Transcript Lecture 25: mirrors

Phy2005
Applied Physics II
Spring 2016
Announcements:
• Test 2 Wednesday, March 23 covers chs. 22-25, sections listed in
syllabus + all material covered in class
• 2 practice tests posted on course Tests page
• Review session in class March 21 in class + March 21 6pm NPB 2205
Last time
The Superhero in Artificial Intelligence
http://www.theguardian.com/technology/2016/feb/16/demis-hassabis-artificial-intelligence-deepmind-alphago
Google DeepMind AlphaGo beat a Go master (Lee, Se-dol, world ranking #2)
4 times out of 5 matches.
Last time
θ
I = I0cos2θ
Intensity after polaroid
Intensity before polaroid
For unpolarized light, a polaroid sheet reduces its intensity
to half (sunglasses).
ACADEMIC HONESTY
Each student is expected to hold himself/herself to a high standard
of academic honesty. Under the UF academic honesty policy.
Violations of this policy will be dealt with severely. There will be
no warnings or exceptions.
Q1 There are two light sources. One is unpolarized and the
other is linearly polarized. When you look at these light
sources through a polaroid while rotating the polaroid axis 360
degree. What do you expect to observe?
1) For both you do not see any changes in the intensity during the
full rotation.
2) The intensity varies during the full rotation for both light
source.
3) One sees the intensity variation only for the unpolarized light.
4) One sees the intensity variation only for the polarized light.
Q2 A polaroid is placed at a 45o angle with respect to a linearly
polarized light beam incident on it. The transmitted beam is
then passed through a second polaroid at 45o with respect to the
first polaroid. What fraction of the initial lightlight intensity is
perpendicular to the initial beam?
1) 0
Remember I = I0cos2θ
?
2) 1/3
3) 1
4) 1/2
5) 1/4
Starting from A, make one touch at the wall and finish at B.
Which path takes the shortest time?
x A
x B
θ
θ
Principle of shortest time determines: relection, refraction
Reflection and Mirrors
qi
qi = qr
qr
Law of reflection
Specular Reflection
Diffuse Reflection
When we talk about an image, start from an ideal point light source.
Every object can be constructed as a collection of point light sources.
VIRTUAL
IMAGE
p
q
Image forms at the point where the light rays converge.
When real light rays converge  Real Image
When imaginary extension of L.R. converge  Virtual Image
Only real image can be viewed on screen placed at the spot.
VIRTUAL
IMAGE
p
q
For plane mirror: p = q
How about left-right?
From now on
Let’s check
p: distance to the object
q: distance to the image
Spherical Mirror
R: radius of curvature
focal Point
f: focal length = R/2
Optical axis
concave
convex
Parallel light rays: your point light source is very far away.
Focal point:
(i) Parallel incident rays converge after reflection
(ii) image of a far away point light source forms
(iii) On the optical axis
Reflected rays do not converge:
Not well-defined focal point
 not clear image
Spherical Aberration
f = R/2 holds strictly for a very
narrow beam.
Parabolic mirror can fix this problem.
Spherical Aberration:
some mirrors were ground wrong by
1/50th of human hair thickness.
Case 1: p > R
p
f
P > q
Real Image
q
Case 2: p = R
p = q
Real Image
Case 3: f < p < R
p < q
Real Image
Case 4: p = f
q = infinite
Case 5: p < f
q <0
Virtual Image
Mirror Equation
1/p + 1/q = 1/f
For a small object, f = R/2 (spherical mirror)
1/p + 1/q = 2/R
Alert!!
Be careful with the sign!!
Negative means that it is inside the mirror!!
p can never be negative (why?)
negative q means the image is formed inside the mirror
VIRTUAL
How about f?
For a concave mirror: f > 0
Focal point inside the mirror
f < 0
1/p + 1/q = 1/f < 0 : q should be negative.
1/p + 1/q = 1/f < 0 : q should be negative.
All images formed by a convex mirror are VIRTUAL.
Magnification, M = -q/p
Negative M means that the image is upside-down.
For real images, q > 0 and M < 0 (upside-down).
Ex. 26.1 An object is placed at the center of curvature of a
Mirror. Where is the image formed? Describe the image?
1/p + 1/q = 1/f
f = R/2
Object is at the center: p = R
1/q = 1/f – 1/p
= 2/R – 1/R
= 1/R
q = R > 0 (Real Image)
M = -q/p = -R/R = -1
No magnification but upside-down
Ex. 26.2 A concave mirror has a 30 cm radius of curvature.
If an object is placed 10 cm from the mirror, where will the
image be found?
Case 5: p < f
f = R/2 = 15 cm, p = 10 cm
1/p + 1/q = 1/f  1/10 + 1/q = 1/15
3/30 + 1/q = 2/30
1/q = -1/30
q = -30 cm
Real or Virtual
M = -q/p = 3
q < 0
Magnified or Reduced
Up-right or Upside-down
Q. An upright image that is one-half as large as an object is
needed to be formed on a screen in a laboratory experiment
using only a concave mirror with 1 m radius of curvature.
If you can make this image, I will give you $10. If you can’t
you should pay me $10. Deal or no deal? Why?
1/p + 1/q = 1/f = 2/R > 0
M = -q/p = ½ > 0
should be a real image: q > 0
M = -q/p cannot be positive, if q > 0.
No deal!!!