OPTICS
I.
IMAGES
A. Definition- An image is formed
where light rays originating from the
same point on an object intersect on
a surface or appear to intersect for
the observer
B. Real Image- formed when light
rays from a common point pass
though an optical system that
causes them to converge and
intersect at a point
1. A real image can be projected on a
screen when placed where the image is
formed.
a. Lenses in a slide projector or a
camera produce real images
C. Virtual Image-formed when the light
rays from a common point pass
through or are reflected by an optical
system that causes them to diverge
and appear to come to a single point.
1. cannot be projected on a screen
because no light from the object actually
reaches the point where the image
appears to be located
a. Example Reflection in a plane
mirror. The rays reaching the observers eyes
actually come from the object. They are
reflected by the mirror in such a way that
they appear to come from the image
EACH RAY IS REFLECTED IN A REGULAR WAY ( MEANING THAT THE ANGLE
OF REFLECTION EQUALS THE ANGLE OF INCIDENCE). IF THE REFLECTED
RAY IS PROJECTED BACK THEY FORM THE LOCATION OF THE IMAGE
BEHIND THE MIRROR, A VIRTUAL IMAGE
II. CONSTRUCTING A RAY DIAGRAM FOR A
PLANE MIRROR.
1. DRAW A RAY FROM THE OBJECT TO THE
MIRROR.
2. THEN CONSTRUCT THE REFLECTED RAY.
3. EXTEND THE REFLECTED RAY BEHIND
THE MIRROR. Use dotted lines
4. REPEAT FOR ONE MORE RAY.
5. WHERE THE RAYS INTERSCET BEHIND
THE MIRROR IS WHERE IS IMAGE IS
LOCATED.
IMAGES PRODUCED BY A
PLANE MIRROR ARE VIRTURAL,
THE SAME SIZE AS THE OBJECT
AND
LOCATED AN EQUAL DISTANCE
BEHIND THE MIRROR AS THE
OBJECT WAS IN FRONT OF THE
MIRROR
III. Images formed by spherical mirrors
A. Definitions
1. Concave mirrors-Generally form
real images
2. Convex Mirror-Always form virtual images
3. Convex Lens-Generally form real
images
4. Concave Lens-Always form virtual
images
►
►
►
1.
Constructing Ray Diagrams for Mirrors
REMEMBER MIRRORS REFLECT LIGHT
Concave mirrors
a. Use the real focal point, located in front
of the mirror
b. A real image is always formed except
when the object is located between F and
the mirror.
c. If the rays do not intersect in front the
mirror extend them behind the mirror to
locate the virtual image
2.
Convex Mirrors
a. A virtual image is always formed
b. Use the virtual focal point (located
behind the mirror)
c. The rays do not intersect in front the
mirror, extend them behind the mirror to
locate the virtual image
Lenses- Refract light
Constructing Ray Diagrams for Lenses
► Convex Lenses
a. Use the real focal point located on the
side opposite of the object.
b. Will always form real images except
when the object is located between F
and the lens
c. If the rays do not intersect behind of the
lens then extend them to the front of the
lens to locate the virtual image
2.
Concave Lens
a. Always form virtual images
b. Use the virtual focal point (located on
the same side of the lens as the object )
c. The rays will not intersect behind the
lens so extend them to in front of the lens
to locate the virtual image
► VI
Mirror/lens Equations
► A. Mathematically relates the locations and
size of the image and the object
B. 1 + 1 = 1
do
di
f
Where
do = distance from object to mirror/lens
d1 = distance from image to mirror/lens
f = focal length
Units- any unit of length all must be the same
For convex mirror and concave lens
C. So = do
Si
di
Where
So = size of object
Si = size of image
do = distance from object to mirror/lens
di = distance from image to mirror/lens
Units- any units of length, all the same
Example:
An object 2.0 cm high is 30.0 cm from a
concave mirror. The focal length of the
mirror is 10.0cm. A. What is the location of
the image? B. What is the size of the
image?
1 + 1 = 1
1 + 1= 1
do
di
f
30cm di 10.0cm
1 = 0.06667 di= 15 cm
di
► B.
S o = do
Si
di
Cross multiply
2X 15 = 30 Si
Si = 1cm
2.0cm = 30.0cm
Si
15cm
Example 2: Find the location and size of the
image of a 2.0cm high object located
5.0cm in front of a convex lens of focal
length 10.0cm.
1 + 1 = 1
1
+ 1 = 1
do
di
f
5.0cm
di
10.0cm
1 = -.10
di
di= -10.0cm negative indicates
virtual image
S o = do
Si
di
2.0cm = 5.0cm
Si
-10.0cm
Cross multiply
-20 = 5 Si
Si = -4 cm
Negative means virtual image
Example 3: calculate the position of the image
of an object located 15cm in front of a
convex mirror of focal length -10.ocm.
1 + 1 = 1
1 +
1
=
1
do
di
f
15cm
di
-10.0cm
1= -0.1667
di
di =-6.0cm negative means
a virtual image
Lens Defects
A. Two Types
1. Chromatic Aberration
a. Occurs because different colors
of light do not focus at the same
point.
b. Lens edges act like a prism,
different wavelengths bent at slightly
different angles.
c. Often occurs in cameras
d. Reduced by using combination of
lenses made of different types of glass.
VIII.
2. Spherical Aberration
a. Occurs because the spherical
shape of the lens is not ideal for
converging light to a single point.
b. Can be reduced by restricting the
light beam so it is incident close to the
center of the lens.
c. accomplished by reducing the size of
the lens opening. (the aperture)
►DIFFRACTION
AND
INTERFERENCE OF LIGHT
A. Diffraction- Bending of waves around a
boundary. Readily seen with water
waves.
B. Young’s Double Slit experiment
1. Young showed that light bends or
diffracts around boundaries.
2. Light was allowed to fall on two
closely spaced narrow slits.
3. The light passing through each slit was
diffracted.
4. The spreading light from the two slits
overlap. (Interfere)
5. When the light from the two slits fell on a
screen a pattern of light and dark bands were
observed. Called interference fringes.
6.Bright bands correspond to constructive
interference. (two crests or two troughs)
7. Dark bands correspond to destructive
interference. (a crest and a trough)
8. Central Band- zero order band
9. bright bands on either side of the central band
are referred to as the first order band,
second- order band and so on.
► J.
Mathematical Relationship
X = λL
d
Where X is the distance between any two bright
bands
.λ = wavelength of light used
L= distance from the slits to the screen.
d.= distance between the slits.
Units of length all must be the same
Example Monochromatic light is incident
on a pair of slits 1.95 X 10-5 m apart. The
distance between the first two bright
bands is 2.11 X 10-2m. If the distance
between the slits and the screen is
0.600m.
A) calculate the wavelength
B) state the color of the light
d= 1.95 X 10-5 m X= 2.11 X 10-2m
L = 0.600m
1.
X = λL
d
2.11 X 10-2m = λ(0.600m)
1.95 X 10-5 m
λ = 2.11 X 10-2m (1.95 X 10-5 m)
0.600m
λ = 6.86 X 10-7m
This is red light
III.
Single Slit Diffraction
A. Diffraction pattern produced differs
from that of a double slit.
1. The central bright band is much
wider than any of the other bright bands
2. The intensity of the central band is
greater than the intensity of any of the
other bright bands.
3. Mathematical Relationship
X= λL
w
Where
X = separation distance between central band
and dark band
.λ = wavelength of light used
L = distance from the slit to the screen.
W = width of the slit.
Units of length all must be the same
Example: Monochromatic orange light of
wavelength 605nm falls on a single slit of
width 0.095mm. The silt is located 85cm
from a screen. How far is the first dark
band?
λ= 605nm = 605 X 10-9m
w = 0.095mm = .000095m
L = 85cm= .85m
4.
X= λL
w
X=605 X 10-9m (.85m)
.000095m
X = .0054m= 5.4mm
IV.
Diffraction Gratings
A. Used in practice to measure the
wavelength of light.
B. Made by scratching very fine lines with
a diamond point on glass.
1. The clear lines on the glass serve as
the slits.
2. There may be as many as 10,000
lines per cm.
C. Interference Pattern similar to double
silt except:
1. The bright bands are narrower and
the dark bands are broader.
2. gives a more precise measurement
than double slit.
D. Mathematical relationship
X = λL
d
Examples:
A good diffraction grating has 2.50 X
103lines/cm. What is the distance between
two lines in the grating?
1/2.50 X 103lines/cm= .00040cm
Example: Using a grating with a spacing of
4.00 X 10-4cm, a red line appears 16.5 cm
from the central line on a screen. The
screen is 1.00m from the grating. What is
the wavelength of the red light.
X = λL
d
16.5cm = λ(100cm)
4.00 X 10-4cm
λ=6.6 X 10-5 cm