Lec. 9: Ch. 3 ‐ Geometrical Optics
We did this
We are here
1.
2.
3.
Virtual images
Spherical mirrors, ray tracing
Spherical lenses, ray tracing
Thin lens approximation
3 formulas
4.
Aberrations of lenses
We covered 32 of these
viewgraphs. We skipped
14 and 15 for now.
To change your iClicker frequency, turn on your iClicker.
Hold down on the power button until the power light blinks.
Then hit DC. You should see a green light indicating that
you have changed the frequency.
1
Real image made by a concave mirror
(demonstration)
Mirror
surface
Axis
Center
Focal
point
A
C
B
Review ray tracing: convex and concave mirrors
Convex mirror: 3 rays:
1. Start parallel to axis, then
bends thru an F
2. Thru C, no bend
3. Starts bent thru an F,
becomes parallel Concave mirror:
(alternatively, arrow could be closer to mirror than F) 3
Demo: blackboard optics
Review ray tracing: convex and concave lenses
Convex lens as a magnifier:
(alternatively, could be a projection lens) Concave lens:
3 rays:
1. Start parallel to axis, then bends thru an F
2. Thru the middle, no bend
3. Starts bent thru an F, becomes parallel (demagnifier) 4
Demo: lenses and vugraf projector
Review: Ray tracings (1 to 4)
5
Review: Ray tracings (5 and 6) 6
Here is one example of how to use the lens eqn
with a converging lens
• Given:
• f = 10 cm
• Object is 15 cm in front of lens: x0 = 15
• Find:
– Where is image and is it real or virtual?
• Solve equation for xi:
– Substitute numbers for letters
– Subtract 1/15 from both sides
– Arithmetic on calculator
– Multiply by xi/0.033
1
1
1
f
10
xo
15
xi
1
1
1
10
15
xi
—=—+—
—-—=—
1
0.033 = —
xi
1
xi = .033
— = 30 cm
Image is 30 cm to right of center of lens
and is real because xi is positive
Here is a sketch to show the previous result
• We can verify our result by ray‐tracing
f = 10
xo = 15
xi = 30
Here is an example of how to use the lens eqn
with a diverging lens (see Fig. 3.28)
• Given:
• f = —5 cm • NOTE, THE FOCAL LENGTH OF A DIVERGING LENS IS NEGATIVE
• Object is 12 cm in front of lens: x0 = 12
• Find:
– Where is image and is it real or virtual?
• Solve equation for xi:
–
–
–
–
Substitute numbers for letters
Subtract 1/12 from both sides
Arithmetic on calculator
Multiply by xi/(‐0.283)
1
1
1
f
-5
xo
12
xi
1
1
1
-5
12
xi
—=—+—
—-—=—
1
-0.283 = —
xi
1
xi = -0.283
— = -3.53 cm
Image is 3.53 cm to left of center of lens
and is virtual because xi is negative
Review: Rules for mirror ray tracing
• Rule 1 for mirrors: All incident rays parallel to the axis are reflected so that they appear to have come from the focal point F.
• We also see that a ray coming in along a radius will have a zero angle with the normal (it IS a normal), so its reflected back on itself.
• Rule 2 for mirrors: Incident rays coming toward the center of curvature, C, are reflected back onto themselves.
• Notice that if I turn the reflected ray around, so that it is an incoming ray, it goes out parallel to the axis (turn around the paraxial ray so it goes out). Let’s call this rule 3:
• Rule 3 for mirrors: Incident rays headed for F are reflected so that they are parallel to the axis. This is simply rule 1 with the rays turned around so the outgoing ray is the incoming ray and vice versa.
10
Review: Rules for ray tracing lenses
• Rule 1. A ray parallel to the axis is deflected through F' (or as though it came from F'). Hint: In either case, the point F' will be on the ruler edge.
• Rule 2. A ray through the center of the lens is not bent.
• Rule 3. A ray from F (extension may be necessary) is deflected parallel to the axis.
Rule 3 is rule 1 with a reversed path, just like for mirrors. You may have to extend the edge of the lens or the rays to apply the rules. 11
Lec. 9: Ch. 3 ‐ Geometrical Optics
1.
2.
3.
We are here
Virtual images
Spherical mirrors, ray tracing
Spherical lenses, ray tracing
Thin lens approximation
3 formulas
4.
Aberrations of lenses
To change your iClicker frequency, turn on your iClicker.
Hold down on the power button until the power light blinks.
Then hit DC. You should see a green light indicating that
you have changed the frequency.
DC
12
Compound Lenses
• Can have less aberration.
• A modern lens can have 16 elements and can “zoom”. “stop”
Reduces aberration
Image plane
13
Fresnel Lens
Used in lighthouses Lighthouse lens
Fresnel stage light
14
http://sandiartfullyyours.com/NewFiles/lighthouse3/images/Ponce%20Fresnell.jpg
15
Web tutorials with Java Applets
•
•
•
•
•
Useful web links on curved mirrors
http://micro.magnet.fsu.edu/primer/java/mirrors/concavemirrors/index.html
http://micro.magnet.fsu.edu/primer/java/mirrors/convexmirrors/index.html
http://micro.magnet.fsu.edu/primer/java/mirrors/concave.html
http://micro.magnet.fsu.edu/primer/java/mirrors/convex.html
•
•
•
•
•
•
Useful web links on lenses
http://micro.magnet.fsu.edu/primer/lightandcolor/lenseshome.html
http://micro.magnet.fsu.edu/primer/java/lenses/simplethinlens/index.html
http://micro.magnet.fsu.edu/primer/java/lenses/converginglenses/index.html
http://micro.magnet.fsu.edu/primer/java/lenses/diverginglenses/index.html
http://micro.magnet.fsu.edu/primer/java/components/perfectlens/index.html
16
Lec. 8: Ch. 3 ‐ Geometrical Optics
1. Virtual images (review)
2. Spherical mirrors
3. Spherical lenses
3 formulas
We are here
4. Aberrations of lenses
17
Aberrations
•
•
•
•
•
Field curvature
Off‐axis aberration
Spherical aberration
Distortion
Chromatic aberration
18
Aberration: field curvature
Image does not lie in one plane
19
Off axis aberration
Edges of images are less clear. 20
Demo with lens and bulb
Spherical aberration
Rays at the edge focus closer to the mirror
21
Demo with lens, not mirror
Aberrations: Distortion
22
Demo with overhead and small lenses
Chromatic Aberration
23
Demo with lens and bulb
Concept Question 1
DC
In this case, the image is:
A) Virtual
B) Real
Eye sees an image here.
24
Concept Question 1
In this case, the image is:
A) Virtual
B) Real
Real because the light rays really go through the image. You can put a screen there to see it.
Eye sees an image here.
25
Concept Question 2
1
screen
2
Two point sources of light are imaged onto a screen by a converging lens. The images are labeled 1 and 2. You slide a mask over the left half of the lens. What happens to the images?
lens
mask
A)
B)
C)
Image 1 vanishes
Image 2 vanishes
Something else happens
26
Concept Question 2
1
screen
2
lens
Two point sources of light are imaged onto a screen by a converging lens. The images are labeled 1 and 2. You slide a mask over the left half of the lens. What happens to the images?
mask
A)
B)
C)
Image 1 vanishes
Image 2 vanishes
Something else happens
The image gets dimmer, but half the lens is still a lens, and it produces a pair of images.
27
Review: The lens equation gives the same results as ray‐tracing but without any rays!
•
1/x0 + 1/xi = 1/f
•
f is focal length of lens
•
xo = positive distance from object to center of lens (when object is left of the lens)
•
xi = distance from image to lens center
•
•
•
xi is a positive number for (real) image on other side of lens from object.
•
xi is a negative number for (virtual) image on same side as object.
Given two, find the third.
Can use the lens equation to find image position if know object position or vice versa, without any rays
(will be positive
number for this case)
Concept Question 3
For a lens,
F = 10 cm
Xo = 20 cm
What is Xi ?
a. 10 cm
b. 20 cm
c. 40 cm
d. 80 cm
e. None of these
Concept Question 3
• Given:
• f = 10 cm
• Object is 15 cm in front of lens: x0 = 15
• Find:
– Where is image and is it real or virtual?
• Solve equation for xi:
– Substitute numbers for letters
– Subtract 1/15 from both sides
– Arithmetic on calculator
– Multiply by xi/0.033
1
1
1
f
xo
xi
0.1
0.2
—=—+—
10 = 5 + 1/xi
5 = 1/xi
1
xi = —
= 20 cm
5
Image is 20 cm to right of center of lens
and is real because xi is positive
Here is a sketch to show the previous result
• We can verify our result by ray‐tracing
xo = 15
xi = 20
Here is an example of how to use the lens eqn with a diverging
lens (see Fig. 3.28)
Also, we keep cm throughout, and don’t convert to m
• Given:
• f = —5 cm • NOTE, THE FOCAL LENGTH OF A DIVERGING LENS IS NEGATIVE
• Object is 12 cm in front of lens: x0 = 12
• Find:
– Where is image and is it real or virtual?
• Solve equation for xi:
–
–
–
–
Substitute numbers for letters
Subtract 1/12 from both sides
Arithmetic on calculator
Multiply by xi/(‐0.283)
1
1
1
f
-5
xo
12
xi
1
1
1
-5
12
xi
—=—+—
—-—=—
1
-0.283 = —
xi
1
xi = -0.283
— = -3.53 cm
Image is 3.53 cm to left of center of lens
and is virtual because xi is negative