Transcript Flat Mirrors - OWU Online | Go OWU

Flat Mirrors
• Consider an object placed in front
of a flat mirror
– Object O is placed a distance p in front
of the mirror (p = object distance)
– Light reflecting from mirror appears to
originate at point I
• An image of object O is formed at a distance q behind the mirror
(q = image distance)
• It is a virtual image (light does not really converge there)
• When light actually converges at an image point, it is a real image
(can be projected on a screen)
– We can use (at least 2) rays to
determine the image orientation and
position
• Notice p = q for a flat mirror
• Image is upright and virtual
Flat Mirrors
• The height h of the object equals the image height h’
• Lateral magnification M:
image height h
M

object height h
– General definition for any type of mirror
– For flat mirrors, M = 1
– “Magnification” can mean enlargement or reduction in size
in optics
• A flat mirror also produces an apparent left – right
reversal
– A waving left hand appears to be a waving right hand in
the mirror
Example Problem #23.1
Does your bathroom mirror show you older
or younger than your actual age? Compute
an order-of-magnitude estimate for the age
difference, based on data that you specify.
Solution:
Class interactive – Solution will be
determined in class
Concave Spherical Mirrors
• Consider light from an object O
striking a spherical concave
mirror
– If rays diverge at small angles,
they all reflect through same
image point
– Large diverging angles mean rays intersect principle axis
at different points,
resulting in a
blurred image
• Geometry of
concave mirrors
h
q
M  
h
p
(from yellow and blue
similar triangles)
Concave Spherical Mirrors
• Additional triangle geometry yields the mirror
equation:
1 1 2
 
p q R
– If an object is very far from the mirror,
then 1 / p ≈ 0 and q ≈ R / 2
– In this case, the image point is called the
focal point F and the image distance q is
called the focal length f = R / 2
– Note that focal point  focus point
– The mirror equation can be written in
terms of focal length:
1 1 1
 
p q f
Convex Spherical Mirrors
• Geometry of a convex mirror
• We can use the same mirror equation as for
concave mirrors, but we just need to use a particular
sign convention for each quantity in the equation
Sign Conventions for Mirrors
Also, R > 0 (R < 0) for concave (convex) mirrors
Ray Diagrams for Concave Mirrors
Ray Diagrams for Convex Mirrors
• Note that the sideview mirror on the passenger side
of a car is a convex mirror (hence the warning given
on the mirror)
Example Problem #23.16
A convex spherical mirror with a radius of
curvature of 10.0 cm produces a virtual image
one-third the size of the real object. Where is
the object?
Solution (details given in class):
10.0 cm in front of the mirror
Example Problem #23.19
A spherical mirror is to be used to form an
image, five times as tall as an object, on a
screen positioned 5.0 m from the mirror.
(a) Describe the type of mirror required.
(b) Where should the mirror be positioned
relative to the object?
Solution (details given in class):
(a) Concave
(b) 1.0 m in front of the mirror
Images Formed by Refraction
• Consider light from object O
refracting at a spherical
surface between 2
transparent media
– From Snell’s law and
geometry:
n1 n2 n2  n1
 
p q
R
h
n1q
M  
h
n2 p
– Note that real images are formed on the side opposite that
of the incident light
Images Formed by Refraction
• If the refracting surface is flat, R   and we get:
n2
q p
n1
• Now the (virtual) image is on
the same side of the surface
as the object
Interactive Example Problem:
Lens Design 101
Animation and solution details given in class.
(PHYSLET Physics Exploration 34.3, copyright Pearson Prentice Hall, 2004)
Example Problem #23.25
A transparent sphere of unknown composition
is observed to form an image of the Sun on its
surface opposite the Sun. What is the
refractive index of the sphere material?
Solution (details given in class):
2.00
Thin Lenses
• A thin lens forms an image by
refraction of light
– Has 2 refracting surfaces
– Examples of thin lenses shown at right
• Lenses in group (a) converge parallel rays to
a focal point on the opposite side of the lens
• Lenses in group (b) diverge parallel rays so
they appear to originate from a focal point on
the same side of the lens
(converging lens)
(diverging lens)
Geometry of Thin Lenses
• Magnification by a
lens
h
q
M  
h
p
(same as for a mirror)
• Thin–lens equation
1 1 1
 
p q f
(same as for a mirror)
• Lens maker’s equation 1  n  1 1  1 


( f = focal length
in air)
f
( n = index of refraction of
 R1 R2 
the lens material)
(R1 (R2) = radius of curvature of front (back) surface)
Sign Conventions for Thin Lenses
Ray Diagrams for Thin Lenses
• Converging lenses
Ray Diagrams for Thin Lenses
• Diverging lenses
Example Problem #23.35
A certain LCD projector contains a single thin
lens. An object 24.0 mm high is to be projected
so that its image fills a screen 1.80 m high. The
object-to-screen distance is 3.00 m. (a)
Determine the focal length of the projection
lens. (b) How far from the object should the
lens of the projector be placed in order to form
the image on the screen?
Solution (details given in class):
(b) 39.5 mm
(a) 39.0 mm
Interactive Example Problem:
Building a Converging Lens
Animation and solution details given in class.
(PHYSLET Physics Problem 35.11, copyright Pearson Prentice Hall, 2004)
Combinations of Thin Lenses
•
If 2 thin lenses are used to form an image, use the
following procedure:
1. The image produced by the first lens is calculated as
though the second lens were not present
2. The image formed by the first lens is treated as the
object for the second lens
3. The image formed by the second lens is the final image
of the system
•
•
The overall magnification is the product of the
magnifications of the separate lenses
This procedure can be extended to 3 or more
lenses
Example Problem #23.40
An object is placed 20.0 cm to the left of a
converging lens of focal length 25.0 cm. A
diverging lens of focal length 10.0 cm is 25.0
cm to the right of the converging lens. Find the
position and magnification of the final image.
Solution (details given in class):
Position = 9.26 cm in front of the 2nd lens
Magnification = +0.370
Example Problem
#23.50 Back of lens
Front of lens
Front of mirror
Back of mirror
The object shown above is midway between the lens and the mirror. The
mirror’s radius of curvature is 20.0 cm, and the lens has a focal length of
–16.7 cm. Considering only the light that leaves the object and travels first
towards the mirror, locate the final image formed by this system. Is the image
real or virtual? Is it upright or inverted? What is the overall magnification of
the image?
Solution (details given in class):
The image is virtual, upright, located 25.3 cm behind the mirror, with an
overall magnification of +8.05.
HW Problems #23.51, 23.59
#23.51:
#23.59:
Aberrations
• Spherical aberration
– Light passing through lens at
different distances from the
principal axis is focused at
different points
– Apertures are used to help
narrow the incoming beam of light
• Chromatic aberration
– Different wavelengths of
light refracted by a lens
focus at different points
– Can be reduced by using
combination of converging
and diverging lenses