the
  situation with p gt f. The demagnification
  will be different though

O
I
F
F
21
Mirrors an overview
type p? image image direction M q f
concave pgtf real inverted Mgt0 M -
concave pltf virtual not inverted Mgt1 M -
convex pgtf virtual not inverted Mlt1 M - -
convex pltf virtual not inverted Mlt1 M - -

• mirror equation 1/p 1/q 1/f
• fR/2 where R is the radius of the mirror
• magnification M-q/p

22
lon-capa

• now do problems 7,8,11 of lon-capa 8

23
question (extra credit)

• a object is placed in front of a concave mirror,
  exactly at the location of the focal point (pf).
  What is the image distance q?
• a) qp
• b) qf
• c) q0
• d) q infinity

24
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 and are thickest at the edges

used in drawings
-
25
lensmakers equation
object
1
2
R2
f focal length of lens n refractive index of
lens R1 radius of front surface R2 radius of back
surface
R1
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)
26
example

• Given R110 cm and R25 cm, what is the focal
  length? The lens is made of glass (n1.5)

object
1
2
R2
R1
R1 is on the right of the curvature, so positive
10 cm R2 is on the left of the curvature, so
negative 5 cm n1.5 1/f0.5(0.1-(-0.2))0.15
f6.67 cm
27
example 2

• Given R15 cm and R210 cm, what is the focal
  length? The lens is made of glass (n1.5)

object
1
2
R1
R2
R1 is on the left of curvature 1 so R1-5 cm R2
is on the right of curvature 2 so R210
cm n1.5 1/f0.5(-0.2-0.1)-0.15 f-6.67 cm
28
example 3

• Given R15 cm and R2?, what is the focal length?
  The lens is made of glass (n1.5)

object
1
2
R1
R2
R1 is on the left of curvature 1, so R1-5 cm R2
is infinity (either positive or
negative) n1.5 1/f0.5(-0.2 0.)-0.1 f-10
cm
29
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

30
converging lens pgtf
O
F
F

A real inverted image is created. The
magnification depends on p M can be lt1, 1 or gt1
31
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
32
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) p20 cm, f12 cm use 1/p 1/q 1/f 1/20
1/q 1/12 solve for q gives q30 cm b)
M-q/p-30/20-1.5 The image is inverted (M
negative). The image is magnified (Mgt1)
33
converging lens pltf
O
F
F

A virtual non-inverted image is created.
Magnification gt1
34
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) p2 cm, f3 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)/23 The
image is not inverted (M positive). The image is
magnified (Mgt1) The image is virtual
35
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

screen
36
NOT CORRECT
37
diverging lens pgtf
O
F
F
-
A virtual non-inverted image is created. The
magnification Mlt1
38
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) p5 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)/50.375 The image is
non-inverted (M positive). The image is
demagnified (Mlt1)
39
diverging lens pltf
F
F
O
-
A virtual non-inverted image is created. The
magnification Mlt1 similar to case with pgtf
40
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) p2 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)/20.6 The image is non-inverted
(M positive). The image is demagnified (Mlt1)
41
lenses, an overview
type p? image image direction M q f
converging pgtf real inverted Mgt0 M -
converging pltf virtual not inverted Mgt1 M -
diverging pgtf virtual not inverted Mlt1 M - -
diverging pltf virtual not inverted Mlt1 M - -

• mirror equation 1/p 1/q 1/f
• magnification M-q/p
• lens makers equation 1/f(n-1)(1/R1-1/R2)

42
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 after correction
43
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
44
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
5cm
5 cm
15 cm
45
answer

-
O
5cm
3cm
15 cm
5 cm
46
lon-capa

• now do problems 9,10,12 of lon-capa 8