Today 2/10
Multiple Lens Systems 26.9
Curved Mirrors 25.5-6
Lab: Mirrors and Thin Lenses
HW:
“2/10 Two Lenses”
Due Thursday 2/12
Exam I Thursday, Feb 13
Reflection (Mirrors)
i
r
When a ray of light reflects from a mirror the angle
of incidence equals the angle of reflection.
An object is placed off to the side of a plane mirror.
Where is the image? Ray Tracing
The image is the same size as the
object, same orientation, and the same
distance behind the mirror.
This is a “virtual” image.
Flat (Plane) Mirrors
Find the image with a ray diagram.
Only the top half!
How much mirror do I need to see my entire image?
Finding images for a concave (converging) mirror.
R stands for the “radius of curvature” of the mirror
R
Center of curvature
Focal point = R/2
Finding images for a concave (converging) mirror.
Any ray parallel to the axis will be reflected
through the focal point.
Bend the rays at the dotted
line to be consistent with
the mirror equations. (We
did the same thing when we
when we bent the ray at the
middle of the lens.)
C
f
Finding images for a concave (converging) mirror.
Any ray through the focal point will be reflected
parallel to the axis. Light is “reversible.”
Any ray passing through
the focal point will be
reflected parallel by the
mirror.
C
f
Finding images for a concave (converging) mirror.
Where would an image be for a far away object?
Image is smaller than the object, inverted, and near
f. This is a “real” image.
Object
Image
C
f
If the object were at infinity, the image would be at the focal
point but would have zero size.
Note this is also true for converging lenses.
Finding images for a concave (converging) mirror.
Bring the object closer.
Image is still smaller than the object (larger than
before), inverted, and farther from f. Still a “real”
image.
Object
Image
C
f
Finding images for a concave (converging) mirror.
Bring the object closer yet.
Image is now larger than the object, inverted, and
farther yet from f. Still a “real” image.
Note the difference in image
position when the rays bend
at the mirror.
Object
C
f
Caution: You must bend the ray at the dotted line
to match the math.
Finding images for a concave (converging) mirror.
What if we place the object at the focus?
Image is now at infinity, infinitely large, and really
doesn’t make much sense.
Object
C
f
This is the inverse of the situation when the object
was at infinity.
Finding images for a concave (converging) mirror.
What if we place the object inside the focus?
But these rays do not intersect, similar to the flat
mirror case.
Ray drawn as if it came
from f.
Object
C
Image
f
Now the image is larger than the object, upright,
and well behind the mirror. This is a virtual image.
Finding images for a concave (converging) mirror.
Last time, very close to the mirror.
This is getting very much like the flat mirror case.
Object
C
Image
f
Now the image is slightly larger than the object,
upright, and nearly the same distance behind the
mirror. This is a virtual image.
Mirror Equations: (look familiar?!)
Image location
Magnification
1
1
1


do di
f
hi
di
m

ho
do
Minus signs are important!
Example: do = 23, f = 4 di = +4.8
?
m = ? m = -di/do = -0.2 meaning inverted and
smaller
Object
Image
C
f
Example: f = + 6 cm and - 18 cm
lenses are 1 cm apart do = + 12 cm
Find the image due to the first lens,
ignore the second for now.
do
Need another ray-from the center of
the diverging lens
along the first image.
do
Now ask what happens to these
rays at the second lens.
Example: f = + 6 cm and - 18 cm
lenses are 1 cm apart do = + 12 cm
Math
First lens do = 12, f = 6 di = ? di = 12
Second lens do = -11, f = -18 di = ? di = 28.3
What about Magnification?
First image m1 = -12/12 = -1
Second image m2 = -28.3/-11 = 2.6
do
do
di
di
Total m = -1(2.6) = -2.6
Inverted and real
Example: f = + 6 cm and - 18 cm
lenses are 1 cm apart do = + 12 cm
Math
First lens do = 9, f = 6 di = ?
First image m1 = -18/9 = -2
di = 18
Second lens do = -9, f = -18 di = ?
Second image m2 = -18/-9 = +2
do
di = 18
Total m = -2(2) = -4
Inverted and real
do
di
di