Physics 212 Lecture 25
Physics 212 Lecture 25, Slide 1
Music Who is the Artist? A) B) C) D) E)
The Band Bob Seger CCR Eagles Steve Miller Band
Why? SOMETIMES PERSISTENCE PAYS OFF…. “CCR CCR CCR CCR!!!! it means Creedence Clearwater Revival BTW! and they are the best band ever! anyone with style would play this in lecture”
Physics 212 Lecture 24
Your Comments “Finally, something tangible!” “Can you link polarization in this unit to the polarization from the previous unit? Also, We’ll go over the key steps which beam are we considering polarized in the water example, the refracted or the reflected?” “confused about the polarization and E fields of the waves when reflected/ refracted...” “I'm not sure that I quite understand how the light becomes polarized from the reflection off of a surface.” “the swimming pool question is interesting:)” “obviously, the checkpoints”
We will definitely discuss the pool.. a small demo, even
Note that the Reflection, Refraction and Lenses HW is now available (due Nov. 29)
Hour Exam 3: 2 weeks from tomorrow (Wed. Nov. 30) • covers L19-26 inclusive (LC circuits to lenses) •
Sign up for conflicts, etc. no later than Fri. Nov. 18 at 10:00 p.m.
“PLEASE ANSWER THIS IN LECTURE!!!!! everyone has been asking if this class is curved at all. none of the TA's know so will you let us know if it is?! THANKS :)” 05
Physics 212 Lecture 25, Slide 3
Let’s start with a summary:
from n1 to n2
from n1 to n2
Physics 212 Lecture 25, Slide 4
The speed of light in a medium is slower than in empty space:
since ε0 < ε
vmedium = c / nmedium
Remember: κ is dielectric constant (capacitors) Physics 212 Lecture 25, Slide 5
A ray of light passes from air into water with an angle of incidence of 30 degrees.
Checkpoint 1a
Which of the following quantities does not change as the light enters the water? A. wavelength B. frequency C. speed of propagation “Well it stated in the prelecture the speed's change, and the index is frequency dependent so, I'm choosing wavelength.”
“Frequency will eb constant as wavelength and speed change.”
“if wavelength changes then frequency changes therefore it must be speed of propagation.” Physics 212 Lecture 25, Slide 6
A ray of light passes from air into water with an angle of incidence of 30 degrees.
Checkpoint 1a
Which of the following quantities does not change as the light enters the water? A. wavelength B. frequency C. speed of propagation
What about the wave must be the same on either side ??? Observers in both media must agree on the frequency of vibration of the molecules
Physics 212 Lecture 25, Slide 7
Reflection
Physics 212 Lecture 25, Slide 8
Refraction: Snell’s Law
D sin θ 2 D sin θ1 = c / n2 c / n1
n2 sin θ 2 = n1 sin θ1
Physics 212 Lecture 25, Slide 9
Think of a day at the beach… What's the fastest path to the ball knowing you can run faster than you can swim?
This one is better Not the quickest route…
Physics 212 Lecture 25, Slide 10
A
x1 l1
Same Principle works for Light !!
y1
D
y2
l2 B
x2
Time from A to B : To find minimum time, differentiate t wrt x1 and set = 0 How is x2 related to x1? Setting dt/dx1 = 0
l l t= 1+ 2 = v1 v2
x12 + y12 x22 + y22 + v1 v2
dt x1 x2 dx2 = + 2 2 dx1 v1 x1 + y1 v2 x22 + y22 dx1 x2 = D − x1
x1 x − 2 =0 v1l1 v2l2
dx2 = −1 dx1 sin θ1 sin θ 2 = v1 v2
v = c/n
n1 sin θ1 = n2 sin θ 2
Physics 212 Lecture 25, Slide 11
The path of light is bent as it passes from medium 1 to medium 2.
Checkpoint 2a
Compare the indices of refraction in the two media. A. n1 > n2 B. n1 = n2
C. n1 < n2
“Snell's law demonstrates that when the angle of incidence is smaller, the index of refraction is larger. ”
“the index of refraction is usually greater as it passes through different mediums other than air. ” “more refraction in n2, thus it is bigger” Physics 212 Lecture 25, Slide 12
The path of light is bent as it passes from medium 1 to medium 2.
Checkpoint 2a
Compare the indices of refraction in the two media. A. n1 > n2 B. n1 = n2
C. n1 < n2
Snell’s Law: n1sinθ1 = n2sinθ2 n decreases fl θ increases Physics 212 Lecture 25, Slide 13
Total Internal Reflection
NOTE: n1 > n2 implies θ2 > θ1
θ1 > θc
BUT: θ2 has max value = 90o !!
Total Internal Reflection Physics 212 Lecture 25, Slide 14
A light ray travels in a medium with n1 and completely reflects from the surface of a medium n2.
Checkpoint 2b
The critical angle depends on: A. n1 only B. n2 only
C. both n1 and n2
“It does not go through n2 at all.”
“The light is only refracting off N2 so thats what it depends on. ”
“if n2 is too large, no light is refracted; if n1 is too small, no light is refracted” Physics 212 Lecture 25, Slide 15
A light ray travels in a medium with n1 and completely reflects from the surface of a medium n2.
Checkpoint 2b
The critical angle depends on: A. n1 only B. n2 only
C. both n1 and n2
θc clearly depends on both n2 and n1 Physics 212 Lecture 25, Slide 16
Intensity
Anything looks like a mirror if light is just glancing off it.
If two materials have the same n then its hard to tell them apart. Physics 212 Lecture 25, Slide 17
Polarization
56.3
o
θ1 + θ 2 = 90
sin θ 2 = sin( 90 − θ1 ) = cos θ1
Snell’s Law: n2 sin θ 2 = n2 cosθ1 = n1 sin θ1
n2 tanθ1 = n1 Physics 212 Lecture 25, Slide 18
A ray of light passes from air into water with an angle of incidence of 30 degrees.
Checkpoint 1b
Some of the light also reflects off the surface of the water. If the incident light is initially unpolarized, the reflected light will be A. unpolarized B. somewhat horizontally polarized C. somewhat vertically polarized “The light will be unpolarized until the Brewster's angle is reached where the light becomes horizontally polarized.”
“since the virtical componet will be smaller ”
“In order to reflect back, there must be some vertical element to it.” Physics 212 Lecture 25, Slide 19
A ray of light passes from air into water with an angle of incidence of 30 degrees.
Checkpoint 1b
Some of the light also reflects off the surface of the water. If the incident light is initially unpolarized, the reflected light will be A. unpolarized B. somewhat horizontally polarized C. somewhat vertically polarized
o = horizontal
Physics 212 Lecture 25, Slide 20
A ball sits in the bottom of an otherwise empty tub at the front of the room. Suppose N people sit high enough to see the ball (N =
).
Physics 212 Lecture 25, Slide 21
A ball sits in the bottom of an otherwise empty tub at the front of the room. Suppose N people sit high enough to see the ball (N = ).
Suppose I fill the tub with water but the ball doesn’t move. Will more or less people see the ball? A) More people will see the ball B) Same # will see the ball C) Less people will see the ball
??
θA θw
Snell’s Law: ray bent away from normal going from water to air Physics 212 Lecture 25, Slide 22
A light is shining at the bottom of a swimming pool (shown in yellow in the figure). A person is standing at the edge of the pool.
Checkpoint 3
Can the person standing on the edge of the pool be prevented from seeing the light by total internal reflection at the water-air surface? A. yes B. no “There is a critical angle where the light is perfectly reflected back at the surface of water” “From experience I think no, but from prelecture, I think yes.” “The light will span the entire area over the pool, so even if the line that connects the man to the light will be below the critical angle, there will be another beam that is able to reach the man.” “not all off the light from the light will be at the correct angle for total internal reflection” Physics 212 Lecture 25, Slide 23
A light is shining at the bottom of a swimming pool (shown in yellow in the figure). A person is standing at the edge of the pool.
Checkpoint 3
Can the person standing on the edge of the pool be prevented from seeing the light by total internal reflection at the water-air surface? A. yes B. no
The light would go out in all directions, so only some of it would be internally reflected. The person would see the light that escaped after being refracted.
DRAW SOME RAYS Physics 212 Lecture 25, Slide 24
Example: Refraction at water/air interface • Diver’s illusion
97º
Diver sees all of horizon refracted into a 97°cone
θair = 90
sinθwater =
na n 1 sin90 = a = nw nw 1.33
θwater = 48.5 Physics 212 Lecture 25, Slide 25
Exercise A meter stick lies at the bottom of a rectangular water tank of height 50cm. You look into the tank at an angle of 45o relative to vertical along a line that skims the top edge of the tank. What is the smallest number on the ruler that you can see?
45o
nwater = 1.33 50 cm
0
20
40
60
80
100
Conceptual Analysis: - Light is refracted at the surface of the water Strategy: - Determine the angle of refraction in the water and extrapolate this to the bottom of the tank.
Physics 212 Lecture 25, Slide 26
45o
Exercise A meter stick lies at the bottom of a rectangular water tank of height 50cm. You look into the tank at an angle of 45o relative to vertical along a line that skims the top edge of the tank. What is the smallest number on the ruler that you can see?
nwater = 1.33 50 cm
0
20
40
60
80
100
If you shine a laser into the tank at an angle of 45o, what is the refracted angle θR in the water ? A) θR = 28.3o
B) θR = 32.1o
C) θR = 38.7o
Snell’s Law: nairsin(45) = nwatersin(θR) sin(θR) = nairsin(45)/nwater = 0.532 θR = sin-1(0.532) = 32.1o Physics 212 Lecture 25, Slide 27
45o
Exercise A meter stick lies at the bottom of a rectangular water tank of height 50cm. You look into the tank at an angle of 45o relative to vertical along a line that skims the top edge of the tank. What is the smallest number on the ruler that you can see?
nwater = 1.33 θR
50 cm
0
20
40
d
60
80
100
θR = 32.1o
What number on the ruler does the laser beam hit ?
A) 31.4 cm
B) 37.6 cm
C) 44.1 cm
tan(θR) = d/50 d = tan(32.1) x 50cm = 31.4cm
Physics 212 Lecture 25, Slide 28
45o
Follow-Up A meter stick lies at the bottom of a rectangular water tank of height 50cm. You look into the tank at an angle of 45o relative to vertical along a line that skims the top edge of the tank.
50 cm
nwater = 1.33
0
20
40
60
80
100
If the tank were half full of water, what number would the laser hit? (When full, it hit at 31.4 cm) A) 25 cm
B) 31.4 cm
C) 32.0 cm
D) 40.7 cm
E) 44.2 cm
Physics 212 Lecture 25, Slide 29
45o
45o
nwater = 1.33 θR
50 cm
50 cm d = 50 cm
d = 31.4 cm 0
20
40
60
80
100
0
20
40
60
80
100
θR = 32.1o 45o
50 cm
nwater = 1.33 θR 0
20
d = 40.7 cm 40
25 cm + (31.4/2) cm
60
80
100 Physics 212 Lecture 25, Slide 30
More Practice A monochromatic ray enters a slab with n1 = 1.5 at an angle θb as shown θb n=1
TOP
n1 = 1.5 n=1
BOTTOM
θb
(A) Total internal reflection at the top occurs for all angles θb, such that sinθb < 2/3
(B)
Total internal reflection at the top occurs for all angles θb, such that sinθb > 2/3
(C) There is no angle θb (0 < θb < 90o) such that total internal reflection occurs at top.
Snell’s law:
n1 sin θ1 = n2 sin θ 2
nsinθ is “conserved”
Ray exits to air with same angle as it entered !!
Physics 212 Lecture 25, Slide 31
Follow-Up A ray of light moves through a medium with index of refraction n1 and is incident upon a second material (n2) at angle θ1 as shown. This ray is then totally reflected at the interface with a third material (n3). Which statement must be true? θ1 n1 n2
(A)
(B) n1 < n3 ≤ n2
n3 < n1
(C)
n3 ≥ n2
n3 θ1
If n1 = n3
n1 n2 n3
Want larger angle of refraction in n3
θ1
n3 < n1 Physics 212 Lecture 25, Slide 32
“Does gravity affect reflection/refraction? Would the reflection of the ray depend on say where you are in space? ex: standing on jupiter etc?”
“Gravity affects the path taken by light, but small effect: Einstein ring”
Physics 212 Lecture 25, Slide 33
