Lecture 14: 13-07-15
Essential Physics I
E
英語で物理学の
エッセンス I
Lecture 14: 13-07-15
Exam
Next Lecture!
(2 weeks)
All lecture slides on course website:
http://astro3.sci.hokudai.ac.jp/~tasker/teaching/ep1
Remember your calculator!
Dictionary is OK!
(Phone is not)
SHOW ALL WORKING!
Next week’s exam
10 multiple choice questions
(A) .....
(B) .....
(C) .....
(D) .....
~6 classical mechanics
~2 oscillations, waves & fluids
~2 optics
Homework
Attendance / clickers
Exam
Total
40 %
20 %
40 %
100 %
Pass > 60 %
Next week’s exam
This is a question.
(A) .....
(B) .....
(C) .....
(D) .....
1 2
x = x0 + ut + at
2
p
u ± u2 4(a/2)(x0 x)
t=
a
p
12 ± 144 4(9.81/2)(17)
=
9.81
If this is wrong....
But parts of this are right...
You will get marks!
SHOW YOUR WORKING!
Last week:
Mirrors & Lenses
Image position can be found by drawing 2
light rays from points on the object.
Rays touch a real image
Last week:
Mirrors & Lenses
Image position can be found by drawing 2
light rays from points on the object.
Rays touch a real image
Rays only appear to touch a virtual image
h0
Magnification: M =
=
h
Mirror/Lens equation:
(thin lenses)
Inside a lens:
s0
s
1
1
1
+ 0 =
s s
f
n1
n2
n2 n1
+ 0 =
s
s
R
f
h
s
h’
s’
Optics
Interference & Diffraction
So far...
We assumed:
x
Light travels in a straight line: a ray
(Geometrical optics)
<< x
x
But, sometimes the wave nature of light is very important.
Interference and diffraction
Lecture 10: waves add
wave 1 + wave 2 = resulting wave
wave 1 + wave 2 cancel
Destructive interference
waves coincide
Constructive interference
Light waves are the same.
Coherence
For continuous interference, waves must be coherent.
e.g.
always constructive or
always destructive or
always ….
Coherence
For continuous interference, waves must be coherent.
t1
t2
t3
If phase between waves changes, wave sum changes
incoherent
Coherent waves
Length where the wave phase is
constant: coherence length
Lasers: long coherence length
Destructive and constructive
Even for lasers, hard to keep coherence
x1
Split a single light source
x2
Light travels different paths
Recombine
If x1 = (m )x2 , constructive interference.
✓
◆
1
x2 ,
If x1 = m +
2
destructive interference patterns.
Double-Slit interference
One method of splitting a source is to pass
through two narrow slits.
Produces 2 coherent sources.
cylindrical wavefronts interfere
Double-Slit interference
illuminating screen
Constructive interference
bright on screen
Destructive interference
dark on screen
interference fringes
Double-Slit interference
Constructive interference
Difference in path: m
If L > d , rays ~ parallel
Difference in path: d sin ✓
Bright fringes (constructive interference):
d sin ✓ = m
m = 0, 1, 2, ...
fringe order
Double-Slit interference
m=1 m=0 m=1
Bright fringes
Double-Slit interference
Destructive interference
Difference in path:
✓
dark fringes:
✓
◆
1
d sin ✓ = m +
2
1
m+
2
◆
m = 0, 1, 2, ...
Typically L ⇠ 1m , d < 1mm and
ybright
y
Therefore: sin ✓ ' tan ✓ =
L
✓
◆
1
L
L
ydark = m +
=m
2
d
d
< 1µm
Double-Slit interference
Quiz
Constructive interference of 2 coherent waves will occur if their
path difference is...
(A)
(B)
3
2
(C)
5
2
(D)
2
Double-Slit interference
Example
2 slits 0.075 mm apart are located 1.5 m from a screen.
The 3rd-order bright fringe is 3.8 cm from the screen centre.
What is
?
m=3
Use:
ybright
L
=m
d
ybright d
=
mL
(0.038m)(0.075 ⇥ 10
=
(3)(1.5m)
= 633nm
3
m)
Double-Slit interference
Quiz
If you increase the slit separation in a 2 slit system, how does the
spacing of the interference fringes change?
(A)
They become further apart
(B)
The spacing does not change
(C)
They become closer together
ybright
L
=m
d
ydark =
✓
1
m+
2
◆
L
d
Double-Slit interference
Quiz
A double-slit experiment has a slit spacing 0.12 mm.
If the bright fringes are 5 mm apart when the slits are illuminated
with 633-nm wavelength laser light....
What is the slit-to-screen distance, L?
(A) 0.95cm
Use:
(B) 95cm
ybright
L
y=1
d
L
=m
d
L
0
= 5mm
d
(C) 150cm
L=
(D) 15cm
yd
(0.12mm)(5.0mm)
=
= 95cm
633nm
Multiple-Slit interference
What if we add more slits?
Constructive interference
All 3 waves must be in phase
Difference in paths: m
For evenly spaced slits:
d sin ✓ = m
(same as for 2 slits)
Destructive interference ... more complicated
All waves must add to zero.
For 3 waves, each wave must out of phase with by 1/3.
d sin ✓ =
✓
✓
1
m+
3
m+
2
3
◆
◆
general
m
d sin ✓ =
N
slit no.
m is integer but not
integer multiple of N
Multiple-Slit interference
✓
m+
1
3
◆
✓
2
m+
3
◆
2 minima between primary
maxima m and (m + 1)
secondary maxima
not fully constructive or
fully destructive interference.
Diffraction grating
What if we add even more slits?
Diffraction grating: ~1000s slits / cm
If using multi-wavelength light:
since: d sin ✓ = m
for m > 0 , angular position ✓
depends on
maxima occur in different
places for different wavelengths
Many slits gives very precise locations
for the wavelengths (clear separations)
Diffraction grating
Example
Light from glowing Hydrogen contains H↵ (hydrogen-alpha) 656.3nm
and H
(hydrogen-beta) 486.1 nm.
Find the 1st order angular separation between these wavelengths
when using a grating of 6000 slits /cm.
m=1
1
d=
cm = 1.667µm
6000 slits / cm
6000
✓ ◆
✓
◆
0.6563µm = 23.2
1
1
✓↵ = sin
= sin
d
1.667µm
✓ ◆
✓
◆
0.4861µm
1
1
= 17
✓ = sin
= sin
d
1.667µm
✓ = 6.2
Diffraction grating
Quiz
Green light at 520 nm is diffracted by a grating with 3000 lines / cm.
Through what angle is the light diffracted in 5th order?
(A) 0.07
(B) 9.0
maxima at: ✓ = sin
1
d=
3000cm
(C) 51
✓ = sin
(D) 0.05
1
✓
1
1
✓
m
d
◆
= 3.33µm
5 ⇥ 520nm
3.33µm
◆
= 51
Diffraction
Why does the slit cause cylindrical waves?
light bending as it
passes object
Diffraction
Really need Maxwell’s equations but....
Huygens’ Principal:
(pronounce: her-genz)
All points on wave front are spherical wave sources.
add for
resultant wave
Diffraction
Spherical waves blocked by the barrier; only some pass through slit.
Sum of the remaining spherical waves
wave bends
(diffraction)
Diffraction
If the slit width ~ wavelength
slit = single wave source
But....
If slit width is larger
Each point in slit is a wave source
Interference between waves in slit
Diffraction
Consider 5 equal spaced sources:
1
Path length for ray 1 and 3 differ by a sin ✓
2
Destructive interference if
or
1
1
a sin ✓ =
2
2
a sin ✓ =
But if ray 1 and 3 interfere destructively, so do 2 and 4...
a
All rays in lower half of slit will interfere destructively with ray
2
above it.
If you look at the slit at ✓ such that a sin ✓ =
, you will see no light
Diffraction
Similarly
1
Ray 1 and 2 differ by a sin ✓
4
Destructive interference if
or
1
1
a sin ✓ =
4
2
a sin ✓ = 2
But is ray 1 and 2 interfere destructively, so do 3 and 4...
a
All rays in lower 3/4 of slit will interfere destructively with ray
4
above it.
If you look at the slit at ✓ such that a sin ✓ = 2 , you will see no light
Diffraction
Generally:
(divide slit into 6, 7, 8 etc sources)
a sin ✓ = m
Destructive interference, single-slit diffraction
for m = 1, 2, 3, ...
for m = 0 no destructive interference
at the central maximum.
Diffraction Limit
Diffraction creates a limit for how close two object can be....
... and still be resolved.
2 sources illuminate the slit
Their waves reach the slit at different angles
(assume sources are incoherent, no regular
interference pattern)
2 single slit diffraction patterns
As the sources get closer
Central maxima begins to overlap
Diffraction Limit
Diffraction patterns merge
2 peak structure disappears
2 peak visible if
Central maxima
of peak 1
=
1st minima
of peak 2
Rayleigh criterion
when true, 2 sources are just resolved
Diffraction Limit
The angular separation, ✓ , between the peaks
=
The angular separation, ✓ , between the sources
1st minima: sin ✓ =
since
<< a
✓min =
a
small angle approximation: sin ✓ ' ✓
(Rayleigh criterion, slit)
a
If using a circular aperture, not slit (e.g. a camera):
✓min
1.22
=
D
(Rayleigh criterion, circular aperture)
aperture diameter
Diffraction Limit
Example
An astroid 20 ⇥ 106 km away appears on a collision course with
the Earth!
What is the minimum size for the astroid that could be resolved
with the 2.4 m diameter diffraction-limited Hubble Space Telescope?
diffraction is what prevents the resolved imaged
(not atmosphere .... etc)
circular aperture
Diffraction Limit
Example
An astroid 20 ⇥ 106 km away appears on a collision course with
the Earth!
What is the minimum size for the astroid that could be resolved
with the 2.4 m diameter diffraction-limited Hubble Space Telescope?
✓min
1.22
=
D
l
1.22
=
L
D
1.22 L
l=
= 5.6km
D
l
✓
L
Opposite ends of the asteroid = 2 peaks
l
(small angle approximation)
✓'
L
Diffraction Limit
Example
An astroid 20 ⇥ 106 km away appears on a collision course with
the Earth!
What is the minimum size for the astroid that could be resolved
with the 2.4 m diameter diffraction-limited Hubble Space Telescope?
✓min
1.22
=
D
l
1.22
=
L
D
1.22 L
l=
= 5.6km
D
big problem!
l
✓
L
Opposite ends of the asteroid = 2 peaks
l
(small angle approximation)
✓'
L
Diffraction Limit
Quiz
What is the longest wavelength you could use to resolve an
object with angular diameter 0.44 mrad, using a microscope with
aperture 1.2 mm in diameter?
(A) 528nm
(B) 220nm
(C) 130nm
(D) 430nm
✓min
1.22
=
D
✓min D
(0.44mrad)(1.2mm)
=
=
= 430nm
1.22
1.22
Real optics
movie
Real optics
Quiz
What has been created at the VLBA (Very Large Telescope Array)?
(A) The biggest mirror on a telescope
(B)
An artificial star
(C) A new type of laser
(D) A defense shield
Real optics
Quiz
Why is this useful?
(A) The laser reflects of stars to find their distance
(B) We can communicate with extra-terrestrial life
(C) Allows the ‘adaptive optics’ technique to be used anywhere
in the sky.
(D) It tracks communication satellites around the world
Real optics
How big is the mirror on the telescope that the laser is
launched from?
(A) 10 m
(B) 8.2 m
(C) 3 m
(D) 5 m
Quiz
Real optics
What height is the “star” created?
(A) 90 km
(B) 10 km
(C)
The same as our nearest star
(D) The same as the moon
Quiz
Real optics
Quiz
How faint (weak / hard to see) is the artificial star?
(A)
20 x fainter than the faintest star that can be seen with
the telescopes
(B)
20 x fainter than the brightest star that can be seen by
eye
(C)
20 x fainter than the faintest star that can be seen by eye
(D) 20 x fainter than the brightest star that can be seen by
eye
Real optics
Quiz
What limits a ground-based telescope’s image sharpness?
(how good the image is)
(A)
Diffraction
(B)
Atmospheric turbulence
(C)
Moon light
(D) Sun light
Real optics
Quiz
How does adaptive optics solve this?
(A)
It uses a bigger mirror
(B)
The laser finds the object’s location more accurately
(C)
A flexible mirror corrects the image
(D) The laser freezes the image
Real optics
Quiz
Adaptive optics needs reference star. Why isn’t a real star used?
(A)
A suitable star isn’t always in the sky area you want to
observe
(B)
A real star’s light is too variable (it twinkles)
(C)
The reference star must be red
(D) The moon is too bright
Real optics
How does the laser make the star?
(A)
Shoots a beam of light into the air
(B)
Causes sodium atoms in the atmosphere to glow
(C)
The laser reflects of a real star, making it brighter
(D) The laser reflects off the moon
Quiz
Real optics
Quiz
What do scientists hope to observe with this new technique?
(A)
Black holes forming in other galaxies
(B)
‘high redshift’ (young) galaxies
(C)
Our galactic centre
(D) All the above
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