Course Updates http://www.phys.hawaii.edu/~varner/PHYS272-Spr10/physics272.html
Reminders:
1) Assignment #12 Æ due Monday
2) Start Optics (Chapter 33)
3) Last HW (#13 posted) Æ due Monday, May 3rd θi =θr
n1 sinθ1 = n2 sinθ 2
2 Index of Refraction
• Speed of light, c, in vacuum is 3x108m/s
• Speed of light, v, in different medium can be v < c.
• index of refraction, n = c/v.
• frequency, f, does not change in wave eqn. of v = f λ,
• wavelength, λ, depends on medium, λ = v/f = c/nf = λ0/n
• In some media, n, depends on f, this is called dispersion.
3 Waves, wave front, rays (Y&F section 33.2) Plane waves moving in +x direction r E(x, y, z,t)= E0 yˆ cos(kx − ωt) ()r () B x, y, z,t = B0zˆ cos kx − ωt
Wave front is a surface of constant phase
E field Wave front (y-z plane surface of constant phase)
y Ray propagation
x 4 z Huygen’s Principle (Y&F section 33.7) • Huygen’s principle; a wave front can be a source of secondary wavelets the spread out in all directions at the speed of propagation in the medium. The envelope of leading edges forms a wave front. • This principle was stated by Huygen in 1678, it is derivable from Maxwell’s eqn. It is a geometrical description of ray propagation.
vt Plane wave example;
Secondary wavelets create another wave front (plane)
5 Reflection from Huygen’s Principle
Consider wave fronts, separated by vt, the incident wave fronts in contact with the surface will create a wavelets according to Huygen’s Principle and leads to another “reflected” wave front.
The result is θi = θr
reflected wave front Incident wave front
vt vt vt ray diagram vt θ θ r i θi θr
θi = θr Reflection Law 6 Refraction from Huygen’s Principle Now the speed changes, from medium a to medium b, so the Speed may change and the wavefront spacing differs.
L sinθa = vat = ct/na vat t L sinθb = vbt= ct/nb on vat fr e av Medium a, va=c/na W LL θa θ Medium b, b vbt vbt v =c/n nt b b e fro Wav
na sinθa = nb sinθb Snell’s Law 7 Snell’s Law (law of refraction)
In c id t e normal n n fro t e r a av y W θa Medium a, v =c/n a a θa R θb Medium b, vb=c/nb e Angles are t f fron r e act Wav θ usually given ted b rac ed relative to Ref
r normal to plane a y
na sinθa = nb sinθb Snell’s Law 8 Question 1
A ray of light passes from air into water with an angle of incidence of 30o. Which of the following quantities does not change as the light enters the water.
a) Wavelength b) frequency c) speed of propagation d) direction of propagation.
9 Question 1
A ray of light passes from air into water with an angle of incidence of 30o. Which of the following quantities does not change as the light enters the water.
a) Wavelength b) frequency c) speed of propagation d) direction of propagation.
10 Question 2 • Which of the following ray diagrams could represent the passage of
light from air through glass and back to air? (nair=1 and nglass=1.5) (a) (b) (c) air air air glass glass glass air air air
11 Question 2 • Which of the following ray diagrams could represent the passage of
light from air through glass and back to air? (nair=1 and nglass=1.5) (a) (b) (c) θ1 air air air glass glass glass θ2 air air air
• The behavior of these rays is determined from Snell’s Law:
n1 sinθ1 = n2 sinθ 2 • Since n(glass) > n(air), sinθ (glass) < sinθ (air) . • Therefore, moving from air to glass, ray will bend toward normal. • this eliminates (a). • Moving from glass to air, ray will bend away from normal. • this eliminates (c). • As a matter of fact, the final angle in air must be equal to the initial angle in air!! 12 θa EXAMPLE, from air into glass;
Suppose we have light in air (n=1) incidence on glass (n=1.55) at an θb angle θa=45 deg. What is the angle of the refracted light, θb?
nair sinθa = nglass sinθb (1)sin()45° = (1.55)sin b (1)sin 45()° 1 sinθb = =θ 1.55 1.55 2 θb = 27°
13 EXAMPLE, from glass into air; θa Suppose we have light in glass (n=1.55) incident into air at an angle θa=30 deg. What is the angle of the refracted light, θb? θb nglass sinθa = nair sinθb (1.55)sin()30° = (1)sin b ()1.55 sinθb = (1.55)sin 45θ° = 2 θb = 51°
14 Typical balloon The ANITA Concept field of regard
~4km deep ice!
15 Refraction limits ANITA !
U E S
• ANITA at float (123Kft) – Seen through amateur telescope from the South Pole – Size of the Rose Bowl! – (thanks to James Roth) 16 “Mini-Antarctica” at Stanford Linear Accelerator
17 SLAC T486
θb
θa
• 2 mile long accelerator
•28 Billion electron Volts
•If you’ve driven I-280 between San Francisco 18 and San Jose Æ • Problem!!!
θb air
θa ice
nice sinθa = nair sinθb
• There is no (1.8)sin()34° = (1)sin b escape!!! () sinθ b = (1.8)sin 34° =1.0 θ θb = 90°
19 • Problem
air θb nice sinθa = nair sinθb
(1.8)sin()34° = (1)sin b sin = (1.8)sin 34()° =1.0 θ θ b a θ = 90° θ ice b
nice sinθa = nair sinθb
(1.8)sin()27° = (1)sin b () sinθ b = (1.8)sin 27° = 0.82 • Solution = cut! θ θb = 54.8° θ =?? ~7degrees air b θb air
θ θb a θa ice ice
20 Mini-Antarctica at Stanford Linear Accelerator
Cut ~7degrees! (jackhammers, chainsaws, mauls not shown)21 10 tons ice
22 23 Let there be light! (OK, radio)
24 Suppose you are stranded on an island with no food. You see a fish in the water. Where should you aim your spear to hit the fish?
ANSWER; do not aim directly at the apparent position of the fish. Aim at the inside of the fish.
25 Suppose in the previous question instead of a spear you had a high power laser to simultaneously kill and cook the fish (in the water). Where should you aim the laser??
ANSWER; aim directly at apparent fish position as the laser beam will refract to the correct fish position.
26 For next time
• Homework #12 posted Æ due Monday
• The home stretch: optics/optical phenomenon
27 Key Instrument pieces
Battery CSBF box CIP
Instrument box
28 Launch: December 15, 2007 (after almost 2 agonizing weeks of waiting)
Courtesy Kim Palladino
• A flawless launch – CSBF truly professional – After day after day after day of false starts, we were really ready to go 29